tag:blogger.com,1999:blog-44999944123193384352024-03-18T04:07:09.488-07:00GeoDomEmmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.comBlogger159125tag:blogger.com,1999:blog-4499994412319338435.post-56627580072510466542024-02-28T07:29:00.000-08:002024-02-28T07:29:49.868-08:00A family of trigonometric formulas for the roots of quadratic equations<p> <span style="font-size: large;">This note presents alternative trigonometric formulas for finding the roots of quadratic equations where $a$, $b$, and $c$ are non-zero real numbers.</span></p><p><span style="font-size: large;">Before the era of calculators, trigonometric formulas were favored for computing quadratic roots due to their time and labor-saving benefits. However, it's not advisable nowadays to rely on these formulas over the traditional quadratic formula for root calculations. Nonetheless, the trigonometric formulas showcased in this note have hinted at certain <a href="https://geometriadominicana.blogspot.com/2024/02/solving-impossible-integrals-with-new.html" target="_blank">identities that appear useful for integrating trigonometric functions</a>. Thus, the sole purpose of this note is to document the origins of these identities.</span></p><p><span style="font-size: large;"><b>Theorem 1</b>. <i>Let $a$, $b$, and $c$ be non-zero real numbers. For the quadratic equation $ax^2+bx+c=0$, the roots are given by:</i></span></p><p><span style="font-size: large;"><i>$$x_{1}= ie^{i\alpha}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}},\tag{1}$$</i></span></p><p><span style="font-size: large;"><i>where $\alpha=\sin^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.</i></span></p><p><span style="font-size: large;"><b>Proof</b>. Consider the quadratic equation $a^2+bx+c=0$. Multiplying both sides by $a$ and making the substitutions $ax=p$ and $ac=q^2$ yields</span></p><p><span style="font-size: large;">$$p^2+bp+q^2=0.$$</span></p><p><span style="font-size: large;">Let's make the substitution $\sin{\alpha} = \frac{b}{2q}$. Consequently,</span></p><p><span style="font-size: large;">$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sin{\alpha}+q^2\\&=p^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)+4qp\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+q^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 + \left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 -i^2\left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2 .\end{aligned}$$</span></p><p><span style="font-size: large;">Factorizing the difference of squares and then factorizing again, one of the factor of the quadratic equation can be express as follows</span></p><p><span style="font-size: large;">$$p\left(\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}\right).$$</span></p><p><span style="font-size: large;">Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain </span></p><p><span style="font-size: large;">$$\begin{aligned}x_1&=-\left(\frac{\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}}{\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\sin{\alpha}-i\cos{\alpha})\sqrt{\frac{c}{a}}\\&=ie^{i\alpha}\sqrt{\frac{c}{a}}. \end{aligned}$$</span></p><p><span style="font-size: large;">The other factor is given by</span></p><p><span style="font-size: large;">$$p\left(\sin{\frac{\alpha}{2}}-i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}-i\sin{\frac{\alpha}{2}}\right).$$</span></p><p><span style="font-size: large;">And similarly,</span></p><p><span style="font-size: large;">$$x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}}.$$ </span></p><p><span style="font-size: large;"><b>Theorem 2</b>. <i>For the quadratic equation $ax^2+bx+c=0$ with non-zero real numbers $a$, $b$, and $c$, the roots are given by:</i></span></p><p><span style="font-size: large;"><i>$$x_{1}=-\tan{\frac{\beta}{2}}\sqrt{\frac{c}{a}}\qquad \text{and} \qquad x_{2}=-\cot{\frac{\beta}{2}}\sqrt{\frac{c}{a}},\tag{2}$$</i></span></p><p><span style="font-size: large;"><i>where $\beta=\csc^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.</i></span></p><p><span style="font-size: large;"><b>Proof</b>. Multiplying by $a$ the quadratic equation $ax^2+bx+c=0$ and then substituting $ax=p$ and $ac=q^2$, we have</span></p><p><span style="font-size: large;">$$p^2+bp+q^2=0.$$</span></p><p><span style="font-size: large;">Use the substitution $\csc{\beta}=\frac{b}{2q}$, then</span></p><span style="font-size: large;">$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\csc{\beta}+q^2\\&=\csc{\beta}\left(p^2\sin{\beta}+2qp+q^2\sin{\beta}\right)\\&=2p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+2qp\left(\sin^2{\frac{\beta}{2}}+\cos^2{\frac{\beta}{2}}\right)+2q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&=p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+qp\sin^2{\frac{\beta}{2}}+ qp\cos^2{\frac{\beta}{2}}+q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&= p\sin{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)+q\cos{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\\&= \left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\left(p\sin{\frac{\beta}{2}}+q\cos{\frac{\beta}{2}}\right).\end{aligned}$$</span><p><span style="font-size: large;">By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain the desired formulas in $(2)$.</span></p><p><span style="font-size: large;"><b>Theorem 3</b>. <i>Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:</i></span></p><p><span style="font-size: large;"><i>$$x_{1}= -e^{i\gamma}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}},\tag{3}$$</i></span></p><p><span style="font-size: large;"><i>where $\gamma=\cos^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.</i></span></p><p><span style="font-size: large;"><b>Proof</b>. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields</span></p><p><span style="font-size: large;">$$p^2+bp+q^2=0.$$</span></p><p><span style="font-size: large;">By making the substitution $\cos{\gamma} = \frac{b}{2q}$, it follows that</span></p><p><span style="font-size: large;">$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\cos{\gamma}+q^2\\&=p^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)+4qp\left(\cos^2{\frac{\gamma}{2}}-\sin^2{\frac{\gamma}{2}}\right)+q^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2+\cos^2{\frac{\gamma}{2}}(p+q)^2\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2-i^2\cos^2{\frac{\gamma}{2}}(p+q)^2.\end{aligned}$$</span></p><p><span style="font-size: large;">By factoring the difference of squares first and then applying further factorization, one of the factors of the quadratic equation can be represented as follows</span></p><p><span style="font-size: large;">$$p\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right)+q\left(-\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$</span></p><p><span style="font-size: large;">Setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain</span></p><p><span style="font-size: large;">$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\cos{\gamma}+i\sin{\gamma})\sqrt{\frac{c}{a}}\\&=-e^{i\gamma}\sqrt{\frac{c}{a}}.\end{aligned}$$</span></p><p><span style="font-size: large;">The other factor is given by</span></p><p><span style="font-size: large;">$$p\left(\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}\right)-q\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$</span></p><p><span style="font-size: large;">And similarly,</span></p><p><span style="font-size: large;">$$x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}}.$$</span></p><p><span style="font-size: large;"><b>Theorem 4</b>. <i>Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by: </i></span></p><p><span style="font-size: large;"><i>$$x_{1,2}=\frac{\tan{\frac{\delta}{2}\pm1}}{\tan{\frac{\delta}{2}\mp1}}\sqrt{\frac{c}{a}},\tag{4}$$</i></span></p><p><span style="font-size: large;"><i>where $\delta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.</i></span></p><p><span style="font-size: large;"><b>Proof</b>. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields</span></p><p><span style="font-size: large;">$$p^2+bp+q^2=0.$$</span></p><p><span style="font-size: large;">Use the substitution $\sec{\delta}=\frac{b}{2q}$, then</span></p><p><span style="font-size: large;">$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sec{\delta}+q^2\\&=\sec{\delta}\left(p^2\cos{\delta}+2qp+q^2\cos{\delta}\right)\\&=\cos{\delta}(p^2+q^2)+2qp\\&= \cos{\delta}(p+q)^2-2qp\cos{\delta}+2qp\\&= \left(\cos^2{\frac{\delta}{2}}-\sin^2{\frac{\delta}{2}}\right)(p+q)^2-2qp\left(1-2\sin^2{\frac{\delta}{2}}\right)+2qp\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p+q)^2 +4qp\sin^2{\frac{\delta}{2}}\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}\left((p+q)^2 -4qp\right)\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p-q)^2\\&= \left(\cos{\frac{\delta}{2}}(p+q) +\sin{\frac{\delta}{2}}(p-q)\right) \left(\cos{\frac{\delta}{2}}(p+q) -\sin{\frac{\delta}{2}}(p-q)\right).\end{aligned}$$</span></p><p><span style="font-size: large;">Expanding and then factorizing again, one of the factors of the quadratic equation is given by</span></p><p><span style="font-size: large;">$$p\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right).$$ </span></p><p><span style="font-size: large;">By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain</span></p><p><span style="font-size: large;">$$\begin{aligned}x_1&=\frac{\sin{\frac{\delta}{2}}+\cos{\frac{\delta}{2}}}{\sin{\frac{\delta}{2}}-\cos{\frac{\delta}{2}}}\sqrt{\frac{c}{a}}\\&= \frac{\tan{\frac{\delta}{2}+1}}{\tan{\frac{\delta}{2}-1}}\sqrt{\frac{c}{a}}. \end{aligned}$$</span></p><p><span style="font-size: large;">The other factor is given by</span></p><p><span style="font-size: large;">$$p\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right).$$</span></p><p><span style="font-size: large;">Similarly, </span></p><p><span style="font-size: large;">$$x_2=\frac{\tan{\frac{\delta}{2}-1}}{\tan{\frac{\delta}{2}+1}}\sqrt{\frac{c}{a}}.$$</span></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-25115243914813637962024-02-15T03:29:00.000-08:002024-02-15T04:28:01.735-08:00Integrals yielding $e^{\pi}$ or $e^{-\pi}$<p><span style="font-size: large;"> Lately, <a href="https://mathoverflow.net/questions/463459/solving-impossible-integrals-with-a-new-trick" target="_blank">I've been playing a lot with integrals</a>, and coincidentally (with a bit of algebraic manipulation), I've come across these two beauties:</span></p><p><span style="font-size: large;">$$\int_{0}^{1} \left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right) - 1\right) \, dx = e^{\pi},\tag{1}$$</span></p><p><span style="font-size: large;">$$\int_{0}^{1} \left( -\frac{5}{2\left( x - \sqrt{x^2 - 1}\right)^{2i}} - \frac{5x^4}{2} + 1 \right) \, dx = e^{-\pi}.\tag{2}$$</span></p><p><span style="font-size: large;">$e^{\pi}$ is known as <a href="https://en.wikipedia.org/wiki/Gelfond%27s_constant" target="_blank">Gelfond's constant</a>.</span></p><p><span style="font-size: large;">I have provided these integrals as a response to a <a href="https://math.stackexchange.com/questions/122693/is-there-a-definite-integral-that-yields-e-pi-or-e-pi-in-a-non-trivial" target="_blank">question on MathSE</a>. The proofs are left as exercises for the reader.</span></p><p><span style="font-size: large;"><br /></span></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-15492364065223499412024-02-07T15:23:00.000-08:002024-02-29T05:27:26.276-08:00Solving 'impossible' integrals with a new trick<div style="text-align: center;"><span><span style="background-color: white; color: #0c0d0e; font-family: Georgia, Cambria, "Times New Roman", Times, serif; text-align: left;"><i></i></span></span></div><blockquote><div style="text-align: center;"><span><span style="background-color: white; color: #0c0d0e; font-family: Georgia, Cambria, "Times New Roman", Times, serif; text-align: left;"><i>"Complexification formulas are great and it seems like this simplifies the right away."</i></span></span></div><div style="text-align: center;"><i style="color: #0c0d0e; font-family: Georgia, Cambria, "Times New Roman", Times, serif;">Ninad Munshi</i></div></blockquote><div style="text-align: center;"><br /></div><span style="font-size: large;">The following identities have been suggested based on formulas in <a href="https://math.stackexchange.com/questions/4844145/trigonometric-formula-for-solving-quadratic-equations" target="_blank">a previous question of mine</a>.<br /><br /></span><span style="font-size: large;">If complex $\theta_1=\cos^{-1}(p)$ and $\theta_2=\sec^{-1}(p)$, where $p\in(-1, 0) \cup (1, \infty)$, then the following relation holds:<br /><br />$$e^{i\theta_1}=\frac{1-\tan\frac{\theta_2}{2} }{1+\tan\frac{\theta_2}{2}}. <br /> \tag{1}\label{463459_1}$$<br /><br />And for $p\in(-\infty, -1)\cup (0, 1)$, we have<br /><br />$$e^{-i\theta_1}=\frac{1-\tan\frac{\theta_2}{2} }{1+\tan\frac{\theta_2}{2}}. \tag{2}\label{463459_2}$$<br /><br />If complex $\theta_1=\sin^{-1}(p)$ and $\theta_2=\csc^{-1}(p)$, where $p\in(-\infty, -1)\cup (0, 1)$, then the following relation holds:<br /><br />$$ie^{i\theta_1}=-\tan\frac{\theta_2}{2}.\tag{3}\label{463459_3}$$ <br /><br />And for $p\in(-1, 0) \cup (1, \infty)$ we have<br /><br />$$ie^{-i\theta_1}=\tan\frac{\theta_2}{2}.\tag{4}\label{463459_4}$$</span><span style="font-size: large;"><br />There are several variants that we can obtain by equating (and simplifying) the trigonometric formulas for quadratic equations from <a href="https://math.stackexchange.com/questions/4844145/trigonometric-formula-for-solving-quadratic-equations" target="_blank">my previous question</a>.<br /><br />I have noticed that for certain trigonometric integrals defined over permissible intervals of $p$, the evaluation simplifies considerably. For instance, consider the following definite integral:<br /><br />$$\int_2^5 \sqrt{\tan\left(\frac{\csc^{-1}(x)}{2}\right)} \,dx.\tag{5}$$<br /><br /><a href="https://www.integral-calculator.com/#expr=sqrttan%28arccscx%2F2%29&lbound=2&ubound=5" target="_blank">This integral calculator</a> returns the following (<a href="https://www.wolframalpha.com/input?i=Integrate%5BSqrt%5BTan%5BArcCsc%5Bx%5D%2F2%5D%5D%2C+%7Bx%2C+2%2C+5%7D%5D" target="_blank">although Wolfram Alpha solves it</a>):<br /><br /><i><span style="color: red;"><b>No antiderivative could be found within the given time limit, or all supported integration methods were tried unsuccessfully. Note that many functions don't have an elementary antiderivative.</b></span></i><br /><br />But it gives an approximation of $1.178881841955109.$<br /><br />Given that the interval $[2, 5]$ is within the permissible values of $p$, we can use $e^{i\cos^{-1}(p)}=\tan\frac{\csc^{-1}(p)}{2}$ (derived from identities $(1)$ and $(4)$, valid for $p\in[-1, 0) \cup [1, \infty)$, to convert $(5)$ into<br /><br />$$\int_2^5 \sqrt{e^{i\cos^{-1}(x)}}\,dx.\tag{6}$$<br /><br /><a href="https://www.integral-calculator.com/#expr=%28e%5E%28iarccosx%29%29%5E%281%2F2%29&lbound=2&ubound=5" target="_blank">The same calculator</a> provides the same answer but now displaying the steps as well. As a second example, <a href="https://mathoverflow.net/questions/463459/identities-that-simplify-tedious-integrals?noredirect=1#comment1203657_463459" target="_blank"><i>Mathematica</i> is unable to solve this integral</a>:</span><div><span style="font-size: large;"><br /></span><div><span style="font-size: large;">$$\int_{2}^{3} \frac{{1 - \tan\frac{{\sec^{-1}x}}{2}}}{{1 + \tan\frac{{\sec^{-1}x}}{2}}}\sqrt{\tan\frac{\csc^{-1}x}{2}}\,dx\tag{7}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Neither <a href="https://www.integral-calculator.com/#expr=%28%281-tan%28%28arcsecx%29%2F2%29%29%2F%20%281%2Btan%28%28arcsecx%29%2F2%29%29%20%29%2Asqrt%28tan%28%28arccscx%29%2F2%29%29&lbound=2&ubound=3" target="_blank">this integral calculator</a>. Although both the calculator and <a href="https://www.wolframalpha.com/input?i=integrate+%28%281-tan%28%28arcsecx%29%2F2%29%29%2F+%281%2Btan%28%28arcsecx%29%2F2%29%29+%29*sqrt%28tan%28%28arccscx%29%2F2%29%29+from+2+to+3&lang=es" target="_blank">Wolfram Alpha can give you a numerical approximation</a>. However, thanks to this new trick, you can convert $(7)$ into</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int_{2}^{3} e^{\frac{3i\arccos(x)}{2}}\,dx\tag{8}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Note that <a href="https://www.integral-calculator.com/#expr=e%5E%28i%2Aarccosx%29%2Ae%5E%280.5%2Ai%2Aarccosx%29&lbound=2&ubound=3" target="_blank">this integral calculator</a> has no problem solving (elegantly!) integral $(8)$.</span></div></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Other examples of integrals that at least Wolfram Alpha is not capable of solving but that can be evaluated using the transformations described in this blog (each integral is linked to its solution in the integral calculator):</span></div><div><span style="font-size: large;"><div><br /></div><div><a href="https://www.integral-calculator.com/#expr=lnxsqrt%28e%5E%28i%2Aarccosx%29&lbound=2&ubound=3" target="_blank">$$\int_{2}^{3} \ln(x) \sqrt{\tan\left(\frac12\csc^{-1}x\right)} \, dx\tag{9}$$</a></div><div><br /></div><div><a href="https://www.integral-calculator.com/#expr=sin%28e%5E%28i%2Aarccosx%29%29&lbound=2&ubound=3" target="_blank">$$\int_{2}^{3} \sin\left(\tan\frac12\csc^{-1}{x}\right) \, dx\tag{10}$$</a></div><div><br /></div><div><a href="https://www.integral-calculator.com/#expr=cos%28e%5E%28i%2Aarccosx%29&lbound=2&ubound=3" target="_blank">$$\int_{2}^{3} \cos\left(\tan\frac12\csc^{-1}{x}\right) \, dx\tag{11}$$</a></div><div><br /></div><div>So far I have considered integrals involving $\tan{\left(\frac12\csc^{-1}x\right)}$. However, this technique can be applied to a countless number of cases that surpass my initial expectations. For example, consider the integral <a href="https://www.wolframalpha.com/input?i=integrate+sqrt%28tanx%29+from+pi%2F4+to+pi%2F20&lang=es" target="_blank">$\int_{\frac{\pi}{4}}^{\frac{\pi}{20}}\sqrt{\tan{x}}\,dx$</a>. This can be solved by letting $x=\frac12\csc^{-1}t$, where $\,dx=-\frac{1}{2t\sqrt{t^2-1}}\,dt$, transforming the original integral into <a href="https://www.integral-calculator.com/#expr=-0.5%2Asqrt%28e%5E%28i%2Aarccost%29%29%2A%281%2F%28tsqrt%28t%5E2-1%29&intvar=t&lbound=csc%28pi%2F2%29&ubound=csc%28pi%2F10%29" target="_blank">$-\frac12\int_{\csc(\frac{\pi}{2})}^{\csc(\frac{\pi}{10})}e^{\frac12i\arccos(t)} \left(\frac{1}{t\sqrt{t^2-1}}\right) \, dt$</a>. Certainly, there will be instances where employing these transformations might seem overly intricate, akin to cutting bread with a saw. However, what I aim to emphasize is the remarkable versatility of this technique.</div><div><br /></div><div><b>Related material</b></div><div><a href="https://en.wikipedia.org/wiki/Integration_using_Euler%27s_formula" target="_blank">Integration using Euler's formula</a></div></span></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-16166162989090012232024-01-13T11:33:00.000-08:002024-01-19T11:25:29.335-08:00Trigonometric formula for solving quadratic equations<span style="font-size: large;">In this note, we provide an alternative trigonometric formula for solving <a href="https://en.m.wikipedia.org/wiki/Quadratic_equation?fbclid=IwAR27kNZLhYdXZ4Hc1KNJp1QOsURfCmMpLv-FGK4hRav1TxR2jv11-qZc54E" target="_blank">quadratic equations</a> where $a$, $b$, and $c$ are non-zero real numbers.<br /><br /></span><div><span style="font-size: large;">If $ax^2+bx+c=0$, then </span><div><span style="font-size: large;">$$x_{1,2}=\left(1\pm\frac{2}{\tan{\frac{\theta}{2}\mp1}}\right)\sqrt{\frac{c}{a}},$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">where $\theta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Proof</b>. Multiplying by $a$ the quadratic equation we have</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$(ax)^2+b(ax)+ac=0.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Let's make the substitution $\sec{\theta}=\frac{b}{2\sqrt{ac}}$, then</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\begin{aligned}0&=(ax)^2+b(ax)+ac\\&=(ax)^2+2\sqrt{ac}(ax)\sec{\theta}+ac\\&=\sec{\theta}\left((ax)^2\cos{\theta}+2\sqrt{ac}(ax)+ac\cos{\theta}\right)\\&=\cos{\theta}((ax)^2+ac)+2\sqrt{ac}(ax)\\&= \cos{\theta}(ax+\sqrt{ac})^2-2\sqrt{ac}(ax)\cos{\theta}+2\sqrt{ac}(ax)\\&= \left(\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\right)(ax+\sqrt{ac})^2-2\sqrt{ac}(ax)\left(1-2\sin^2{\frac{\theta}{2}}\right)+2\sqrt{ac}(ax)\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 +4\sqrt{ac}(ax)\sin^2{\frac{\theta}{2}}\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}\left((ax+\sqrt{ac})^2 -4\sqrt{ac}(ax)\right)\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}(ax-\sqrt{ac})^2\\&= \left(\cos{\frac{\theta}{2}}(ax+\sqrt{ac}) +\sin{\frac{\theta}{2}}(ax-\sqrt{ac})\right) \left(\cos{\frac{\theta}{2}}(ax+\sqrt{ac}) -\sin{\frac{\theta}{2}}(ax-\sqrt{ac})\right)\\&=\left(ax\left(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right) \right) \left(ax\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\left(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\right) \right).\end{aligned}$$</span></div><div><br /></div><div><span style="font-size: large;">Now, by setting the factors equal to zero and solving for x, we obtain,</span></div><div><span style="font-size: large;"><br /></span></div><span style="font-size: large;">$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}}\right)\sqrt{\frac{c}{a}}\\&= \left(1+\frac{2}{\tan{\frac{\theta}{2}-1}}\right)\sqrt{\frac{c}{a}}. \end{aligned}$$</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Similarly we obtain, </span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\begin{aligned}x_2= \left(1-\frac{2}{\tan{\frac{\theta}{2}+1}}\right)\sqrt{\frac{c}{a}} \end{aligned}.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><div style="text-align: right;"><span style="font-size: x-large;">$\square$</span></div><div style="text-align: left;"><span style="font-size: large;"><br /></span></div><div style="text-align: left;"><span style="font-size: large;">The formula appears to be new, at least on the internet. <a href="https://en.wikipedia.org/wiki/Quadratic_equation#Trigonometric_solution" target="_blank">Wikipedia</a> presents a trigonometric method, but it is different from mine. Stuart Simons offers a method that does seem to be related to my approach, but he only provides direct formulas for complex roots. I was able to independently derive Simons' formulas using $\cos{\theta}$ in the substitution instead of $\sec{\theta}$, and this was before coming across the Wikipedia article that references Simons' paper: <a href="https://www.cambridge.org/core/journals/mathematical-gazette/article/abs/9304-alternative-approach-to-complex-roots-of-real-quadratic-equations/3C86AC38022EB371998C12462E02BE6A" target="_blank">Simons, Stuart, 'Alternative approach to complex roots of real quadratic equations,' <i>Mathematical Gazette</i> 93, March 2009, 91–92.</a></span></div></div></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><span style="font-size: large;">More discussion about this formula can be found on the <a href="https://math.stackexchange.com/questions/4844145/trigonometric-formula-for-solving-quadratic-equations?noredirect=1#comment10323646_4844145" target="_blank">MathSE</a> forum.</span></div><div style="text-align: left;"><span style="font-size: large;"><br /></span></div><div style="text-align: left;"><span style="font-size: large;"><b>Remark</b>. <a href="https://geometriadominicana.blogspot.com/2023/12/the-half-angle-formulas-are-central.html" target="_blank">The half-angle formulas are ubiquitous</a>, and this alternative formula for quadratic equations is another piece of evidence for that.</span></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-59004407830415072822024-01-04T20:58:00.000-08:002024-01-07T14:54:52.399-08:00A generalization of Burlet's theorem to cyclic quadrilaterals<p><span style="font-size: large;"> The <a href="https://pt.m.wikipedia.org/wiki/Teorema_de_Burlet?fbclid=IwAR27ynAJMgfKKIoUFr4ojMRHaO9JCtVpGptX5aDQ9LDIblF8crdgXFlTpSw" target="_blank">Burlet's theorem</a> is a result in Euclidean geometry, which can be formulated as follows:</span></p><p><span style="font-size: large;"><b>Theorem 1</b>. Consider triangle $ABC$ with $\angle{BCA}=\gamma$. Let $P$ be the point where the incircle touches side $AB$, and denote the lengths $AP$ and $BP$ as $m$ and $n$, respectively. Then</span></p><p><span style="font-size: large;">$$\Delta_0=mn\cot{\frac{\gamma}{2}},\tag{1}$$</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhl6O5kCkFgpP7iNbVU-i6TMJvf9zbSwoy6-earbgCfCW36IG1Tz5FRfMWQgBYfBrVzhr9zOd9Cq1-ApZwC9A-NuRwZsI16qCP7fkwengDvCEjlOd7NedPSFtbUnL79-_d7J_P4u7gVddogJ-yzEUgPZ6vryC4DHoKRr4zcTgED9rilC0xy9pzuab6mLXaF/s1998/Generalized%20Burlet's%20Theorem3.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="1811" data-original-width="1998" height="363" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhl6O5kCkFgpP7iNbVU-i6TMJvf9zbSwoy6-earbgCfCW36IG1Tz5FRfMWQgBYfBrVzhr9zOd9Cq1-ApZwC9A-NuRwZsI16qCP7fkwengDvCEjlOd7NedPSFtbUnL79-_d7J_P4u7gVddogJ-yzEUgPZ6vryC4DHoKRr4zcTgED9rilC0xy9pzuab6mLXaF/w400-h363/Generalized%20Burlet's%20Theorem3.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">A triangle $ABC$ with $AP=m$ and $BP=n$.</td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><span style="font-size: large;">where $\Delta_0$ denotes the area of $ABC$. Note that when $\gamma=\frac{\pi}{2}$, then $\Delta_0=mn$.</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">We adopt unconventional notation to reduce the relation of Theorem 2 into the one referenced in $(1)$. Additionally, we denote $s$ the semiperimeter, applicable to both triangles and cyclic quadrilaterals.<br /></span><p></p><p><span style="font-size: large;"><b>Proof</b>. Let $AB=a$, $BC=b$ and $AC=c$. By <a href="https://en.wikipedia.org/wiki/Heron%27s_formula" target="_blank">Heron's formula</a>, we have</span></p><span style="font-size: large;">$$\begin{aligned}\Delta_0&=\sqrt{s(s-a)(s-b)(s-c)}\\&=\sqrt{s(s-a)(s-b)(s-c)}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}.\end{aligned}$$</span><p><span style="font-size: large;">However, using <a href="https://geometriadominicana.blogspot.com/2020/06/another-proof-for-two-well-known.html" target="_blank">half-angle formulas for a triangle</a>, we can express $\cot{\frac{\gamma}{2}}$ as $\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$, where $m=(s-b)$ and $n=(s-c)$. This establishes the relationship mentioned in $(1)$.</span></p><p style="text-align: right;"><span style="font-size: large;">$\square$</span></p><p style="text-align: left;"><span style="font-size: large;"><b>Theorem 2 (Generalization)</b>. Consider a cyclic quadrilateral $ABCD$ with side lengths $AB=a$, $BC=b$, $CD=c$, and $DA=d$. Additionally, let $\angle DAB = \alpha$. In this context, the following relation is valid:</span></p><p style="text-align: left;"><span style="font-size: large;">$$\Delta_1=(s-a)(s-d)\cot{\frac{\alpha}{2}},\tag{2}$$</span></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhK_dVInlJz3lO59l_O6NTGZYRVqbJkderX_rEBt0R807Ra76Ft7DyxvNu6WdGv2srYs3-_lrt1s4TMhiB2Es9rFWgBb1jqtjmeXN4yOl5_jvjKEHct9x5unBMxF42t3qDERJJsaD0ul-cw6bxtegE-HoDXFWHF-6_ToEH7aC8JGCN3NdzgpwWsmsQC90G9/s5202/Generalized%20Burlet's%20Theorem.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="5202" data-original-width="4999" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhK_dVInlJz3lO59l_O6NTGZYRVqbJkderX_rEBt0R807Ra76Ft7DyxvNu6WdGv2srYs3-_lrt1s4TMhiB2Es9rFWgBb1jqtjmeXN4yOl5_jvjKEHct9x5unBMxF42t3qDERJJsaD0ul-cw6bxtegE-HoDXFWHF-6_ToEH7aC8JGCN3NdzgpwWsmsQC90G9/w385-h400/Generalized%20Burlet's%20Theorem.png" width="385" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">A cyclic quadrilateral $ABCD$.</td></tr></tbody></table><p style="text-align: left;"><span style="font-size: large;">where $\Delta_1$ denotes the area of $ABCD$.</span></p><p style="text-align: left;"><span style="font-size: large;"><b>Proof.</b> Similar to how we proceeded in theorem 1, using the <a href="https://en.wikipedia.org/wiki/Brahmagupta%27s_formula" target="_blank">Brahmagupta's formula</a>, we have</span></p><p style="text-align: left;"><span style="font-size: large;">$$\begin{aligned}\Delta_1&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\color{red}{\sqrt{\frac{(s-a)(s-d)}{(s-a)(s-d)}}}\\&=(s-a)(s-d)\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}.\end{aligned}$$</span></p><p style="text-align: left;"><span style="font-size: large;">Now, invoking the <a href="https://geometriadominicana.blogspot.com/2020/07/killing-three-birds-with-one-stone.html" target="_blank">half-angle formulas for cyclic quadrilaterals</a>, we have that $\cot{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}$, from which the relation $(2)$ is deduced.</span></p><p style="text-align: right;"><span style="font-size: large;">$\square$</span></p><span style="font-size: large;">Let $\angle{BCD}=\gamma$ and assume that $d=0$. We have that $\alpha=\pi-\gamma$, so substituting in $(2)$, </span></div><div><span style="font-size: large;"><br /></span></div><div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9ufRS1rIsE2gmBdKPdHdqIoxfL6-KiWxCO8qVG0ZzE-kcbDOLuToqs81YnMosDwMVK7Yw7HGZGOPBYUjK1Q02Ga8UcvHqCbhzdvRTlOcgY0d3Ld_EdYp7Tv5m6RieX8ntZzg8yU-psBssjtpP_fKORDDD6CMMdair8XTJ4qVC2Q4m7vQilbVmvg8BeMFJ/s1998/Generalized%20Burlet's%20Theorem2.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="1811" data-original-width="1998" height="363" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9ufRS1rIsE2gmBdKPdHdqIoxfL6-KiWxCO8qVG0ZzE-kcbDOLuToqs81YnMosDwMVK7Yw7HGZGOPBYUjK1Q02Ga8UcvHqCbhzdvRTlOcgY0d3Ld_EdYp7Tv5m6RieX8ntZzg8yU-psBssjtpP_fKORDDD6CMMdair8XTJ4qVC2Q4m7vQilbVmvg8BeMFJ/w400-h363/Generalized%20Burlet's%20Theorem2.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">When $d=0$, by the alternate segment theorem, $\alpha=\pi-\gamma$.</td></tr></tbody></table><br /><span style="font-size: large;">$$\begin{aligned}\Delta_1&=(s-a)(s-0)\cot{\left(\frac{\pi-\gamma}{2}\right)}\\&=s(s-a)\tan{\frac{\gamma}{2}}\\&=s(s-a)\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\\&=(s-b)(s-c)\cot{\frac{\gamma}{2}},\end{aligned}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">which is Burlet's theorem for a triangle since $m=(s-b)$ and $n=(s-c)$.</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Remark</b>. </span><span style="font-size: large;">A. Burlet (Dublin) presented this theorem as a problem in the <i><a href="https://www.google.com/books/edition/Nouvelles_annales_de_math%C3%A9matiques/RxUFAAAAQAAJ?hl=en&gbpv=1&pg=PA290&printsec=frontcover" target="_blank">Nouvelles annales de mathématiques, Vol. 15, 1856, p. 290</a></i>. Consequently, the theorem might have derived its name from him.</span></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-21898580678712983952023-12-19T07:41:00.000-08:002023-12-19T07:45:40.633-08:00The half-angle formulas are central!<p><span style="font-size: large;"> As the title suggests, the half-angle formulas are <i>central</i>. Even more central than the law of cosines, which is nothing more than the half-angle formulas in disguise. Virtually, every metric relationship characterizing the triangle can be derived from the triangular half-angle formulas. The mind map below can give you an idea of the pivotal role that the half-angle formulas play in relation to the other most important metric relationships in classical geometry. I have also written an essay titled <a href="https://drive.google.com/file/d/1jPrWxUhB1Tnsd5RH5UiMhwmkp2jCv11v/view?usp=sharing" target="_blank">"The Theoretical Importance of Half-Angle Formulas"</a> where you can see the details of the proofs, as well as the <i>new</i> generalizations I have managed to derive from this novel approach.</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHEaBgtHePt7GCqm0CriWEGKzE5AE4tk-_iNI0LbY42ZKUJ0n661agPhP54DQiykoUAwXhm8an408FZkzckihDgSrPD-i6DDDvksULyn0v3qVwmQ6JZ3HJx0tzxHH4h-c1SojNOkwC8earKgdmuJSO01dmNJdLsSkewDJ2oPfg9WDfiQe5lyUO8xG_OWnh/s4997/mapmind3.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="2316" data-original-width="4997" height="296" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHEaBgtHePt7GCqm0CriWEGKzE5AE4tk-_iNI0LbY42ZKUJ0n661agPhP54DQiykoUAwXhm8an408FZkzckihDgSrPD-i6DDDvksULyn0v3qVwmQ6JZ3HJx0tzxHH4h-c1SojNOkwC8earKgdmuJSO01dmNJdLsSkewDJ2oPfg9WDfiQe5lyUO8xG_OWnh/w640-h296/mapmind3.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Click on the image for a better view</td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;"><span style="font-size: large;">Identifying the central theorems of an area is important because it helps streamline the process of understanding it. By knowing the basic principles, you can, with logic and a bit of ingenuity, understand (by proving) the rest of the discipline without having to memorize as much.</span></div><p></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-40598317385782164722023-08-22T07:36:00.007-07:002023-08-28T04:10:53.725-07:00 “Matemáticos” Dominicanos que Publican en Revistas de Dudosa Reputación<p><span style="font-size: large;">En un escenario matemático que desafía toda lógica y sentido común, un grupo de investigadores dominicanos, respaldados por el <a href="https://mescyt.gob.do/programas/fondocyt/" target="_blank">Fondo Nacional de Innovación y Desarrollo Científico y Tecnológico (FONDOCYT)</a> ha sido atrapado en un laberinto editorial de dudosa reputación. ¿El lugar de los hechos? La revista <a href="https://www.mdpi.com/journal/mathematics" target="_blank"><i>Mathematics</i></a>, publicada por la misteriosa <a href="https://es.wikipedia.org/wiki/MDPI#Controversias" target="_blank">MDPI (Multidisciplinary Digital Publishing Institute)</a>, una editorial que ha estado en el ojo del huracán académico en los últimos años. Pero, ¿qué está ocurriendo realmente en el mundo de las publicaciones matemáticas? Prepárense para sumergirse en un cuento de matemáticas y manipulación editorial que pondrá a prueba su fe en la integridad académica.</span></p><p><span style="font-size: large;">El escándalo salió a la luz gracias al matemático de la <a href="https://en.m.wikipedia.org/wiki/University_of_California,_Los_Angeles" target="_blank">Universidad de California, Los Angeles (UCLA)</a>, <a href="https://en.m.wikipedia.org/wiki/Igor_Pak?fbclid=IwAR3FpfniF6wn9zb0u4p0sRW2QTYU65MKlF1B8bhrfiTqglHF13-G224uSzQ" target="_blank">Igor Pak</a>. Un hombre que ha dedicado su vida a la investigación matemática y cuya carrera está salpicada de logros académicos. ¿Quién mejor para arrojar luz sobre esta sombría situación? Échele un ojo a su artículo <a href="https://igorpak.wordpress.com/2022/01/?fbclid=IwAR2gTS5LSN_raLJHx6izk0FUv9IPvdXuAhpdhstEjUMsvTg1bKcWCun0yYg" target="_blank">The insidious corruption of open access publishers</a>.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>¿Qué es MDPI?</b></span></p><p><span style="font-size: large;">MDPI es una editorial de revistas en línea de "acceso abierto" con fines de lucro. ¿Son legítimas o depredadoras? Esa es una buena pregunta. El mundo académico parece estar un poco perplejo al respecto. MDPI publica más de 200 revistas, la mayoría de las cuales tienen títulos de una sola palabra que parecen salidos de un generador aleatorio de temas: <a href="https://www.mdpi.com/journal/data" target="_blank">Data</a>, <a href="https://www.mdpi.com/journal/diseases" target="_blank">Diseases</a>, <a href="https://www.mdpi.com/journal/diversity" target="_blank">Diversity</a>, <a href="https://www.mdpi.com/journal/dna" target="_blank">DNA</a>, etc. Si eres investigador, seguro que estas revistas han estado inundando su bandeja de entrada de correo electrónico, incluso más que los príncipes nigerianos. ¿No ha revisado su carpeta de spam últimamente? MDPI muy probablemente le ha enviado invitaciones para ser editor invitado en campos tan variados como <a href="https://www.mdpi.com/journal/sustainability" target="_blank">Sustainability</a> y <a href="https://www.mdpi.com/journal/symmetry" target="_blank">Symmetry</a>, desde <a href="https://www.mdpi.com/journal/entropy" target="_blank">Entropy</a> hasta <a href="https://www.mdpi.com/journal/axioms" target="_blank">Axioms</a>. Al parecer, solo necesita estar "vivo" y poder "hablar" para ser editor invitado en todas ellas.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>¿Qué es <i>Mathematics</i>?</b></span></p><p><span style="font-size: large;">Ahora, centrémonos en la revista <i>Mathematics</i>, que ha estado en el centro de esta tormenta editorial. Fundada en 2013, ha publicado más de 7,600 artículos. Aparentemente, no está revisada por <a href="https://en.wikipedia.org/wiki/MathSciNet" target="_blank">MathSciNet</a> ni <a href="https://es.wikipedia.org/wiki/Zentralblatt_MATH" target="_blank">ZbMath</a>, lo que normalmente sería una señal de advertencia. Sin embargo, sorprendentemente, su factor de impacto parece indicar que es una de las revistas matemáticas más importantes, rivalizando con nombres establecidos como <a href="https://en.wikipedia.org/wiki/Duke_Mathematical_Journal" target="_blank"><i>Duke</i></a>, <a href="https://es.wikipedia.org/wiki/American_Journal_of_Mathematics" target="_blank"><i>Amer. Jour. Math</i></a> y <a href="https://en.wikipedia.org/wiki/Journal_of_the_European_Mathematical_Society" target="_blank"><i>JEMS</i></a>. ¿Cómo puede ser esto? Es una pregunta que aún no tiene respuesta.</span></p><p><span style="font-size: large;">Pak revela que <i>Mathematics</i> tiene un consejo editorial que parece más grande que una multitud en un concierto de rock. Con 11 "editores en jefe" y 814 "editores", <i>Mathematics</i> parece estar lista para cubrir todo el espectro matemático, o eso parece. En comparación, <a href="https://en.wikipedia.org/wiki/Transactions_of_the_American_Mathematical_Society" target="_blank"><i>Trans. AMS</i></a> tiene alrededor de 25 miembros en su consejo editorial. ¿Quiénes son estos editores? Pak descubrió que algunos de ellos son matemáticos respetados, pero no mencionan <i>Mathematics</i> en sus currículos. ¿Por qué? ¿Están avergonzados de estar asociados con esta revista? ¿O simplemente no están lo suficientemente avergonzados como para pedir que se elimine su nombre de la lista? Las preguntas sin respuesta abundan.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>El Misterio del Atractivo de <i>Mathematics</i></b></span></p><p><span style="font-size: large;">Entonces, ¿por qué <i>Mathematics</i> es tan popular? Algunos sugieren que es porque aceptan todo tipo de trabajos, incluso aquellos que son, en el mejor de los casos, cuestionables. <i>Mathematics</i> se enorgullece de tomar decisiones rápidas. Al parecer, eligen a los revisores al azar de su base de datos, a menudo sin experiencia en el tema del artículo. Esto, combinado con un plazo de dos semanas para las revisiones, da como resultado una evaluación superficial en el mejor de los casos.</span></p><p><span style="font-size: large;">Pak señala que las razones detrás de esta popularidad son claras: <i>Mathematics</i> acepta cualquier cosa. Los autores pueden obtener una decisión en días, en lugar de esperar un año o más en otras revistas de alto prestigio. Esto plantea un dilema para los autores que se sienten presionados por el "publica o perece". Pero, ¿qué hay detrás de esta eficiencia aparente?</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>La Estrategia de MDPI para Hacer Dinero: Vendiendo "Aire" a Escala Masiva</b></span></p><p><span style="font-size: large;">La estrategia maestra de MDPI parece ser, a simple vista, una fórmula efectiva pero perturbadora: cobrar tarifas de procesamiento de artículos (APC) y hacerlo a gran escala. Mientras que revistas de renombre como <i>Trans. AMS</i>, <i>Ser. B</i> pueden cobrar miles de dólares en APC por artículo, MDPI opta por un enfoque diferente, solicitando alrededor de 1,960 dólares suizos, lo que equivale a aproximadamente 2,000 dólares estadounidenses, por cada artículo. Sin embargo, lo que hace que esta estrategia sea verdaderamente impresionante es la sorprendente cantidad de artículos que publican. En esencia, están vendiendo "aire", un producto que a primera vista puede parecer insustancial, pero que al parecer ha cautivado a un gran número de personas.</span></p><p><span style="font-size: large;">La premisa es simple: al cobrar tarifas de APC, MDPI obtiene ingresos significativos de cada artículo que procesan. Lo que hace que esta estrategia sea altamente rentable es la voluminosa cantidad de artículos que publican a lo largo del tiempo. A pesar de que sus tarifas son relativamente bajas en comparación con las de otras revistas de prestigio, su enfoque en la cantidad sobre la calidad les ha permitido generar un flujo constante de ingresos.</span></p><p><span style="font-size: large;">Es como si estuvieran vendiendo una ilusión de reconocimiento académico: un lugar donde los investigadores pueden ver sus trabajos publicados rápidamente y, en teoría, obtener ese anhelado "factor de impacto" que tanto importa en el mundo académico. En un entorno donde la presión por publicar es intensa y el tiempo es un recurso valioso, esta oferta puede ser atractiva para muchos.</span></p><p><span style="font-size: large;">La pregunta que surge es si este modelo de negocio sacrifica la integridad académica en favor de las ganancias financieras. ¿Está MDPI más interesada en llenar sus arcas que en mantener los estándares rigurosos de revisión por pares y asegurar que solo se publiquen investigaciones de alta calidad? La respuesta a esta pregunta es esquiva y genera controversia, ya que algunos argumentan que MDPI está desdibujando las líneas entre la publicación legítima y la publicación cuestionable en su búsqueda de ingresos.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>¿Predadora o No Predadora?</b></span></p><p><span style="font-size: large;">La gran pregunta es si <i>Mathematics</i> y MDPI son depredadores o simplemente astutos comerciantes. Pak argumenta que <i>Mathematics</i> es más parasitaria que depredadora. Los depredadores suelen engañar a los autores para que paguen por publicar en sus revistas de aspecto dudoso. Los autores son víctimas de un engaño, pensando que están obteniendo reconocimiento científico. En contraste, <i>Mathematics</i> parece no engañar a sus autores, quienes aparentemente están felices con el resultado.</span></p><p><span style="font-size: large;">Entonces, ¿quiénes son las víctimas aquí? Las bibliotecas universitarias y los organismos de financiamiento están gastando grandes sumas de dinero en productos de calidad inferior. Las investigaciones y la educación podrían beneficiarse mucho más de este dinero en otros lugares. En resumen, <i>Mathematics</i> corrompe todo el proceso de revisión por pares al monetizarlo hasta el punto en que el costo del APC se convierte en la consideración principal en lugar de la contribución matemática real del artículo.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>El Futuro de las Revistas Matemáticas</b></span></p><p><span style="font-size: large;">Entonces, ¿qué nos depara el futuro de las revistas matemáticas? La respuesta parece clara según Pak y su sutil ironía: las bibliotecas deben dejar de pagar por el acceso abierto. Los organismos de financiamiento deben prohibir que las subvenciones se utilicen para pagar publicaciones. Los matemáticos deben huir en la dirección opuesta cada vez que alguien mencione el dinero. Simplemente digan no.</span></p><p><span style="font-size: large;">El modelo correcto, según Pak, es el modelo de superposición de <a href="https://es.wikipedia.org/wiki/ArXiv" target="_blank">arXiv</a>, que es económico y accesible. No hay necesidad de que las bibliotecas paguen por la publicación si el artículo ya está disponible de forma gratuita en arXiv. El mensaje es claro: el dinero que las universidades gastan en APC de <i>Mathematics</i> sería mejor invertido en apoyar la investigación y la educación reales.</span></p><p><span style="font-size: large;">Si usted es miembro del consejo editorial de <i>Mathematics</i>, Pak tiene un mensaje para usted: renuncie y nunca revele que estuvo allí. Ya obtuvo lo que quería, su artículo se publicó, su nombre está en la portada de alguna edición especial (las imprimen para los autores). Quizás la vergüenza funcione en este caso.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>El Caso de Idriss Aberkane y la Conjetura de Collatz</b></span></p><p><span style="font-size: large;">En medio del torbellino editorial que rodea a la revista <i>Mathematics</i> y su peculiar enfoque en la publicación de artículos matemáticos, se encuentra un caso que ha dejado perplejos a matemáticos de todo el mundo. Hablamos del controvertido <a href="https://en.wikipedia.org/wiki/Idriss_Aberkane" target="_blank">Idriss Aberkane</a>, un orador público y ensayista francés, cuyas afirmaciones y publicaciones han desencadenado una serie de reacciones, que van desde la admiración hasta la crítica más feroz.</span></p><p><span style="font-size: large;">La biografía de Aberkane es un compendio de controversia. Conocido por sus escritos y conferencias sobre desarrollo personal, en 2016 publicó un exitoso ensayo titulado "¡Libera tu mente!". Sin embargo, su carrera se vio ensombrecida por acusaciones de inflar artificialmente su currículum y de hablar sobre ciencias que no estaban dentro de su área de experiencia. Algunos investigadores cuestionaron la precisión científica de sus afirmaciones y publicaciones, lo que le valió la etiqueta de "antivacunas" y "conspirador" debido a sus opiniones sobre la COVID-19 y las vacunas.</span></p><p><span style="font-size: large;">Pero lo que ha dejado a la comunidad matemática perpleja es la afirmación de Aberkane de haber resuelto la <a href="https://es.wikipedia.org/wiki/Conjetura_de_Collatz" target="_blank">Conjetura de Collatz</a>. Esta conjetura, que ha desconcertado a matemáticos durante décadas, trata sobre una secuencia numérica aparentemente simple, pero aún no se ha encontrado una prueba definitiva para confirmar su veracidad. Aberkane publicó un artículo titulado <a href="https://www.mdpi.com/2227-7390/10/11/1835" target="_blank"><i>Collatz Attractors Are Space-Filling</i></a> en la revista <i>Mathematics</i>, que afirma haber resuelto este enigma.</span></p><p><span style="font-size: large;">Lo que hace que este caso sea aún más intrigante es que <i>Mathematics</i>, la misma revista que ha estado bajo escrutinio por sus prácticas editoriales, publicó el artículo de Aberkane. Esta revista, que ha sido objeto de debate debido a la aparente falta de revisión y los cuestionamientos sobre su integridad, sorprendió a muchos al respaldar la afirmación de Aberkane.</span></p><p><span style="font-size: large;">Aberkane no solo afirma haber resuelto la Conjetura de Collatz, sino que también se atreve a compararse con <a href="https://es.wikipedia.org/wiki/Terence_Tao" target="_blank">Terence Tao</a>, un renombrado matemático galardonado con la <a href="https://es.wikipedia.org/wiki/Medalla_Fields" target="_blank">Medalla Fields</a> (equivalente al Nobel de las matemáticas). Según Aberkane, su resultado supera al de Tao. Esta audaz afirmación ha generado escepticismo y críticas de la comunidad matemática, que encuentra difícil aceptar que alguien sin una formación matemática sólida pueda superar a un genio como Tao.</span></p><p><span style="font-size: large;">La reacción de la comunidad matemática no se hizo esperar. Expertos en matemáticas, como <a href="https://www.echosciences-hauts-de-france.fr/articles/fabien-durand-elu-president-de-la-societe-mathematique-de-france" target="_blank">Fabien Durand</a>, profesor de Matemáticas en la <a href="https://fr.wikipedia.org/wiki/Universit%C3%A9_de_Picardie_Jules-Verne" target="_blank">Université de Picardie Jules-Verne</a> y presidente de la Sociedad Matemática Francesa, han analizado las publicaciones de Aberkane y han señalado importantes errores en sus trabajos. Según Durand, la mayoría de las pruebas presentadas por Aberkane están al nivel de un estudiante de secundaria o de primer año de universidad.</span></p><p><span style="font-size: large;">El escepticismo y las críticas se han multiplicado, y la pregunta sigue sin respuesta: ¿Aberkane realmente ha resuelto la Conjetura de Collatz o se trata de una afirmación infundada? </span></p><p><span style="font-size: large;">El misterio que rodea a Idriss Aberkane y su aparición en la revista <i>Mathematics</i> es un capítulo adicional en la saga de esta revista, que ha estado en el centro de la controversia editorial. Queda por verse si Aberkane ha logrado lo que tantos matemáticos han intentado durante décadas, o si su afirmación es simplemente una pieza más en el complicado rompecabezas de las publicaciones académicas.</span></p><p><span style="font-size: large;"><br /></span></p><p><span style="font-size: large;"><b>Investigadores Dominicanos: Atrapados en la Telaraña de <i>Mathematics</i></b></span></p><p><span style="font-size: large;">Pero esta historia no estaría completa sin abordar el papel de los investigadores dominicanos atrapados en esta telaraña editorial. Entre los nombres de los investigadores que han contribuido a esta revista cuestionable se encuentran <a href="https://scholar.google.es/citations?user=RGdlbz0AAAAJ&hl=es" target="_blank">Pedro A. Solares Hernandez</a> (que incluso cuenta con su propio <a href="https://www.diariolibre.com/actualidad/ciencia/ingenieros-dominicanos-aportan-avances-en-el-estudio-de-los-numeros-primos-y-el-triangulo-de-pascal-IN18627852" target="_blank">reportaje en Diario Libre</a> donde se cita el artículo <a href="https://www.mdpi.com/2227-7390/8/2/254" target="_blank"><i>"Divisibility Patterns within Pascal Divisibility Networks"</i></a> publicado por <i>Mathematics</i>), <a href="https://scholar.google.com/citations?user=Uk7yZGYAAAAJ&hl=es" target="_blank">Juan Hernández</a> y <a href="https://scholar.google.com/citations?user=576Rq84AAAAJ&hl=es" target="_blank">Juan Toribio Milané</a>. Estos académicos dominicanos, a través de sus artículos, han quedado inmersos en una trama editorial llena de interrogantes.</span></p><p><span style="font-size: large;">Uno de los artículos que lleva la firma del investigador dominicano Juan Hernández es <a href="https://www.mdpi.com/2227-7390/11/8/1956" target="_blank"><i>"Sequentially Ordered Sobolev Inner Product and Laguerre–Sobolev Polynomials"</i></a>. Este trabajo se suma a la lista de investigaciones publicadas en <i>Mathematics</i>, una revista que ha sido objeto de debate en el mundo académico debido a sus prácticas editoriales cuestionables.</span></p><p><span style="font-size: large;">Otro artículo, titulado <a href="https://www.mdpi.com/2227-7390/11/15/3420" target="_blank">"Differential Properties of Jacobi-Sobolev Polynomials and Electrostatic Interpretation"</a>, lleva la firma de Héctor Pijeira-Cabrera, Javier Quintero-Roba y Juan Toribio-Milane. En este caso, vemos a otro investigador dominicano, Juan Toribio-Milane, involucrado en la publicación de un trabajo en <i>Mathematics</i>.</span></p><p><span style="font-size: large;">La pregunta que surge es si estos investigadores estaban plenamente conscientes de las controvertidas prácticas editoriales de MDPI y <i>Mathematics</i> al enviar sus investigaciones. ¿O fueron arrastrados por la promesa de una publicación rápida y un factor de impacto aparentemente alto? ¿O simplemente se encontraron en una encrucijada editorial donde las alternativas eran limitadas?</span></p><p><span style="font-size: large;">Independientemente de las circunstancias que llevaron a estos investigadores dominicanos a elegir <i>Mathematics</i> como el destino de sus investigaciones, su participación en esta trama editorial plantea cuestionamientos sobre las decisiones que toman los académicos en un entorno académico cada vez más complejo y desafiante.</span></p><p><span style="font-size: large;">En última instancia, estos nombres permanecerán en la narrativa de la controvertida relación entre los investigadores y las revistas académicas de reputación dudosa. Sus historias sirven como recordatorio de la importancia de la transparencia y la evaluación crítica en el mundo de la publicación académica.</span></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-72742195096735582752023-08-10T14:19:00.003-07:002023-08-11T04:05:31.808-07:00Solution to Problem 1315 of Gogeometry<p><span style="font-size: large;"> The following problem was proposed by my esteemed friend, Ángel Mejía. However, this is <a href="https://gogeometry.com/school-college/4/p1315-triangle-incircle-tangent-chord-congruence.htm" target="_blank">problem 1315 from Gogeometry</a>. Here I give a trigonometric proof.</span></p><span style="font-size: large;">The figure shows a triangle $ABC$ with the inscribed circle $O$ ($D$, $E$, and $T$ are the tangency points). $OB$ cuts chord $DE$ and arc $DE$ at $M$ and $F$, respectively. $AF$ and $CF$ cut chord $DE$ at $G$ and $N$, respectively. Prove that $DG = MN$.</span><div><span style="font-size: large;"><br /></span></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwOGF_SGa6qLr1uOUSEVkpPYWfxso7XHF1tUP5UATBNH6eLzYGL_e8naxrN3C0_DsB1yH_N0G6x6Bk4GjblvKTtu3FyDHZ90n5qu0EFn1ShTmwSI6jb_B7WghXmheee53f7i8YE41dE2BxGe-7fOTCb0L5kuKhNUq3iPFYMs6SzicPyM6suFjzENXZcdMM/s456/geogeometry.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="398" data-original-width="456" height="349" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwOGF_SGa6qLr1uOUSEVkpPYWfxso7XHF1tUP5UATBNH6eLzYGL_e8naxrN3C0_DsB1yH_N0G6x6Bk4GjblvKTtu3FyDHZ90n5qu0EFn1ShTmwSI6jb_B7WghXmheee53f7i8YE41dE2BxGe-7fOTCb0L5kuKhNUq3iPFYMs6SzicPyM6suFjzENXZcdMM/w400-h349/geogeometry.png" width="400" /></a></div><br /><span style="font-size: large;"><b>Proof</b>. Let's denote $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$, $\angle{ACB}=\gamma$, $\angle{AFM}=\angle{AFO}=\delta$, and $\angle{DFA}=\epsilon$. Applying the sine law in triangles $GFM$ and $DFG$, we obtain</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$GM = FG \cdot \sin{\delta},\tag{1}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$DG = \frac{FG \cdot \sin{\epsilon}}{\sin{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}.\tag{2}$$<br /></span><p><span style="font-size: large;">Dividing $(1)$ by $(2)$ and then squaring, we get</span></p><span style="font-size: large;">$$\left(\frac{GM}{DG}\right)^2=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\cdot{\sin^2{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\left(\frac{1-\sin{\frac12\beta}}{2}\right).\tag{3}$$</span></div><div><span style="font-size: large;">Now, applying the sine law once more, this time in triangles $AFO$ and $ADF$, we obtain</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\sin{\delta}=\frac{AO\cdot{\sin{\frac12(\alpha+\beta)}}}{AF},\tag{4}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\sin{\epsilon}=\frac{AD\cdot{\sin{\frac14(3\pi+\beta})}}{AF}.\tag{5}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Dividing $(4)$ by $(5)$ and then squaring, we have</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\frac{\sin^2{\delta}}{\sin^2{\epsilon}}=\frac{\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}}{\sin^2{\frac14(3\pi+\beta})}=\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}\left(\frac{2}{1-\sin{\frac12\beta}}\right).\tag{6}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Substituting $(6)$ in $(3)$ and taking square roots,</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\frac{GM}{DG}=\frac{\cos{\frac12\gamma}}{\cos{\frac12\alpha}}.$$</span></div><div><span style="font-size: large;"><br /></span></div><span style="font-size: large;">Analogously,</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\frac{MN}{EN}=\frac{\cos{\frac12\alpha}}{\cos{\frac12\gamma}}.$$</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">This implies $\frac{GM}{DG}=\frac{EN}{MN}$ and, since $DM=ME$, we conclude that $DG=MN$.</span></div><div><span style="font-size: large;"><br /></span></div><div style="text-align: right;"><span style="font-size: large;">$\square$</span></div></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-72453998512376667322023-07-31T08:38:00.017-07:002023-08-05T09:13:53.722-07:00Another Identity on Triangle Areas Arising from Reflection<p><span style="font-size: large;">In a <a href="https://geometriadominicana.blogspot.com/2022/12/van-kheas-areal-problem.html" target="_blank">previous post</a>, we proved an identity about triangle areas arising from reflection. In this occasion, we prove another interesting identity associated with triangle areas.</span></p><p><span style="font-size: large;"><b>Theorem</b>. Let $ABC$ be a triangle. Denote $D$, $E$ and $F$ arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $A'$ be the reflection of $A$ with respect to $D$. Define $B'$ and $C'$ similarly. Denote $A''$ the reflection of $A'$ with respect to the midpoint of $BC$. Define $B''$ and $C''$ similarly. Then </span></p><p><span style="font-size: large;">$$[A'B'C']-[A''B''C''] = 3[ABC].$$</span></p><p><span style="font-size: large;">The brackets $[\, ]$ represent the area of the enclosed figure.</span></p><p></p><p><span style="font-size: large;"><b>Lemma 1</b>. $[A'B'C]=[ABA''B'']$, $[A'C'B]=[ACA''C'']$ and $[B'C'A]=[BCB''C'']$.</span></p><p><span style="font-size: large;"><b>Proof</b>. Observe that the diagonals of $BCB'C'$ and $BCEF$ are in a 2:1 ratio, and the angle between them remains unchanged. Hence, $[BCB'C']=4[BCEF]$. This is easily deduced from the formula for the area of a quadrilateral $K=\frac12pq\sin{\theta}$, where the lengths of the diagonals are $p$ and $q$, and the angle between them is $\theta$.</span></p><p><span style="font-size: large;">Also, observe that $AF$ and $BF$ are medians of triangles $\triangle{ACC'}$ and $\triangle{BCC'}$, respectively. It follows that $[ACF]=[AC'F]$ and $[BCF]=[BC'F]$, and then $[ABC]=[ABC']$. Analogously, $[ABC]=[ACB']=[BCA']$. Now, it turns out that</span></p><p><span style="font-size: large;">$$[AB'C']=[BCB'C']-3[ABC]=4[BCEF]-3[ABC]=4[ABC]-4[AEF]-3[ABC]=[ABC]-4[AEF].\tag{1}$$</span></p><p><span style="font-size: large;">Let $H$ and $I$ be the midpoints of $AC$ and $AB$, respectively. Then</span></p><p><span style="font-size: large;">$$[ABC]=4[AHI]=4[AEF]+4[HEFI].\tag{2}$$</span></p><p><span style="font-size: large;">From $(1)$ and $(2)$ follows that $[AB'C']=4[HEFI]$. Now, observe that $BC$, $BB''$, and $CC''$ are the homothetic images with a scale factor of 2 of $IH$, $EH$, and $FI$ with respect to $A$, $B'$, and $C'$, respectively. It follows that $BCC''B''$ and $HEFI$ are homothetic with $[BCC''B'']=4[HEFI]$, therefore,</span></p><p><span style="font-size: large;">$$[AB'C']=[BCC''B''].$$</span></p><p style="text-align: right;"><span style="font-size: large;">$\square$</span></p><span style="font-size: large;">A similar reasoning must show that $[A'B'C]=[ABB''A'']$, and for the case of $\triangle{A'C'B}$ and $ACA''C''$ where $F$ and $D$ are on opposite sides of the line $GI$, where $G$ is the midpoint of the side $BC$, only a minor adjustment will be required.</span><div><span style="font-size: large;"><br /></span><div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFjR_jCd2KSmhtyFi1cd27xiIMk1Uf7A7JXuajOnpNWBgpFg8ON-VXwuWDDe-KaMjVxoMoSRvrn5vC6OltG642ClevnAReHEFRdr00vvuTosfmK5Y_0t3x8PrfktpgYnLVSoHhGtHvX097p-kZpSO3UKjO6JSWOgzXMusx4saWaks9eHW2ahif5cKahnaL/s2999/Another%20areal%20theorem.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="2537" data-original-width="2999" height="542" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFjR_jCd2KSmhtyFi1cd27xiIMk1Uf7A7JXuajOnpNWBgpFg8ON-VXwuWDDe-KaMjVxoMoSRvrn5vC6OltG642ClevnAReHEFRdr00vvuTosfmK5Y_0t3x8PrfktpgYnLVSoHhGtHvX097p-kZpSO3UKjO6JSWOgzXMusx4saWaks9eHW2ahif5cKahnaL/w640-h542/Another%20areal%20theorem.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">$\color{blue}{[A'B'C]=[ABA''B'']}$, $\color{red}{[A'C'B]=[ACA''C'']}$ and $\color{green}{[B'C'A]=[BCB''C'']}$.</td></tr></tbody></table><br /><span style="font-size: large;"><b>Remark</b>. Notice lemma 1 gives us an alternative way to construct a triangle with an area equal to a given quadrilateral, as long as it is not a rectangle or a square.</span><br /><p><span style="font-size: large;"><b>Back to the main problem</b></span></p><p><span style="font-size: large;">Notice that </span></p><p><span style="font-size: large;">$$[A'B'C']=[A'C'B'C]-[A'B'C]=4[ABC]+[AB'C']+[BA'C']-[A'B'C].$$</span></p><p><span style="font-size: large;">But $[AB'C']=[ABC]-4[AEF]$ and similarly $[BA'C']=[ABC]-4[BDF]$ and $[A'B'C]=4[CDE]-[ABC]$, so</span></p><p><span style="font-size: large;">$$[A'B'C']= 7[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{3}$$</span></p><span style="font-size: large;">As $HIFE$ and $BCC''B''$ are inversely homothetic with scale factor $-2$, then so are segments $EF$ and $B''C''$ and similarly for $ED$ and $A''C''$ and $FD$ and $A''C''$, meaning that triangles $\triangle{DEF}$ and $\triangle{A''B''C''}$ are also inversely homothetic with factor $-2$ and then</span><p><span style="font-size: large;">$$[A''B''C'']=4[DEF]=4[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{4}$$</span></p><p><span style="font-size: large;">Substracting $(4)$ from $(3)$ we get </span></p><p><span style="font-size: large;">$$[A'B'C']-[A''B''C'']=3[ABC].$$</span></p><p style="text-align: right;"><span style="font-size: large;">$\square$</span></p></div></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-6602445368824644092023-07-18T09:28:00.002-07:002023-07-19T08:50:47.927-07:00A curious family of integrals that give rational multiples of $\pi$<p><span style="font-size: large;"> I have noticed experimentally that:</span></p><p><span style="font-size: large;">$$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$</span></p><p><span style="font-size: large;">$$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$</span></p><p><span style="font-size: large;">$$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$</span></p><p><span style="font-size: large;">So slight variations in the numerator <i>always</i> seem to produce something like $n\pi$, where $n$ is a rational number.</span></p><p><span style="font-size: large;">At <a href="https://math.stackexchange.com/questions/4736889/a-curious-family-of-integrals-that-give-pi/4736909?noredirect=1#comment10046418_4736909" target="_blank">MathSE</a> I have asked what the exact relationship is between $n$ and the numerator of the integrand, to which Quanto has responded with a general formula:</span></p><span style="font-size: large;">$$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx=\frac\pi8\color{green}{(c+b-a)}+\frac\pi{3\sqrt3}\color{green}{(a+c)}.\tag{4}$$</span><p><span style="font-size: large;">However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula $(34)$ in this <a href="https://mathworld.wolfram.com/PiFormulas.html" target="_blank">list of $\pi$ formulas</a>:</span></p><p><span style="font-size: large;">$$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$</span></p><p><span style="font-size: large;">Notice that the denominator is different. But again, a slight variation in the numerator and it <i>still</i> produces something like $n\pi$:</span></p><p><span style="font-size: large;">$$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$</span></p><p><span style="font-size: large;">More generally, </span></p><p><span style="font-size: large;">$$\int_{0}^{1} \frac{ax^2+bx+c}{x^4-2x^3+4x-4}\,dx=\frac{\pi}{16}\color{green}{(2a-c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt32}\color{green}{(b+c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt8}\color{green}{a}.\tag{7}$$</span></p><span style="font-size: large;">From $(7)$, we can deduce that the integral will yield rational multiples of $\pi$ when $b$ and $c$ are opposite to each other and $a=0$. However, this formula is still far from being the ultimate generalization since it does not take into account the coefficients of the denominator. The fact that the denominator is not the same in $(4)$ and $(7)$ suggests that a further generalization is possible.</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>A more intimidating integral</b></span></div><div><span style="font-size: large;">Doing some arithmetic with the integrands, we can obtain more intimidating integrands that still yield $\pi$. The following one comes from adding integrands (2) and (5) with different denominators:</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int_{0}^{1} \frac{2(1-x)(x^5-5x^4-10x^3-4x^2+8x-8)}{x^8-2x^6-2x^5+9x^4-2x^3-16x^2+12x-4}=\pi.$$<br /></span><p><span style="font-size: large;"><b>Addendum</b>. I wonder if it is possible to characterize the integrand $\frac{P(x)}{Q(x)}$ in such a way that by simple inspection we can say $I=n\pi$? Or in other words, what should be the relationship between the coefficients of the numerator and the denominator for the integral to yield $n\pi$?</span></p></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-26613495544774132862023-07-13T11:39:00.009-07:002023-07-15T18:47:38.030-07:00An integral involving the golden ratio<p><span style="font-size: large;"> Several interesting integrals involving the <a href="https://en.wikipedia.org/wiki/Golden_ratio" target="_blank">golden ratio</a> can be found at <a href="https://math.stackexchange.com/questions/1653979/is-there-any-integral-for-the-golden-ratio?page=1&tab=scoredesc#tab-top" target="_blank">MathSE</a>. Here I present another one that I haven't seen anywhere else on the web. I'm referring to this one:</span></p><p><span style="font-size: large;">$$\int_\frac{\pi}{2}^{\pi} \frac{\sqrt{5}}{\sin{x}-\cot{x}}\,dx=4\ln\phi.\tag{1}$$</span></p><span style="font-size: large;">I discovered $(1)$ serendipitously while exploring the advantages of using the <a href="https://geometriadominicana.blogspot.com/2023/07/sine-half-angle-substitution.html" target="_blank">sine half-angle substitution</a>. </span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Proof</b>. Let's start by evaluating the indefinite integral:</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int \frac{1}{\sin{x}-\cot{x}}\,dx=\int \frac{\sin{x}}{\sin^2{x}-\cos{x}}\,dx=\int -\frac{\sin{x}}{\cos^2{x}+\cos{x}-1}\,dx.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Substitute $u=\cos{x} \rightarrow du=-\sin{x}\,dx$.</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int \frac{1}{\sin{x}-\cot{x}}\,dx=\int \frac{1}{u^2+u-1}\,du.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Factoring the denominator and performing partial fraction decomposition,</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\begin{aligned}\int \frac{1}{u^2+u-1}\,du&=\frac{2}{\sqrt{5}} \int \frac{1}{2u-\sqrt{5}+1}\,du- \frac{2}{\sqrt{5}} \int \frac{1}{2u+\sqrt{5}+1}\,du\\&= \frac{1}{\sqrt{5}}\left(\ln{(2u-\sqrt{5}+1)}-\ln{(2u+\sqrt{5}+1)}\right)+C.\end{aligned}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Undoing the substitution $u=\cos{x}$, taking into account that $2\phi=\sqrt{5}+1$ and $-\sqrt{5}+1=2-2\phi$, and applying properties of logarithms,</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\sqrt{5}\int \frac{1}{\sin{x}-\cot{x}}\,dx=\ln{\frac{\left|\cos{x}+1-\phi\right|}{\cos{x}+\phi}}+C.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">Now, evaluating the definite integral,</span></div><div><span style="font-size: large;"><br /></span></div><span style="font-size: large;">$$\begin{aligned}\sqrt{5}\int_\frac{\pi}{2}^{\pi} \frac{1}{\sin{x}-\cot{x}}\,dx&=\ln{\frac{\left|\cos{x}+1-\phi\right|}{\cos{x}+\phi}}\\&=\ln{\frac{\phi}{\left|1-\phi\right|}}-\ln{\frac{\left|1-\phi\right|}{\phi}}\\&=\ln{\left(\frac{\phi}{\phi-1}\right)}-\ln{\left(\frac{\phi-1}{\phi}\right)}\\&=2\ln{\left(\frac{\phi}{\phi-1}\right)}.\end{aligned}$$</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">But <a href="https://en.wikipedia.org/wiki/Golden_ratio#Golden_ratio_conjugate_and_powers" target="_blank">$\phi-1=\frac{1}{\phi}$</a>, thus</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int_\frac{\pi}{2}^{\pi} \frac{\sqrt{5}}{\sin{x}-\cot{x}}\,dx=4\ln{\phi}.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Other examples</b>:</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{5}}{\csc{x}-\tan{x}}\,dx=\ln{\phi}.$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">The golden ratio is a famous ratio, but have you heard of the <a href="https://en.wikipedia.org/wiki/Silver_ratio#:~:text=In%20mathematics%2C%20two%20quantities%20are,larger%20quantity%20(see%20below)." target="_blank">silver ratio</a>, $\phi_s$? Well, the following integral is related to it. I invite you to prove it for yourself.</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\int_{0}^\frac{3\pi}{2} \frac{\sqrt{2}}{\sin{x}-\cos{x}}\,dx=2\ln{\phi_s}.$$</span></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-88467288114694364012023-07-09T14:37:00.001-07:002023-07-09T14:37:40.517-07:00Sine half-angle substitution<p><span style="font-size: large;"> If you are a student of integral calculus, it is highly likely that you have come across or will come across the famous <a href="https://en.wikipedia.org/wiki/Tangent_half-angle_substitution" target="_blank">Weierstrass substitution</a>, which is very useful for converting rational expressions involving trigonometric functions into ordinary rational expressions involving $t$, where $t=\tan{\frac12{x}}$. The general transformation formula is as follows:</span></p><p><span style="font-size: large;">$$\int f(\sin{x},\cos{x})\,dx = \int f\left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}\right)\frac{2dt}{1+t^2}.$$</span></p><p><span style="font-size: large;">In this note, we introduce another substitution as a companion to the Weierstrass substitution that can transform certain rational expressions of trigonometric functions into simpler ordinary rational expressions than the Weierstrass substitution. For instance, it would be useful when a common linear factor of $\sin{x}$ appears in the numerator or denominator of the integrand. The general transformation formula is given by:</span></p><p><span style="font-size: large;">$$\int f(\sin{x},\cos{x})\,dx = \int f(2\sqrt{s-s^2}, 1-2s)\frac{ds}{\sqrt{s-s^2}}.$$</span></p><p><span style="font-size: large;">Here $s=\sin^2{\frac12x}$. </span></p><p><span style="font-size: large;"><b>Derivation</b></span></p><span style="font-size: large;">Using the double-angle formulas and the Pythagorean identity, one gets<br />$$\sin{x}=2\sin{\frac12x}\cos{\frac12x}=2\sin{\frac12x}\sqrt{1-\sin^2{\frac12x}}=2\sqrt{\sin^2{\frac12x}-\sin^4{\frac12x}}=2\sqrt{s-s^2},$$<br />$$\cos{x}=1-2\sin^2{\frac12x}=1-2s.$$<br />Finally, since $s=\sin^2{\frac12x}$, differentiation rules imply<br />$$ds=\sin{\frac12x}\cos{\frac12x}\,dx=\frac{\sin{x}}{2}\,dx,$$<br />and thus,<br />$$dx=\frac{ds}{\sqrt{s-s^2}}.$$</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Example 1</b></span></div><div><span style="font-size: large;">By applying the sine half-angle substitution and simplifying,</span></div><div><span style="font-size: large;">$$\int \frac{\sin{x}}{\sin^2{x}+2\cos{x}}\,dx=\int \frac{1}{1-2s^2}\,ds.$$</span></div><div><span style="font-size: large;">Using <a href="https://en.wikipedia.org/wiki/Hyperbolic_functions#Standard_integrals" target="_blank">standard integrals</a>,</span></div><span style="font-size: large;">$$\int \frac{1}{1-2s^2}\,ds=\frac{\tanh^{-1}{(\sqrt{2}s)}}{\sqrt{2}}+C=\frac{\tanh^{-1}{(\sqrt{2}\sin^2{\frac12x})}}{\sqrt{2}}+C.$$</span><div><span style="font-size: large;">The advantage of this substitution is evident when comparing it to the solution provided in this <a href="https://www.integral-calculator.com/#expr=sinx%2F%28sin%5E2x%2B2cosx" target="_blank">integral calculator</a> (which solution, by the way, is equivalent to the one given here)</span><span style="font-size: x-large;"> </span><span style="font-size: large;">or when using the Weierstrass substitution.</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Example 2 </b></span></div><div><span style="font-size: large;">By applying the sine half-angle substitution,</span></div><div><span style="font-size: large;">$$\int \frac{1}{\sin^3{x}}\,dx=\frac18\int \frac{1}{(s-s^2)^2}\,ds.$$</span></div><div><span style="font-size: large;">Using partial fraction decomposition, </span></div><div><span style="font-size: large;">$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac18\int \left(\frac{1}{s}+\frac{1}{s^2}-\frac{2}{s-1}+\frac{1}{(s-1)^2}\right)\,ds=\frac14\ln{\left|\frac{s}{1-s}\right|}+\frac18\left(\frac{1}{1-s}-\frac{1}{s}\right)+C.$$</span></div><div><span style="font-size: large;">Substituting $s$ by $\sin^2{\frac12x}$, we have</span></div><div><span style="font-size: large;">$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac14\ln{\left|\tan^2{\frac12x}\right|}+\frac18\left(\frac{1}{\cos^2{\frac12x}}-\frac{1}{\sin^2{\frac12x}}\right)+C=\frac12\left(\ln{|\tan{\frac12x}|}-\frac{\cos{x}}{\sin^2{x}}\right)+C.$$</span></div><span style="font-size: large;">A solution using integration by parts is given at <a href="https://www.youtube.com/watch?v=q55UjdCe9tg">Integrals For You</a>. </span>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-11744047328465805192023-05-17T11:43:00.003-07:002023-06-10T09:31:56.208-07:00A Generalization of the Law of Sines<p> <span style="font-size: large;">The following is an extension of the <a href="https://en.wikipedia.org/wiki/Law_of_sines" target="_blank">law of sines</a> for cyclic quadrilaterals that comes to accompany the <a href="https://geometriadominicana.blogspot.com/2022/01/generalization-of-mollweides-formulas.html" target="_blank">generalization of Mollweide's formula (rather Newton's)</a> and the <a href="https://geometriadominicana.blogspot.com/2022/03/a-generalization-of-law-of-tangent-for.html" target="_blank">generalization of the law of tangent</a>.</span></p><p><span style="font-size: large;"><b>Generalization</b>. Consider a cyclic quadrilateral, $ABCD$, with side lengths $AB=a$, $BC=b$, $CD=c$, and $DA=d$. Let $\angle{DAB}=\alpha$, $\angle{ABD}=\beta$, $\angle{BCD}=\gamma$, and $\angle{CDA}=\delta$. Then the following identity holds</span></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPdomukM3wk9jHpWeAd3EP9H2Y2cNlBTSOXeCXo-keXrYzkb1Xt134rO5qsnF7aFAVBxVcW7uCwpyFjaxBafDu7PFH_PbiB5FWLvTMRxVotFEzKSHQpkZ7XtgeWfW6v3iAhg97_1XRnq6SlVJ5ln8ja-m1Nodwn2Zxgf3fJryno66w3DB0GR6PBE5h8A/s1186/extension%20law%20of%20sine.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1186" data-original-width="1186" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPdomukM3wk9jHpWeAd3EP9H2Y2cNlBTSOXeCXo-keXrYzkb1Xt134rO5qsnF7aFAVBxVcW7uCwpyFjaxBafDu7PFH_PbiB5FWLvTMRxVotFEzKSHQpkZ7XtgeWfW6v3iAhg97_1XRnq6SlVJ5ln8ja-m1Nodwn2Zxgf3fJryno66w3DB0GR6PBE5h8A/w400-h400/extension%20law%20of%20sine.png" width="400" /></a></div><br /><span style="font-size: large;">$$\frac{ab+cd}{\sin{\alpha}}=\frac{ad+bc}{\sin{\beta}}=\frac{ab+cd}{\sin{\gamma}}=\frac{ad+bc}{\sin{\delta}}.$$</span><p></p><p><span style="font-size: large;">At first, I was reluctant to publish this result because the proof is very straightforward (hence, I leave it as an exercise to the reader). However, I have shared it on my social media platforms (see <a href="https://www.facebook.com/groups/matematicos/permalink/10159982556989319/" target="_blank">here</a>, <a href="https://twitter.com/vanclides/status/1475861551651471364" target="_blank">here</a> and <a href="https://mathstodon.xyz/@vanclides88/109416244186274201?fbclid=IwAR3BdN3galamNWcQpQP5HDgMmffzrwLSLTgsiEb685HImv3Ikszw7wstkSk" target="_blank">here</a>) and the audience's response has been more favorable than my generalizations of Mollweide's formula and the law of tangents combined.</span></p><p><span style="font-size: large;"><b>Related material</b></span></p><p><span style="font-size: large;"><a href="https://math.stackexchange.com/questions/4490209/generalizing-lamis-theorem" target="_blank">Generalizing Lami's theorem</a><br /></span></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-37848888235017964192023-04-30T10:22:00.002-07:002023-05-02T10:14:09.847-07:00A new proof of angle sum identity for the sine<p><span style="font-size: large;">Several interesting proofs of angle sum identity are given on <a href="https://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry?fbclid=IwAR1KrPqSCiH6wSr8WWDd-jGPC1TyRIcO07Kukk7AJJM-xPN37CBCnBQAK_E" target="_blank">MathSE</a>. In this note, I will provide another proof that is possible because, although traditionally presented in textbooks as a consequence of the angle sum identity $\sin{(x+y)}$, the double-angle formula for the sine, $\sin{(2x)}=2\sin{(x)}\cos{(x)}$, can be derived independently of it (see <a href="https://math.stackexchange.com/q/460302" target="_blank">the wonderful proof without words of "Start wearing purple"</a>). Then, <a href="https://math.stackexchange.com/questions/4685675/new-derivation-for-the-angle-sum-identity-for-sine-via-the-cosine-rule?noredirect=1#comment9914303_4685675" target="_blank">as Blue has pointed out</a>, the fact that supplementary angles have the same sine is an easy consequence of the double-angle formula for the sine.</span></p><span style="font-size: large;">The following proof is valid for $0\le\alpha\le\pi$, $0\le\beta\le\dfrac{\pi}{2}$, $0\le\alpha+\beta\le\pi$.</span><p><span style="font-size: large;"><b>Theorem</b>. The sum identity for sine states that</span></p><p><span style="font-size: large;">$$\sin{(\alpha+\beta)}=\sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)}.\tag{1}$$</span></p><p><span style="font-size: large;"><b>Proof</b>. Suppose $\triangle{ABC}$ is a triangle with sides $\overline{BC}=a$, $\overline{AC}=b$ and $\overline{AB}=c$. Let $\angle{BAC}=\alpha$, $\angle{CBA}=\beta$ and $\angle{ACB}=\gamma$. Recalling that the area of a triangle can be expressed as $\Delta=\frac12bc\sin{(\alpha)}$ and substituting from the cosine rule on the right-hand side of $(1)$, we have </span><span style="font-size: x-large;"> </span></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9GXH2uBjcbxi3CL_RqQDazM5qoEv78AyjuoRXY-h8dvZ-Fn3CSnlOaAWzpPKfYZPJyNGC6Uydr9bQCfkiIUtWlu8NxxOVxAh-w6g4wr794L4MtYCItZfxUT35TzUEijXWmhe6n128OvsAO18K15dGrMnn9eXvEaANms95k69kaQypsB5WkO0LSEjlLA/s2116/sum%20sine2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1392" data-original-width="2116" height="264" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9GXH2uBjcbxi3CL_RqQDazM5qoEv78AyjuoRXY-h8dvZ-Fn3CSnlOaAWzpPKfYZPJyNGC6Uydr9bQCfkiIUtWlu8NxxOVxAh-w6g4wr794L4MtYCItZfxUT35TzUEijXWmhe6n128OvsAO18K15dGrMnn9eXvEaANms95k69kaQypsB5WkO0LSEjlLA/w400-h264/sum%20sine2.png" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><p></p><span style="font-size: large;">\[\begin{aligned} \sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)} &= \left(\frac{2\Delta}{bc}\right)\cos{(\beta)}+\left(\frac{2\Delta}{ac}\right)\cos{(\alpha)}\\&=\frac{2\Delta}{c}\left(\frac{\cos{(\beta)}}{b}+\frac{\cos{(\alpha)}}{a}\right)\\ &=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin{(\gamma)}\\&=\sin{(\pi-(\alpha+\beta))}\\&=\sin{2\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}\\&=2\color{red}{\sin{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\color{blue}{\cos{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\\&=2\color{red}{\cos{\left(\frac{\alpha+\beta}{2}\right)}}\color{blue}{\sin{\left(\frac{\alpha+\beta}{2}\right)}}\\&= \sin{\left(\alpha+\beta\right)}.\end{aligned}\]</span><div style="text-align: right;"><span style="font-size: large;">$\square$</span></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-11155893172018261282023-04-13T05:33:00.003-07:002023-08-03T08:03:10.571-07:00Breve discurso ante una multitud de estudiantes: el enfoque de medio ángulo (EMA)<span style="font-size: large;">Hace dos semanas, recibí una invitación para dar un breve discurso a estudiantes de secundaria acerca de mis más recientes descubrimientos en geometría plana. A continuación, les comparto lo que dije:</span><div><div class="separator" style="clear: both; text-align: center;"><br /></div><div><span style="font-size: large;"><div><i>«Hola, yo soy Emmanuel y he venido literalmente a presumir, ¿qué les parece? Pero antes de empezar quiero hacerles la siguiente pregunta: ¿saben ustedes qué es el <a href="https://es.wikipedia.org/wiki/Informe_PISA" target="_blank">informe PISA</a>? Pues es un estudio llevado a cabo a nivel mundial que mide el rendimiento académico de los alumnos de 15 años en matemáticas, ciencia y lectura comprensiva. Si usted quiere encontrar a República Dominicana en la larga lista de países que participan en este estudio, usted solo debe dirigirse al final de la <a href="https://eldia.com.do/pruebas-pisa-rd-ocupa-peores-puestos-entre-los-paises-de-america-en-lectura-y-matematicas/" target="_blank">lista</a>, ¡qué mal! Aunque cuando cuento esto a mis estudiantes muchos se echan a reír, es una situación para llorar. ¡Esto es peor que la barrida que nos dieron en el Clásico Mundial de Baseball!</i></div><div><i><br /></i></div><div><i>Pero…a pesar de estos resultados tan alarmantes, aunque usted no lo crea, en República Dominicana se ha descubierto un <a href="https://medium.com/@emmanueljosgarca/the-theoretical-importance-of-the-half-angle-formulas-6142994251f2" target="_blank">nuevo enfoque</a> (al que yo llamo EMA) que me ha permitido <a href="https://mathoverflow.net/questions/419185/a-generalization-of-the-law-of-tangents#:~:text=The%20law%20of%20tangents%20is,lengths%20of%20the%20opposing%20sides.&text=tan12(%CE%B1%E2%88%92%CE%B2,and%20a%20side%2C%20are%20known." target="_blank">generalizar una fórmula de más de 900 años</a>: la ley de la tangente. Lo que quería decir es que luego de casi un año de espera, desempolvando libros y preguntando a matemáticos de todas partes del mundo, recientemente se me informó que mi generalización será publicada en la prestigiosa revista americana <a href="https://www.maa.org/press/periodicals/mathematics-magazine" target="_blank">Mathematics Magazine</a>. Pero la cosa no se queda ahí, este nuevo enfoque me ha permitido también generalizar las <a href="https://drive.google.com/file/d/1PjbZOa3YNzn2kcf0_9MLvAuWiVm8WAEh/view" target="_blank">fórmulas de medio ángulo</a>, la <a href="https://geometriadominicana.blogspot.com/2022/06/a-generalization-of-pythagorean.html" target="_blank">identidad pitagórica</a>, la <a href="https://mathstodon.xyz/@vanclides88/109416244186274201?fbclid=IwAR3BdN3galamNWcQpQP5HDgMmffzrwLSLTgsiEb685HImv3Ikszw7wstkSk" target="_blank">ley de senos</a> (¡me honra que el gran <a href="https://en.wikipedia.org/wiki/John_C._Baez" target="_blank">John Baez</a> haya compartido mi generalización!), el <a href="https://math.stackexchange.com/questions/4490209/generalizing-lamis-theorem" target="_blank">teorema de Lamy</a> y la <a href="https://elnuevodiario.com.do/dominicano-asegura-haber-superado-formula-matematica-de-newton/" target="_blank">fórmula de Newton</a>, además de poder derivar una plétora de fórmulas y teoremas bien conocidos en geometría plana.</i></div><div><i><br /></i></div><div><i>Pero, ¿qué es una generalización? Una generalización es, en palabras llanas, ampliar el alcance de un teorema. Grandes matemáticos tienen sus propias generalizaciones. ¿Qué creen ustedes que es el <a href="https://es.wikipedia.org/wiki/Teorema_del_binomio" target="_blank">binomio de Newton</a> o la famosa <a href="https://es.wikipedia.org/wiki/F%C3%B3rmula_de_Euler" target="_blank">fórmula de Euler</a>? ¡Generalizaciones! Por cierto, no me pregunten las horas que pasé desarrollando todo esto.</i></div><div><i><br /></i></div><div><i>Para finalizar quiero citar al <a href="http://www.supermath.info/" target="_blank">Dr. James Cook</a>, de la Universidad de Alabama, quien escribió lo siguiente sobre mi enfoque:</i></div><div><i><br /></i></div><div><i>Creo que has presentado un caso convincente de que estas fórmulas son bastante básicas. Por supuesto, sospecho que podría derivarse casi todo partiendo de la ley de los cosenos. ¿Qué es la ley de cosenos sino el corazón del producto escalar? Y, ¿qué es el producto escalar? Es la encapsulación algebraica del ángulo. Como mínimo, esto debería aparecer como problemas o una sección de tema adicional en los textos de trigonometría. Parece que esto sería excelente para un curso de honor de escuela de verano para dotados en matemáticas. El hecho de que no se enseñe podría aprovecharse para permitir que los estudiantes lo descubran. »</i></div></span></div></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-88461041721357937612023-03-18T09:17:00.002-07:002023-03-18T09:41:12.809-07:00The Half-Angle Formulas: A Powerful Tool for Trigonometry Students<p><span style="font-size: large;"> Let me tell you about an exciting discovery I made while exploring the theoretical importance of half-angle formulas in trigonometry. These formulas, often overlooked and underappreciated in comparison to the more well-known laws of sines, cosines, and tangents, have a wealth of untapped potential waiting to be discovered.</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAWbn_Yis4e6hLJ5LOP68kWzjPKdvY4-KGOm-tYKksP5FDXVifaGjiHCb9ViPelcPF2V8F_zFj8lcz0bMkUfVTCh3nedVMxitSPXQCiVYTFq9qEwi-S-erbuUoXrZPNp0Egx8QiMs6mLwIz7DqI55xLsU4l18sLDDNbIneHR40GNBrELfHrjzibsucEw/s1185/estandar.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="155" data-original-width="1185" height="84" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAWbn_Yis4e6hLJ5LOP68kWzjPKdvY4-KGOm-tYKksP5FDXVifaGjiHCb9ViPelcPF2V8F_zFj8lcz0bMkUfVTCh3nedVMxitSPXQCiVYTFq9qEwi-S-erbuUoXrZPNp0Egx8QiMs6mLwIz7DqI55xLsU4l18sLDDNbIneHR40GNBrELfHrjzibsucEw/w640-h84/estandar.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The standard half-angle formulas</td></tr></tbody></table><p></p><p><span style="font-size: large;">I'm not talking about the standard version of the half-angle formulas that appear in most introductory trigonometry textbooks. No, I'm referring to a version that relates the sides of a triangle, the perimeter, and its angles. These formulas have been largely ignored by the mathematical community, with even the almighty Wikipedia lacking an article on them.</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsX57_aWZuX-HECsja7m_DV3NUR7SXgM1-Wsow0JazYu0yQCdban649dCufbQP4Qo8Chm7_VB_kgMXLVdUmwJu0sTV5GISFyfdd6sBQTh1Vluk452CLWo2lnBMM0oP02ykrVg5xFcXX8b_Bnm3cLe8DIPYohGjwQG1pxIfz7snsPlQ996aH_P9bPdYoQ/s1154/1.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="158" data-original-width="1154" height="88" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsX57_aWZuX-HECsja7m_DV3NUR7SXgM1-Wsow0JazYu0yQCdban649dCufbQP4Qo8Chm7_VB_kgMXLVdUmwJu0sTV5GISFyfdd6sBQTh1Vluk452CLWo2lnBMM0oP02ykrVg5xFcXX8b_Bnm3cLe8DIPYohGjwQG1pxIfz7snsPlQ996aH_P9bPdYoQ/w640-h88/1.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The half-angle formulas for a triangle</td></tr></tbody></table><p></p><p><span style="font-size: large;">My journey began when I stumbled upon <a href="https://geometriadominicana.blogspot.com/2020/06/another-proof-for-two-well-known.html" target="_blank">(1)</a> while attempting to prove the law of cosines by contradiction. I later discovered that this formula was already known, but that didn't stop me from exploring its potential. I went on to generalize the formula and discovered (2), which applies to cyclic quadrilaterals. When I found out that (2) was also already known, I refused to give up and continued to push the limits of these formulas.</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvFJAyh6VzHGPfITy4-gygmTNIKlOotPfNNb0VKv4JrYqRuE18gizWfPb8JQp9_0qaS5qAQe34HCF4EfHtfmuNsSIecmqWQvxLHxnB7PFZ3I06Enn0qgKj0FsXYeyW5w_SKCxtV3ek6df_0mpnj45-YXFKM901ES2meSBXcHs6GRH_px-Th7pEmJ97pA/s1151/2.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="141" data-original-width="1151" height="78" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvFJAyh6VzHGPfITy4-gygmTNIKlOotPfNNb0VKv4JrYqRuE18gizWfPb8JQp9_0qaS5qAQe34HCF4EfHtfmuNsSIecmqWQvxLHxnB7PFZ3I06Enn0qgKj0FsXYeyW5w_SKCxtV3ek6df_0mpnj45-YXFKM901ES2meSBXcHs6GRH_px-Th7pEmJ97pA/w640-h78/2.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The half-angle formulas for a cyclic quadrilateral</td></tr></tbody></table><p></p><p><span style="font-size: large;">What I discovered next was truly <i>mind-blowing</i>. I managed to generalize (2) even further to (3), which applies to general quadrilaterals. Surprisingly, (3) also <a href="https://geometriadominicana.blogspot.com/2022/06/a-generalization-of-pythagorean.html" target="_blank">generalizes the Pythagorean trigonometric identity</a>! These formulas not only provide a new framework for the Heron-Brahmagupta-Bretschneider development (see <a href="https://drive.google.com/file/d/1PjbZOa3YNzn2kcf0_9MLvAuWiVm8WAEh/view">Two Identities and their Consequences, pp. 5</a>), but they also have the potential to derive a plethora of other formulas and theorems, including the law of cosines, the law of sines, the law of tangents, and Stewart's theorem.</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAsG4h0fFaBPiq0-EQkWLBWnJDD30G3Rd3cPSw2_keRCv4Ohptd0H97rzejZcn_teSSo-nFZLheU3c9k-vwPZRuacD990BLlPpQEm2keXkHKJHPucq8_vTkT6hOcIBuRgqj_GBc3Gqj14k_HfTV3wvaP92mwMBw7F_B9n1cRitSt2V3wEPAgvirR3MFQ/s846/3.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="230" data-original-width="846" height="174" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAsG4h0fFaBPiq0-EQkWLBWnJDD30G3Rd3cPSw2_keRCv4Ohptd0H97rzejZcn_teSSo-nFZLheU3c9k-vwPZRuacD990BLlPpQEm2keXkHKJHPucq8_vTkT6hOcIBuRgqj_GBc3Gqj14k_HfTV3wvaP92mwMBw7F_B9n1cRitSt2V3wEPAgvirR3MFQ/w640-h174/3.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The half-angle formulas for a general quadrilateral</td></tr></tbody></table><p></p><p><span style="font-size: large;">And it doesn't stop there. Using these half-angle formulas, I also managed to derive a <a href="https://drive.google.com/file/d/1Zu5e2Xp_jQg4XNFgFmF_DVeNsyg90b4D/view" target="_blank">generalization of Mollweide's formula</a> and a <a href="https://mathoverflow.net/questions/419185/a-generalization-of-the-law-of-tangents?_gl=1*1awd6qn*_ga*ODQzNDkxNjMwLjE2Njg3Njk2ODg.*_ga_S812YQPLT2*MTY3OTE1NTk3OS4xMTkuMS4xNjc5MTU1OTgwLjAuMC4w" target="_blank">generalization of the law of tangents</a>. The possibilities are endless, and I'm only scratching the surface of what these formulas can do.</span></p><p><span style="font-size: large;">But what excites me the most about these formulas is their potential to <i>revolutionize</i> the way we teach trigonometry to high school students. By introducing the half-angle formulas, we can help students better understand the development of the formulas that derive from them. This is where the half-angle formulas truly shine and can make a significant impact on math education.</span></p><p><span style="font-size: large;">In conclusion, the theoretical importance of the half-angle formulas cannot be overstated. They have the potential to unlock a whole new world of mathematical discoveries and have already proven to be a powerful tool for deriving a wide range of formulas and theorems. And with the right approach, they can also be a game-changer for teaching trigonometry to the next generation of students. The possibilities are truly endless, and I can't wait to see what else these formulas have in store for us.</span></p><p><span style="font-size: large;">See also <a href="https://medium.com/@emmanueljosgarca/the-theoretical-importance-of-the-half-angle-formulas-6142994251f2" target="_blank">The theoretical importance of the half-angle formulas</a>.</span></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-59676342731967787702022-12-28T18:44:00.008-08:002023-08-03T08:18:06.823-07:00Van Khea's areal problem<p> <span style="font-size: large;">The following problem is a generalization of <a href="https://geometriadominicana.blogspot.com/2022/12/areal-property-of-circumcircle-mid-arc.html" target="_blank">my previous problem</a> conjectured by Van Khea. We will give a proof.</span></p><p><span style="font-size: large;">We will be using standard notation: $\lvert{BC}\rvert=a$, $\lvert{AC}\rvert=b$, $\lvert{AB}\rvert=c$; $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$ and $\angle{BCA}=\gamma$. If $X$, $Y$ and $Z$ are the vertices of a triangle, we denote its area $[XYZ]$.</span></p><p><span style="font-size: large;"><b>Problem</b>. <i>Let $ABC$ be a triangle and $P$ any point on the plane of $ABC$. Let $X$, $Y$ and $Z$ be arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $D$ be the reflection of $P$ around $X$. Similarly, define $E$ and $F$. Denote $U$, $V$ and $W$ the midpoints of sides $BC$, $AC$ and $AB$, respectively. Let $D'$ be the reflection of $D$ around $U$. Similarly, define $E'$ and $F'$ (see figure below). Prove that</i></span></p><p><span style="font-size: large;"><i>$$[DEF]+[D'E'F']=[ABC].$$</i></span></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVbc63m6bDCrA0EUhho9pQRXM5O_qM6UDhvNaVP9s9e1wxeWdENUjVEuMJZBxBaw1SplQFWorEZOveXlTUvkUNRy3QRMwHkrgFyhWB9vWFtmJFQ468MMRuFBGBNj5oajBM4p1mJgpf8YND1Jwy-XoKaS784WTgVDS_fBoLm-t3Q8dM5NbNn3brD-78nw/s4928/Van%20Khea%C2%B4s%20generalization.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2904" data-original-width="4928" height="378" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVbc63m6bDCrA0EUhho9pQRXM5O_qM6UDhvNaVP9s9e1wxeWdENUjVEuMJZBxBaw1SplQFWorEZOveXlTUvkUNRy3QRMwHkrgFyhWB9vWFtmJFQ468MMRuFBGBNj5oajBM4p1mJgpf8YND1Jwy-XoKaS784WTgVDS_fBoLm-t3Q8dM5NbNn3brD-78nw/w640-h378/Van%20Khea%C2%B4s%20generalization.png" width="640" /></a></div><div class="separator" style="clear: both; text-align: left;"><span style="font-size: large;"><b><br /></b></span></div><div class="separator" style="clear: both; text-align: left;"><span style="font-size: large;"><b>Proof.</b> </span><span style="font-size: large;">Denote $AZ=g$, $BZ=h$, $BX=j$, $CX=k$, $CY=l$ and $AY=m$. The area of triangle $XYZ$ can be expressed as follows</span></div><div class="separator" style="clear: both;"><span style="font-size: large;">$$[XYZ]=\frac12bc\sin{\alpha}-\frac12gm\sin{\alpha}-\frac12hj\sin{\beta}-\frac12kl\sin{\gamma}.$$</span></div><div><div><span style="font-size: large;">Since triangles $XYZ$ and $DEF$ are homothetic with scale factor $2$, it follows that $[DEF]=4[XYZ]$. Thus, we have</span></div><div><span style="font-size: large;">\[{[DEF]}=2bc\sin{\alpha}-2gm\sin{\alpha}-2hj\sin{\beta}-2kl\sin{\gamma}.\tag{1}\]</span></div><div><span style="font-size: large;">Dividing both sides of $(1)$ by $[ABC]$ we have</span></div><div><span style="font-size: large;">\[\frac{[DEF]}{[ABC]}=4\left[1-\left(\frac{gm}{bc}+\frac{hj}{ca}+\frac{kl}{ab}\right)\right].\tag{2}\]</span></div><div><span style="font-size: large;">Segments $UX$ and $PD'$ are homothetic with center at $D$ and scale factor $2$. It follows that</span></div><div><span style="font-size: large;">$$PD'=2(j-\frac12a)=2j-a.$$</span></div><div><span style="font-size: large;">Similarly, we get $PE'=b-2l$ and $PF'=c-2g$. Since $UX$ and $PD'$ are homothetic segments, then $UX$ and $PD'$ are parallel and so are $VY$ and $PE'$. Hence $\angle{D'PE'}=\gamma$. Similarly, $\angle{D'PF'}=\beta$. So the area of triangle $D'E'F'$ is given by the expression</span></div><div><span style="font-size: large;">$$\begin{aligned}{[D'E'F']}&=[D'PE']+[D'PF']-[E'F'P]\\&=\frac{(2j-a)(b-2l)\sin{\gamma}}{2}+\frac{(2j-a)(c-2g)\sin{\beta}}{2}-\frac{(b-2l)(c-2g)\sin{(\beta+\gamma)}}{2}.\end{aligned}$$</span></div><div><span style="font-size: large;">Taking into account that $\sin{(\beta+\gamma)}=\sin{(\pi-\alpha)}=\sin{\alpha}$ and dividing by $[ABC]$ we obtain</span></div><div><span style="font-size: large;">\[\frac{[D'E'F']}{[ABC]}=\frac{(2j-a)(b-2l)}{ab}+\frac{(2j-a)(c-2g)}{ca}-\frac{(b-2l)(c-2g)}{bc}.\tag{3}\]</span></div><div><span style="font-size: large;">Adding equations $(2)$ and $(3)$, expanding and factorizing,</span></div><span style="font-size: large;">$$\frac{[DEF]}{[ABC]}+\frac{[D'E'F']}{[ABC]}=\frac{ca(4l+b)-4ga(m+l-b)-4j(b(g+h-c)+lc)-4klc}{abc}.$$</span><div><span style="font-size: large;">But $b=m+l$, $c=g+h$ and $a=j+k$, so</span></div><div><span style="font-size: large;">$$\begin{aligned}\frac{[DEF]}{[ABC]}+\frac{[D'E'F']}{[ABC]}&=\frac{ca(4l+b)-4jlc-4klc}{abc}\\&=\frac{ca(4l+b)-4cl(j+k)}{abc}\\&=\frac{ca(4l+b)-4cla}{abc}\\&=1.\end{aligned}$$</span></div><div><span style="font-size: large;">Therefore, </span></div><div><span style="font-size: large;">$$[DEF]+[D'E'F']=[ABC].$$</span></div></div><div><div style="text-align: right;"><span style="font-size: x-large;">$\square$</span></div><div style="text-align: left;"><span style="font-size: large;"><b>Note</b>: The point $P$ may cross the side lines of the triangle $ABC$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.</span></div><div style="text-align: left;"><span style="font-size: large;"><br /></span></div><div style="text-align: left;"><span style="font-size: large;"><b>Remark</b>: This theorem remains valid if $P$ is an arbitrary point in three-dimensional space.</span></div><div style="text-align: left;"><span style="font-size: large;"><br /></span></div><div style="text-align: left;"><span style="font-size: large;"><b>See also</b></span></div><div style="text-align: left;"><span style="font-size: large;"><a href="https://geometriadominicana.blogspot.com/2023/07/another-identity-on-triangle-areas.html" target="_blank">Another Identity on Triangle Areas Arising from Reflection</a><br /></span></div><p></p></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-74178799245934370302022-12-23T10:29:00.006-08:002023-12-24T07:19:06.592-08:00Areal property of the circumcircle mid-arc triangle<p><span style="font-size: large;"> The following theorem is a property of the <a href="https://mathworld.wolfram.com/CircumcircleMid-ArcTriangle.html" target="_blank">circumcircle mid-arc triangle</a> which appears to be unknown.</span></p><p><span style="font-size: large;">We will be using standard notation: $\lvert{BC}\rvert=a$, $\lvert{AC}\rvert=b$, $\lvert{AB}\rvert=c$; $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$ and $\angle{BCA}=\gamma$; $s$ for the semiperimeter; $R$ for the circumradius and $r$ for the inradius. If $X$, $Y$ and $Z$ are the vertices of a triangle, we denote its area $[XYZ]$.</span></p><p><span style="font-size: large;"><b>Problem</b>. <i>Let $ABC$ be a triangle and $I$ its Incenter. Denote $\omega$ the circumcircle of $ABC$. Let $AI$ intersect $\omega$ again at $D$. Define $E$ and $F$ cyclically. Let $U$, $V$ and $W$ be the midpoints of $DE$, $EF$ and $FD$, respectively. Let $A'$ be the reflection of $A$ with respect to $V$. Define $B'$ and $C'$ cyclically (see figure 1). Then</i></span></p><p><span style="font-size: large;"><i>$$[DEF]=[ABC]+[A'B'C'].\tag{1}$$</i></span></p><p><span style="font-size: large;"></span></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg34lVF8dD1JyQ44NZyg3nPqpPAeKeBMVviLE2710RSVJobrSK0ApRnhOb51doBOhTICzCvBGqlWtesZx1d2BAcwpSTkyPXW71ws-160Km8kUqDfm2sV6eL9wb6y7IfQdZBS_pr9tC_8qtDNGpgSk9hEs3_e8LouIgT6W7FXEy61YEo5iGcT6aHhR3sGw/s5032/Teorema%20de%20areas.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="5032" data-original-width="4848" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg34lVF8dD1JyQ44NZyg3nPqpPAeKeBMVviLE2710RSVJobrSK0ApRnhOb51doBOhTICzCvBGqlWtesZx1d2BAcwpSTkyPXW71ws-160Km8kUqDfm2sV6eL9wb6y7IfQdZBS_pr9tC_8qtDNGpgSk9hEs3_e8LouIgT6W7FXEy61YEo5iGcT6aHhR3sGw/w385-h400/Teorema%20de%20areas.png" width="385" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 1</td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><span style="font-size: large;"></span></div><p></p><span style="font-size: large;"><div><br /></div></span><div><span style="font-size: large;"><b>Lemma 1</b>. If $ABC$ is a triangle and $DEF$ is its circumcircle mid-arc triangle, then</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">$$\boxed{\frac{[ABC]}{[DEF]}=\frac{2r}{R}.}\tag{2}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Proof</b>. By property of inscribed angles, $\angle{BAF}=\angle{ACF}=\frac12\gamma$ and $\angle{CAE}=\frac12\beta$. It follows that</span></div><span style="font-size: large;">$$\begin{aligned}&\angle{EAF}=\alpha+\frac12\beta+\frac12\gamma\\&=\pi-\beta-\gamma+\frac12\beta+\frac12\gamma\\&=\pi-\frac12(\beta+\gamma).\end{aligned}$$</span><div><span style="font-size: large;">By the law of sines, </span></div><div><span style="font-size: large;">$$\frac{EF}{\sin{(\pi-\frac12(\beta+\gamma))}}=\frac{EF}{\sin{\frac12(\beta+\gamma)}}=\frac{AF}{\sin{\frac12\gamma}}.\tag{3}$$</span></div><div><span style="font-size: large;">Since $\angle{AFC}=\beta$ and, again, by the law of sines, </span></div><div><span style="font-size: large;">$$\frac{AF}{\sin{\frac12\gamma}}=\frac{b}{\sin{\beta}}.\tag{4}$$</span></div><div><span style="font-size: large;">From $(3)$ and $(4)$ we get</span></div><div><span style="font-size: large;">$$EF=\frac{b\sin{\frac12(\beta+\gamma)}}{\sin{\beta}}=2R\sin{\frac12(\pi-\alpha)}=2R\cos{\frac12\alpha}.\tag{5}$$</span></div><div><span style="font-size: large;">Analogously, we can find that</span></div><div><span style="font-size: large;">$$DF=2R\cos{\frac12\beta}\qquad ED=2R\cos{\frac12\gamma}.\tag{6}$$</span></div><span style="font-size: large;">Note that $\angle{DFC}=\frac12\alpha$ and $\angle{EFC}=\frac12\beta$, so $\angle{DFE}=\frac12(\alpha+\beta)$. Then the area of $DEF$ is given by <br />$$[DEF]=\frac{EF\cdot{DF}\sin{\frac12(\alpha+\beta)}}{2}=2R^2\cos{\frac12\alpha}\cos{\frac12\beta}\cos{\frac12\gamma}.\tag{7}$$<br /></span><span style="font-size: large;">Elsewhere we have proved that $s=4R\cos{\frac12\alpha}\cos{\frac12\beta}\cos{\frac12\gamma}$ (for a proof, see section 3, c), <a href="https://geometriadominicana.blogspot.com/2020/06/another-proof-for-two-well-known.html" target="_blank">here</a>). As the area of $ABC$ can also be written as $[ABC]=rs$, we can rewrite it like this</span><div><span style="font-size: large;">$$[ABC]=4Rr\cos{\frac12\alpha}\cos{\frac12\beta}\cos{\frac12\gamma}.\tag{8}$$</span></div><div><span style="font-size: large;">Dividing $(8)$ by $(7)$ we get</span></div><div><span style="font-size: large;">$$\frac{[ABC]}{[DEF]}=\frac{2r}{R}.$$</span></div><span style="font-size: large;"><div style="text-align: right;">$\square$</div><div style="text-align: left;"><b>Lemma 2</b>. Let $ABC$ be an triangle and $H$ its orthocenter. Let $l_a$, $l_b$ and $l_c$ be the perpendicular bisectors of sides $BC$, $AC$ and $AB$, respectively. Denote $H_1$, $H_2$ and $H_3$ the reflections of $H$ around $l_a$, $l_b$ and $l_c$, respectively. Then</div><div style="text-align: left;"> $$\boxed{\frac{[H_1H_2H_3]}{[ABC]}=\frac{(a^2-b^2)(a^2-c^2)}{b^2c^2}+\frac{(a^2-c^2)(b^2-c^2)}{a^2b^2}-\frac{(a^2-b^2)(b^2-c^2)}{a^2c^2}.}\tag{9}$$</div><div style="text-align: left;"><b>Proof</b>. From figure 2, note that </div></span><div style="text-align: left;"><span style="font-size: large;"><div>$$HH_3=2(\frac{c}{2}-AC_1).\tag{10}$$</div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3wvAN5bzxafjE5QZfIMzB1hQiap49h9CSbLJGDSZ4OWF700Ao8vtjj7MspjvY1vvwnapx4w5W4PpvZQx2RiJm0TPW6NYpDhcypnxw5abb7-6tvRCHY0TqK41qpCm5LpecXBdMD4CIplmX3Ya6Q1SJKBhsIf86CEjld6u4FE1xWTeK4IYGHdHiRAyMHQ/s4992/Lema%20ortocentro.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="4328" data-original-width="4992" height="346" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3wvAN5bzxafjE5QZfIMzB1hQiap49h9CSbLJGDSZ4OWF700Ao8vtjj7MspjvY1vvwnapx4w5W4PpvZQx2RiJm0TPW6NYpDhcypnxw5abb7-6tvRCHY0TqK41qpCm5LpecXBdMD4CIplmX3Ya6Q1SJKBhsIf86CEjld6u4FE1xWTeK4IYGHdHiRAyMHQ/w400-h346/Lema%20ortocentro.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2</td></tr></tbody></table><div><br /></div><div>Let $CH$ intersect $AB$ in $C_1$. Applying the law of sines in the triangle $ACC_1$, we have that</div><div>$$b=\frac{AC_1}{\sin{(\frac{\pi}{2}-\alpha)}}=\frac{AC_1}{\cos{\alpha}}.$$</div><div>So, $AC_1=b\cos{\alpha}$. Substituting in $(10)$ and then substituting from the law of cosines $\cos{\alpha}=\frac{b^2+c^2-a^2}{2bc}$ we have</div><div>$$HH_3=2(\frac{c}{2}-b\cos{\alpha})=c-2b\cos{\alpha}=\frac{a^2-b^2}{c}.$$</div><div>Similarly, we get that $HH_1=\frac{b^2-c^2}{a}$ and $HH_2=\frac{a^2-c^2}{b}$. Now, note that </div><div>$$[H_1H_2H_3]=[HH_2H_3]+[HH_1H_2]-[HH_1H_3].\tag{11}$$</div><div><div>As $HH_3$ and $HH_2$ are perpendicular to $l_c$ and $l_a$, respectively, then $\angle{H_2HH_3}=\alpha$ and similarly $\angle{H_2HH1}=\gamma$. Thus, equation $(11)$ can be written as follows</div></div><div>$$[H_1H_2H_3]=\frac{(a^2-b^2)(a^2-c^2)\sin{\alpha}}{2bc}+\frac{(a^2-c^2)(b^2-c^2)\sin{\gamma}}{2ab}-\frac{(a^2-b^2)(b^2-c^2)\sin{(\alpha+\gamma)}}{2ac}.\tag{12}$$</div><div><div>Dividing both sides of equation $(12)$ by $[ABC]=\frac12bc\sin{\alpha}=\frac12ab\sin{\gamma}=\frac12ac\sin{\beta}$ and taking into account that $\sin{(\alpha+\gamma)}=\sin{(\pi-\beta)}=\sin{\beta}$.</div></div><div>$$\frac{[H_1H_2H_3]}{[ABC]}=\frac{(a^2-b^2)(a^2-c^2)}{b^2c^2}+\frac{(a^2-c^2)(b^2-c^2)}{a^2b^2}-\frac{(a^2-b^2)(b^2-c^2)}{a^2c^2}.$$</div><div style="text-align: right;">$\square$</div><div style="text-align: left;">Note that the terms on the right hand side of the equation $(9)$ are the products of the <a href="https://en.wikipedia.org/wiki/Mollweide%27s_formula" target="_blank">Mollweide's formulas</a>. Substituting from Mollweide's formulas and applying the identity of sine of double angle, we can write the equation $(9)$ as follows</div><div style="text-align: left;">$$\boxed{\frac{[H_1H_2H_3]}{[ABC]}=\frac{\sin{(\alpha-\beta)}\sin{(\alpha-\gamma)}}{\sin{\gamma}\sin{\beta}}+\frac{\sin{(\alpha-\gamma)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\beta}}-\frac{\sin{(\alpha-\beta)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\gamma}}.}\tag{13}$$</div><div style="text-align: left;"><br /></div></span><div><span style="font-size: large;"><div><b>Back to the main problem</b></div><div>It is well-known that the orthocenter of the circumcircle mid-arc triangle is the incenter of the reference triangle. So $IB\perp{FD}$. Moreover, because of properties of inscribed angles, $$\angle{IDF}=\angle{FCB}=\angle{FDB}=\frac12\gamma$$ </div><div>and $$\angle{IFD}=\angle{CAD}=\angle{DAB}=\angle{DFB}=\frac12\alpha.$$ </div><div>Hence, by $ASA$, $DFI\cong{DFB}$. As a consequence, $I$ is the reflection of $B$ around $FD$. As $B'$ is the reflection of $B$ around $W$, it follows that the perpendicular bisector of $FD$ must also bisect $IB'$, meaning that $B'$ is the reflection of the orthocenter of $DEF$, $I$, around the perpendicular bisector of $DF$. Similarly, we conclude that $A'$ and $C'$ are the reflections of $I$ around the perpendicular bisectors of $EF$ and $DE$, respectively. Now our goal will be to show that $\frac{[A'B'C']}{[DEF]}=1-\frac{2r}{R}$. Note that $\angle{EFD}=\frac12(\alpha+\beta)$, $\angle{EDF}=\frac12(\beta+\gamma)$ and $\angle{DEF}=\frac12(\alpha+\gamma)$. Applying formula $(13)$ to triangles $A'B'C'$ and $DEF$ and substituting angles we have</div><div>$$\frac{[A'B'C']}{[DEF]}=\frac{\sin{\frac12(\alpha-\gamma)}\sin{\frac12(\beta-\gamma)}}{\cos{\frac12\beta}\cos{\frac12\alpha}}+\frac{\sin{\frac12(\beta-\gamma)}\sin{\frac12(\beta-\alpha)}}{\cos{\frac12\gamma}\cos{\frac12\alpha}}-\frac{\sin{\frac12(\alpha-\gamma)}\sin{\frac12(\beta-\alpha)}}{\cos{\frac12\gamma}\cos{\frac12\beta}}. \tag{14}$$</div><div>Note that the right hand side of the equation $(14)$ are, again, products of the <a href="https://en.wikipedia.org/wiki/Mollweide%27s_formula" target="_blank">Mollweide's formulas</a>. Thus, we have</div><div>$$\begin{aligned}\frac{[A'B'C']}{[DEF]}&=\frac{(a-c)(b-c)}{ab}+\frac{(b-c)(b-a)}{ac}-\frac{(a-c)(b-a)}{bc}\\&=1-\frac{c(a+b-c)}{ab}+1-\frac{b(a-b+c)}{ac}+1-\frac{a(-a+b+c)}{bc}.\end{aligned}\tag{15}$$</div><div>Substituting from the <a href="https://geometriadominicana.blogspot.com/2020/06/another-proof-for-two-well-known.html" target="_blank">half-angle formulas</a>, </div><div>$$\begin{aligned}\frac{[A'B'C']}{[DEF]}&=3-\frac{2a\cos^2{\frac12\alpha}+2b\cos^2{\frac12\beta}+2c\cos^2{\frac12\gamma}}{s}\\&=1-\frac{a\cos{\alpha}+b\cos{\beta}+c\cos{\gamma}}{s}.\end{aligned}\tag{16}$$</div><div>Substituting from the law of sines, factorizing and applying double-angle identity for sine,</div><div>$$\begin{aligned}\frac{[A'B'C']}{[DEF]}&=1-\frac{R(\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma})}{s}.\end{aligned}\tag{17}$$</div><div>But $\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma}=4\sin{\alpha}\sin{\beta}\sin{\gamma}=\frac{abc}{2R^3}=\frac{2rs}{R^2}$, hence</div><div>$$\frac{[A'B'C']}{[DEF]}=1-\frac{R}{s}\left(\frac{2rs}{R^2}\right)=1-\frac{2r}{R}.\tag{18}$$</div><div>Finally, from $(2)$ and $(18)$,</div><div>$$[ABC]+[A'B'C']=\frac{2r}{R}[DEF]+\left(1-\frac{2r}{R}\right)[DEF]=[DEF].$$</div><div style="text-align: right;">$\square$</div><div style="text-align: left;"><b>Some corollary inequalities:</b></div></span><span style="font-size: large;"><ul style="text-align: left;"><li><span style="font-size: large;">$\frac{(a^2-b^2)(a^2-c^2)}{b^2c^2}+\frac{(a^2-c^2)(b^2-c^2)}{a^2b^2}\geq\frac{(a^2-b^2)(b^2-c^2)}{a^2c^2}.$ (follows from $(9)$)</span></li><li><span style="font-size: large;">$\frac{\sin{(\alpha-\beta)}\sin{(\alpha-\gamma)}}{\sin{\gamma}\sin{\beta}}+\frac{\sin{(\alpha-\gamma)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\beta}}\geq\frac{\sin{(\alpha-\beta)}\sin{(\beta-\gamma)}}{\sin{\alpha}\sin{\gamma}}.$ (follows from $(13)$)</span></li><li><span style="font-size: large;">$\frac{c(a+b-c)}{ab}+\frac{b(a-b+c)}{ac}+\frac{a(-a+b+c)}{bc}\leq3.$ </span>(follows from $(15)$) (this is <a href="https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_2" target="_blank">IMO, 1964/2</a>)</li><li>$a\cos{\alpha}+b\cos{\beta}+c\cos{\gamma}\leq{s}.$ (follows from $(16)$)</li><li>$\sin{2\alpha}+\sin{2\beta}+\sin{2\gamma}\leq\frac{s}{R}.$ (follows from $(17)$)</li><li>$R\geq{2r}.$ (This is <a href="https://mathworld.wolfram.com/EulersInequality.html" target="_blank">Euler's Inequality</a>! It follows from (18))</li></ul></span></div><div><span style="font-size: large;"><div style="text-align: left;">Relation $(1)$ seems like a good companion for property 4 of the Garcia-reflection triangle. See <a href="https://drive.google.com/file/d/18CXRU3f8hoDySUfPLs54iQ4U3PjsC0FX/view" target="_blank">M. Dalcín, S. N. Kiss Some Properties of the García Reflection Triangles 119--126.</a></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><b>Update</b>. Van Khea has generalized my problem. I have given a proof of his nice generalization which is available at <a href="https://geometriadominicana.blogspot.com/2022/12/van-kheas-areal-problem.html" target="_blank">Van Khea's areal problem</a>.</div></span></div></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com1tag:blogger.com,1999:blog-4499994412319338435.post-32897949611012727492022-08-07T13:39:00.002-07:002022-08-10T04:43:05.350-07:00A property of the Spieker point<p><span style="font-size: large;"><b>Problem</b>. Let $ABC$ be a triangle. Let $A'$ be on $BC$ such that $AA'$ is an internal angle bisector. Call $M$ the midpoint of $AA'$. Let $T$ be the point of tangency of $BC$ with the $A$-excircle. Prove that the <a href="https://mathworld.wolfram.com/SpiekerCenter.html" target="_blank">Spieker center</a> of $ABC$ lies on $MT$.</span></p><p><span style="font-size: large;"><b>Proof</b>. Let $D$, $E$ and $F$ be the midpoints of sides $AC$, $AB$ and $BC$, respectively. Let the angle bisector of $\angle{EDF}$ intersect $ET$ in $D'$ and $BC$ in $D''$. Similarly, let the angle bisector of $\angle{DEF}$ intersect $DT$ in $E'$, $AC$ in $E''$ and $BC$ in $E^*$. Call $S$ the intersection of $DD'$ and $EE'$. Since $AB\parallel{DF}$, a simple angle chase lead us to conclude that $\triangle{FDD''}$ is isosceles with $FD=FD''$. Moreover, since $DE\parallel{D''T}$ it follows that $\triangle{EDD'}\sim{\triangle{TD'D''}}$, from which we have</span></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdw3np5dXmyWy85oDAqfivEc43VdSK25b9U2xUPG4aH8oWNLatpm7YBgFoUZX5Xw9U4-vf49amZRYc4nl5mRzz1a8jwdCPzHOR_J7OQMqsCGcS0G0Z6DdoWwTKsU-xhd32ezE4qDiVt6R6alO58hi_19mjJpjPr1gt9y5dcREIKSK_Hc0BLWfp_aLwVg/s1998/New%20triangle.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1445" data-original-width="1998" height="462" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdw3np5dXmyWy85oDAqfivEc43VdSK25b9U2xUPG4aH8oWNLatpm7YBgFoUZX5Xw9U4-vf49amZRYc4nl5mRzz1a8jwdCPzHOR_J7OQMqsCGcS0G0Z6DdoWwTKsU-xhd32ezE4qDiVt6R6alO58hi_19mjJpjPr1gt9y5dcREIKSK_Hc0BLWfp_aLwVg/w640-h462/New%20triangle.png" width="640" /></a></div><p><br /></p><span style="font-size: large;">\[\frac{ED'}{D'T}=\frac{DE}{D''T}=\frac{DE}{D''F+FT}=\frac{\frac{a}{2}}{\frac{c}{2}+\frac{a}{2}-(s-b)}=\frac{a}{b}.\tag{1}\]</span><div><span style="font-size: large;">An analogous reasoning lead us to conclude that $\triangle{DEE''}$ is isosceles (with $DE=DE''$) and $\triangle{DEE'}\sim{\triangle{TE'E^*}}$, from which we obtain</span></div><span style="font-size: large;">\[\frac{TE'}{E'D}=\frac{TE^*}{ED}=\frac{CT+CE^*}{ED}=\frac{CT+CE''}{ED}=\frac{(s-b)+\frac{b}{2}-\frac{a}{2}}{\frac{a}{2}}=\frac{c}{a}.\tag{2}\]</span><div><span style="font-size: large;">Now, since $BC$ is the homothetic image of $ED$ with scale factor $2$, then $M$ must lies on $DE$ and by the Angle Bisector Theorem, </span></div><div><span style="font-size: large;">\[\frac{DM}{ME}=\frac{AD}{AE}=\frac{b}{c}.\tag{3}\]</span></div><div><span style="font-size: large;">We want to show that $EE'$, $DD'$ and $MT$ are concurrent, so by invoking the Ceva's Theorem and substituting from $(1, 2, 3)$,</span></div><div><span style="font-size: large;">$$\frac{ED'}{D'T}\cdot{\frac{TE'}{E'D}}\cdot{\frac{DM}{ME}}=\frac{a}{b}\cdot{\frac{c}{a}}\cdot{\frac{b}{c}}=1.$$</span></div><div><span style="font-size: large;">This means $EE'$, $DD'$ and $MT$ are concurrent at $S$. Hence $S$ lies on $MT$.</span></div><div style="text-align: right;"><span style="font-size: large;">$\square$</span></div><div><p></p></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-45862198484372280792022-08-02T18:43:00.003-07:002022-08-10T04:36:45.545-07:00Solution of a problem by Tran Viet Hung<p> <span style="font-size: large;">This is problem $037$ in <a href="https://drive.google.com/file/d/1uQQZjyvBOZcl9grSc-wjaebHXpihqe5E/view?fbclid=IwAR0352tcprUTRXTLTlWZMhAAmCqmWDtSlxl9uJGvFQ-GNnUqdozbHd_Yja4" target="_blank">this document</a> by the Vietnamese geometer, <a href="https://www.facebook.com/hoang.heo.79" target="_blank">Tran Viet Hung</a>. </span></p><p><span style="font-size: large;"><b>Problem</b>. Let $ABC$ be a triangle and $I$ its Incenter. Denote $D$, $E$ the points of tangency of incircle of $ABC$ with sides $BC$ and $AC$, respectively. Let $P$ be the intersection of the perpendicular of $AI$ at $A$ and $BC$. Construct $Q$ similarly. Denote $M$ and $N$ the midpoints of $AP$ and $BQ$, respectively. Prove that $MN$, $AB$ and $DE$ concur (see figure below). </span></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjYvll1gyA1M98VtBWgAIg0m0JEJC3DJq2NnKbDHeJ22X2tkWuuxAyy3AV9LOaUJknSLYkhWRNhG9G4LvVkMTd4x0w9cAEmDbHirfozjIBljDgOGf2k1NTIHFRkB4qylGmB8NMUDr8Elbr_e3Clb4utQBN78IkHTJ0yN-KvwYgoEYnmbwyFAsLmHCEuNg/s669/Hung1.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="592" data-original-width="669" height="354" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjYvll1gyA1M98VtBWgAIg0m0JEJC3DJq2NnKbDHeJ22X2tkWuuxAyy3AV9LOaUJknSLYkhWRNhG9G4LvVkMTd4x0w9cAEmDbHirfozjIBljDgOGf2k1NTIHFRkB4qylGmB8NMUDr8Elbr_e3Clb4utQBN78IkHTJ0yN-KvwYgoEYnmbwyFAsLmHCEuNg/w400-h354/Hung1.jpg" width="400" /></a></div><br /><span style="font-size: large;"><b>Lemma 1</b>. Let $P'$ be the intersection of $AI$ with $BC$ and $Q'$ the intersection of $BI$ with $AC$. Denote $M'$ and $N'$ the midpoints of $AP'$ and $BQ'$, respectively. Call $X$ the intersection of $AN'$ with $BM'$. If $T'$ is the point of tangency of the $C$-excircle with $AB$, then $T'$, $X$ and $I$ are collinear. </span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Proof</b>. Here we'll be using standard notations, this is $BC=a$, $AC=b$, $AB=c$ and $s$ is semiperimeter. From the <a href="https://proofwiki.org/wiki/Length_of_Angle_Bisector" target="_blank">formula for the length of angle bisector</a> we have</span></div><div><span style="font-size: large;">$$AP'=\frac{2\sqrt{bcs(s-a)}}{b+c}.\tag{1}$$</span></div><div><span style="font-size: large;">Moreover, it is <a href="https://www.cut-the-knot.org/triangle/RelationsInTriangle.shtml#AISQ" target="_blank">well-known</a> that</span></div><div><span style="font-size: large;">$$AI=\sqrt{\frac{bc(s-a)}{s}}.\tag{2}$$</span></div><div><span style="font-size: large;">From $(1)$ and $(2)$ follows that</span></div><div><span style="font-size: large;">$$M'I=AI-\frac{AP'}{2}=\sqrt{\frac{bc(s-a)}{s}}-\frac{\sqrt{bcs(s-a)}}{b+c}=\frac{1}{b+c}\cdot{\sqrt{\frac{bc(s-a)^3}{s}}}.$$</span></div><div><span style="font-size: large;">Similarly, we can derive an expression for $N'I$, </span></div><div><span style="font-size: large;">$$N'I=BI-\frac{BQ'}{2}=\sqrt{\frac{ac(s-b)}{s}}-\frac{\sqrt{acs(s-b)}}{a+c}=\frac{1}{a+c}\cdot{\sqrt{\frac{ac(s-b)^3}{s}}}.$$</span></div><div><span style="font-size: large;">Invoking Ceva's theorem, our goal now is to show that</span></div><div><span style="font-size: large;">$$\frac{AM'}{M'I}\cdot{\frac{N'I}{BN'}}\cdot{\frac{BT'}{AT'}}=1.\tag{3}$$</span></div><div><span style="font-size: large;">Since $AT'=(s-b)$ and $BT'=(s-a)$, equation $(3)$ can be re-written as</span></div><span style="font-size: large;">$$\frac{\sqrt{bcs(s-a)}}{\sqrt{\frac{bc(s-a)^3}{s}}}\cdot{\frac{\sqrt{\frac{ac(s-b)^3}{s}}}{\sqrt{acs(s-b)}}}\cdot{\frac{s-a}{s-b}}=\frac{s}{s-a}\cdot{\frac{s-b}{s}}\cdot{\frac{s-a}{s-b}}=1.$$</span><div><span style="font-size: large;">This means that $AN'$, $BM'$ and $IT'$ are concurrent at $X$. Hence $T'$, $X$ and $I$ must be collinear.<br /></span><div style="text-align: right;"><span style="font-size: large;">$\square$ </span></div><div style="text-align: left;"><span style="font-size: large;"><b>Back to Hung's original problem</b></span></div><div><span style="font-size: large;">Call $I_c$ the $C$-excenter of $ABC$. By property of ex-centers, $\angle{IBI_c}=\angle{IAI_c}=90^\circ$, so $AP$ and $BQ$ must intersect at $I_c$. Denote $Y$ the intersection of $AN$ with $BM$ and $T$ the point of tangency of the incircle with $AB$. Then $Y$, $I_c$ and $T$ must be collinear since this is the extraverted version of Lemma 1. Now, suppose $DE$ and $MN$ intersect $AB$ at $R$ and $R'$, respectively. It is well-known that $AD$, $BE$ and $CT'$ are concurrent at the <a href="https://mathworld.wolfram.com/GergonnePoint.html" target="_blank">Gergonne Point</a>, so by property of harmonic bundles we have</span></div><div style="text-align: left;"><span style="font-size: large;">$$-1=(A, B; T, R)=(A, B; T, R'),$$</span></div><div style="text-align: left;"><span style="font-size: large;">wich means $R=R'$, hence $MN$, $AB$ and $DE$ concur.</span></div><div style="text-align: right;"><span style="font-size: large;">$\square$</span></div><div style="text-align: left;"><span style="font-size: large;"><b>Remark</b>. Lemma 1 has given rise to a new special triangle, namely the <b>Garcia-Moses triangle</b>, published in the <a href="https://faculty.evansville.edu/ck6/encyclopedia/ETCPart26.html" target="_blank">Encyclopedia of Triangle Centers</a>.</span></div><div><div><p></p></div></div></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-43028435657333463892022-07-21T07:47:00.007-07:002022-07-27T10:30:48.008-07:00The jealous engineer problem<p><span style="font-size: large;"> </span></p><div class="separator" style="clear: both; text-align: center;"><span style="font-size: large;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiV5yeuOmhiYmPr141KiM5gucSTqF1P2wh_mjxluXL7Dv4hMUpNqO0uuKuiaswVfCaaEPHkKiReuXsDAPly7veJO-4Vvw4ND-cu1wp8Il-mqH1PDQLrHXMXUIbN4RqKAJ41WuioFP9n9xwhEHylHYW_no9L29voNmLXGer6-on5ESryZhbZpgt5cJBgrg/s500/esposa.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="417" data-original-width="500" height="334" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiV5yeuOmhiYmPr141KiM5gucSTqF1P2wh_mjxluXL7Dv4hMUpNqO0uuKuiaswVfCaaEPHkKiReuXsDAPly7veJO-4Vvw4ND-cu1wp8Il-mqH1PDQLrHXMXUIbN4RqKAJ41WuioFP9n9xwhEHylHYW_no9L29voNmLXGer6-on5ESryZhbZpgt5cJBgrg/w400-h334/esposa.jpg" width="400" /></a></span></div><span style="font-size: large;"><br /></span><p></p><p><span style="font-size: large;">One day Tony woke up and didn't find his wife in the house.</span></p><p><span style="font-size: large;">— Where were you?</span></p><p><span style="font-size: large;">— I wanted to work out and just walked around the block. Look at my pedometer.</span></p><p><span style="font-size: large;">— What distance does the pedometer show?</span></p><p><span style="font-size: large;">— Really Tony? again with your jealousy?</span></p><p><span style="font-size: large;">Angry, Tony snatches the pedometer from her and he read a distance traveled of $805$ meters.</span></p><p><span style="font-size: large;">— How many laps did you do?</span></p><p><span style="font-size: large;">— One lap!</span></p><p><span style="font-size: large;">— East or North?</span></p><p><span style="font-size: large;"> — Are you serious, Tony?</span></p><p><span style="font-size: large;">— East or North?</span></p><p><span style="font-size: large;">— East!</span></p><p><span style="font-size: large;">Tony had been the engineer in charge of paving the block years ago, and he knew that the four streets that made up the block were the same distance from the church where he married his wife. In addition, he remembered that the corners formed the following sequence of angles until he got back to his house: $60°$, $135°$, $85°$, $80°$ and that if his wife went east she must have traveled $200$ meters on the first street before to reach the $60°$- corner, a distance that he also remembered perfectly.</span></p><p><span style="font-size: large;">Was the wife lying? Justify your answer.</span></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-14913742673953394932022-07-21T07:25:00.007-07:002022-07-27T10:31:16.097-07:00El problema del ingeniero celoso<p></p><p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"></span></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAvs7a-9fXfS5--i7UPTb13GYcAr2C8134bH74pKqnTblhiChtyrl5RSBvBay-i_E99BnwYKXP-uRnrxAfDAyLKOrKb6z-7wXE43gEMFguYZUZV0ylE5YdMv1NvyXpuB0DmDvN3dUPzgVYB1Wr7BG_hRF6SA9gmM3kIN2SBkK_bMU9Uch56K3UvJscug/s500/esposa.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="417" data-original-width="500" height="334" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAvs7a-9fXfS5--i7UPTb13GYcAr2C8134bH74pKqnTblhiChtyrl5RSBvBay-i_E99BnwYKXP-uRnrxAfDAyLKOrKb6z-7wXE43gEMFguYZUZV0ylE5YdMv1NvyXpuB0DmDvN3dUPzgVYB1Wr7BG_hRF6SA9gmM3kIN2SBkK_bMU9Uch56K3UvJscug/w400-h334/esposa.jpg" width="400" /></a></div><br /><span style="font-size: large;"><br /></span><p></p><p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">Un día, Tony se despertó y no encontró a su esposa en la casa. <o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿Dónde estabas?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— Quería ejercitarme y solo recorrí la cuadra. Mira mi podómetro. <o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿Qué distancia marca el podómetro?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿En serio, Tony? ¿otra vez con tus celos?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">Airado, Tony le arrebata el podómetro y lee una distancia recorrida
de $805$ metros.<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿Cuántas vueltas diste?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¡Una! ¡una vuelta!<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿Este o norte?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿Hablas en serio, Tony?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¿Este o norte?<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">— ¡Este!<o:p></o:p></span></span></p>
<p class="MsoNormal" style="margin-bottom: 0in;"><span lang="ES-DO"><span style="font-size: large;">Tony había sido el ingeniero a cargo del asfaltado de la cuadra años
atrás y sabía que las cuatro calles que formaban la cuadra estaban a la misma
distancia de la iglesia donde se casó con su esposa. Además, recordaba que las
esquinas formaban la siguiente secuencia de ángulos hasta llegar de vuelta a su
casa: $60°$, $135°$, $85°$, $80°$ y que si su esposa partió hacia el este debió
haber recorrido $200$ metros en la primera calle antes de llegar a la esquina de</span></span> <span style="font-size: large;">$60°$, distancia que también recordaba a la perfección. </span></p><p class="MsoNormal" style="margin-bottom: 0in;"><span style="font-size: large;">¿Mentía la esposa? Justifica tu respuesta.</span></p><p></p>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-13278483470415800302022-07-10T21:23:00.009-07:002022-07-15T21:23:54.553-07:00A generalization of Lami's theorem for 4 forces<p><span style="font-size: large;"> In physics, <a href="https://en.wikipedia.org/wiki/Lami%27s_theorem" target="_blank">Lami's theorem</a> is an equation relating the magnitudes of three coplanar, concurrent and non-collinear vectors, which keeps an object in static equilibrium, with the angles directly opposite to the corresponding vectors. According to the theorem</span></p><p><span style="font-size: large;">$$\frac{F_1}{\sin{\alpha}}=\frac{F_2}{\sin{\beta}}=\frac{F_3}{\sin{\gamma}}.\tag{1}$$</span></p><p><span style="font-size: large;">where $F_1$, $F_2$ and $F_3$ are the magnitudes of the three coplanar, concurrent and non-collinear vectors which keep the object in static equilibrium, and $\alpha$, $\beta$ and $\gamma$ are the angles directly opposite to the vectors (see Figure 1).</span></p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuv7MCYdXPEKSwGZ31oQh9OBW_cBALHLlAVRJSiXxAQPNsIigo5L-WtHXP-47wcifRrjfkLwcXk1wwu8HyME8SraD-O1ZQzmGpxQET25tSjsynNgNsXi7EyRozXpDQbFBIvbbq1MZcCoeEbYx7nomOwzqOk-m4pPoqB3aBqDxQ6tzz5a_J3V56wSQy4g/s520/L1.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="375" data-original-width="520" height="289" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuv7MCYdXPEKSwGZ31oQh9OBW_cBALHLlAVRJSiXxAQPNsIigo5L-WtHXP-47wcifRrjfkLwcXk1wwu8HyME8SraD-O1ZQzmGpxQET25tSjsynNgNsXi7EyRozXpDQbFBIvbbq1MZcCoeEbYx7nomOwzqOk-m4pPoqB3aBqDxQ6tzz5a_J3V56wSQy4g/w400-h289/L1.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b>Figure 1</b>. $\alpha$, $\beta$ and $\gamma$ are the angles directly opposite to the vectors $F_1$, $F_2$ and $F_3$</td></tr></tbody></table><span style="font-size: large;"><br /></span><p></p><p><span style="font-size: large;">Lami's theorem is applied in static analysis of mechanical and structural systems. The theorem is named after Bernard Lamy. The proof of Lami's theorem is essentially based on the law of sines.</span></p><span style="font-size: large;">On the Internet there are hundreds of static equilibrium problems where they apply Lami's theorem to a three-force system, see for instance <a href="https://books.google.com.do/books?id=8Yf0AQAAQBAJ&q=lamis%20theorem&redir_esc=y#v=onepage&q=lamis%20theorem&f=false">Dubey - Engineering Mechanics: Statics and Dynamics, section 3.10</a>. Although Dubey's book is recent (2013) there is not a single equilibrium problem based on a four-force system. Coincidentally, the author of this note has come across <a href="https://www.quora.com/Is-Lamis-theorem-applicable-for-more-than-3-forces">questions</a> on the Internet questioning the possibility of applying Lami's theorem for more than three forces. In this note we give a generalization of Lami's theorem for four forces.</span><div><div><span style="font-size: large;"><b><br /></b></span></div><div><span style="font-size: large;"><b>Theorem 1 (Generalization)</b>. If four coplanar, concurrent and non-collinear forces act upon an object, and the object remains in static equilibrium, then</span></div><div><span style="font-size: large;"><br /></span></div><div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibLe_z-ELQz8vFI6_vO7a8i5Bqtp-OHgNl-bNtS9G3bUofLvLUxmjEwHUYH5tQWWfHINANz0xI4hWULGGBiJmTC0OEr8Ehcr3fMl5KKjP2-jJXSg21C7YpzBmSJHRXPXa_mlQ75zayH76HrykuHsUoHpuSMep4TViNUPVtUxObcobsaDhNvWF3Qforbg/s615/L2.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="395" data-original-width="615" height="258" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibLe_z-ELQz8vFI6_vO7a8i5Bqtp-OHgNl-bNtS9G3bUofLvLUxmjEwHUYH5tQWWfHINANz0xI4hWULGGBiJmTC0OEr8Ehcr3fMl5KKjP2-jJXSg21C7YpzBmSJHRXPXa_mlQ75zayH76HrykuHsUoHpuSMep4TViNUPVtUxObcobsaDhNvWF3Qforbg/w400-h258/L2.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b>Figure 2</b>. An example of the situation described in the theorem 1. </td></tr></tbody></table><br /><span style="font-size: large;"><br /></span></div><span style="font-size: large;">$$AD\sin{\alpha'}+BC\sin{\gamma'}=AB\sin{\beta'}+CD\sin{\delta'}.\tag{2}$$</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">where $A$, $B$, $C$ and $D$ are the magnitudes of the four vectors and $\alpha'$, $\beta'$, $\gamma'$ and $\delta'$ are the angles between them (see Figure 2).</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><div><b>Proof</b>. Consider the quadrilateral formed by the four vectors in such a manner that the head of one touches the tail of another (see Figure 3) and denote $\Delta$ its area. If $\alpha$, $\beta$, $\gamma$ and $\delta$ are the interior angles of the quadrilateral, then its area can be written as</div><div><br /></div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiE5erwxpYijycqalFP_3xrBAtn9mzpkwVAtp2bjaCFpN2NHWSXPWBGBK_hMCd4AStchlI9win12E5_vN-DISDvZ6uaQkBQiQPRgOht7eBijRVMd4pt5afiLC9fnUsSJZ4hH7JTDpkCgqikves1tMjCjD5y9UyMb-s6YSYKESsvQ2nMbRVcolA5d9xJ9w/s682/L3.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="404" data-original-width="682" height="238" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiE5erwxpYijycqalFP_3xrBAtn9mzpkwVAtp2bjaCFpN2NHWSXPWBGBK_hMCd4AStchlI9win12E5_vN-DISDvZ6uaQkBQiQPRgOht7eBijRVMd4pt5afiLC9fnUsSJZ4hH7JTDpkCgqikves1tMjCjD5y9UyMb-s6YSYKESsvQ2nMbRVcolA5d9xJ9w/w400-h238/L3.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b>Figure 3</b>. Notice that if $\alpha$, $\beta$, $\gamma$ and $\delta$ are the interior angles of the quadrilateral formed by the four vectors, then $\alpha$ and $\alpha'$ are supplementary and similarly for $\beta'$, $\gamma'$ and $\delta'$</td></tr></tbody></table><br /><div><div>$$\Delta=\frac12AD\sin{\alpha}+\frac12BC\sin{\gamma}=\frac12AB\sin{\beta}+\frac12CD\sin{\delta}$$ </div><div>and as $\sin{\alpha'}=\sin{(\pi-\alpha)}=\sin{\alpha}$, and similarly for $\beta'$, $\gamma'$ and $\delta'$, the relation in $(2)$ follows.</div><div><br /></div><div style="text-align: right;">$\square$</div><div><br /></div><div>Theorem 1 is a generalization in the sense that if one of the vectors vanishes, the relation we obtain is that of Lami's theorem. Indeed, for instance suppose $C=0$, then the relation $(2)$ reduces to</div><div>$$D\sin{\alpha'}=B\sin{\beta'},$$</div><div>which is Lami's theorem.</div></div><div><br /></div><div><b>Remark</b>. A generalization of Lami's theorem is given by <a href="https://www.irjet.net/archives/V6/i10/IRJET-V6I1085.pdf" target="_blank">H. Shekhar</a>. However, this generalization is different since it only considers cyclic polygons with an odd number of sides.</div></span></div></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-19359828295461174522022-06-27T04:48:00.021-07:002023-04-17T04:25:03.755-07:00A generalization of the Pythagorean trigonometric identity<p style="text-align: center;"><i>"Discovery is seeing what everybody else has seen, and thinking what nobody else has thought."</i></p><p style="text-align: center;"><i><span></span></i></p><p style="text-align: center;"><i>-- Albert Szent-Györgyi (1893 - 1986)</i></p><p><span style="font-size: large;">I don't know how I missed it, but the <a href="https://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity" target="_blank">Pythagorean trigonometric identity</a> is a special case of the generalized half-angle formulas. </span></p><p><span style="font-size: large;">The following is a generalization of the half-angle formulas presented at <a href="http://www.nabla.hr/GE-AppTrigonomB1.htm" target="_blank">Nabla - Applications of Trigonometry</a> for a triangle.</span></p><p><span style="font-size: large;"><b>Generalization</b>. Let $a$, $b$, $c$, $d$ be the sides of a general convex quadrilateral, $s$ is the semiperimeter, and $\alpha$ and $\gamma$ are opposite angles, then</span></p><p><span style="font-size: large;"></span></p><p><span style="font-size: large;">$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d).\tag{1}$$</span></p><p><span style="font-size: large;">For a proof of $(1)$ see pp. 8 in <a href="http://matinf.upit.ro/MATINF6/RevistaMATINF_6.pdf" target="_blank">MATINF</a>.</span></p><p><span style="font-size: large;"><b>The Pythagorean identity as a special case</b></span></p><p><span style="font-size: large;"><b> </b>In $(1)$<b>, </b>consider the case when $a=c$, $b=d$ and $\alpha=\gamma$ so that the quadrilateral is a parallelogram . Then</span></p><span style="font-size: large;">$$ab\sin^2{\frac{\alpha}{2}}+ab\cos^2{\frac{\alpha}{2}}=\frac{-a+b+c+d}{2}\cdot{\frac{a+b+c-d}{2}}=ab.$$</span><p><span style="font-size: large;">Dividing both sides by $ab$ you get</span></p><p><span style="font-size: large;">$$\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}=1.$$</span></p><br /><span style="font-size: large;">Or by making $\frac{\alpha}{2}=\theta$,<br /><br />$$\sin^2{\theta}+\cos^2{\theta}=1,$$<br /><br />which is the Pythagorean trigonometric identity.</span><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;">An interesting discussion about this generalization is available at <a href="https://chat.stackexchange.com/rooms/137645/discussion-on-question-by-emmanuel-jose-garcia-generalizing-the-pythagorean-trig" target="_blank">MathSE</a>.<br /></span><p><span style="font-size: large;">For more implications of $(1)$ I invite you to see <a href="https://geometriadominicana.blogspot.com/2022/05/the-theoretical-importance-of-half.html" target="_blank">The theoretical importance of the half-angle formulas</a>.</span></p></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0tag:blogger.com,1999:blog-4499994412319338435.post-47568466203219057472022-06-25T10:21:00.007-07:002022-06-26T16:33:33.318-07:00Another proof of Euler inequality via the half-angle formulas<span style="font-size: large;">The <a href="https://mathworld.wolfram.com/EulersInequality.html" target="_blank">Euler's inequality</a> is an immediate consequence of <a href="https://en.wikipedia.org/wiki/Euler%27s_theorem_in_geometry" target="_blank">Euler's identity in a triangle</a>,<br />$$OI^2=R^2−2Rr.$$<br />An additional proof of Euler's inequality is given at <a href="https://www.cut-the-knot.org/triangle/EulerInequality.shtml" target="_blank">Elias Lampakis, Am Math Monthly, 122 (9), November 2015, p 892.</a> Now, continuing with <a href="https://math.stackexchange.com/questions/4456365/the-theoretical-importance-of-the-half-angle-formulas" target="_blank">this madness</a> of proving <i><b>everything</b></i> from the <a href="https://geometriadominicana.blogspot.com/2020/06/another-proof-for-two-well-known.html" target="_blank">half-angle formulas</a>, we give an alternative proof of Euler's inequality.</span><div><span style="font-size: large;"><br /><b>Theorem (Euler)</b>: If $R$ and $r$ are the circumradius and, respectively, the inradius of a triangle, then<br />$$R\geq2r.\tag{1}$$</span></div><div><span style="font-size: large;"><br /></span></div><div><span style="font-size: large;"><b>Proof</b>. Let $a$, $b$ and $c$ be the sides of $\triangle{ABC}$. Denote $s$ its semiperimeter and $\angle{BAC}=\alpha$. It is easy to show that </span></div><div><span style="font-size: large;">$$(s-a)\tan{\frac{\alpha}{2}}=r.\tag{2}$$</span></div><div><span style="font-size: large;">Also, it is well-known that</span></div><div><span style="font-size: large;">$$\frac{abc}{4\Delta}=R\qquad and \qquad\Delta=\frac{bc\sin{\alpha}}{2},\tag{3}$$</span></div><div><span style="font-size: large;">where $\Delta$ is the area of $\triangle{ABC}$. Substituting $(2)$ and $(3)$ in $(1)$ we have</span></div><div><span style="font-size: large;">$$\begin{aligned}\frac{abc}{4\Delta}&\geq(-a+b+c)\tan{\frac{\alpha}{2}}\\\frac{abc}{4bc\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}&\geq(-a+b+c)\tan{\frac{\alpha}{2}}\\abc&\geq(-a+b+c)(4bc\sin^2{\frac{\alpha}{2}})\\abc&\geq(-a+b+c)(a-b+c)(a+b-c)\end{aligned}$$</span></div><div><span style="font-size: large;">This is <a href="https://www.cut-the-knot.org/triangle/Padoa.shtml" target="_blank">Padoa's inequality</a>, so the proof is complete.</span></div>Emmanuel José Garcíahttp://www.blogger.com/profile/02365244240748674744noreply@blogger.com0