A family of trigonometric formulas for the roots of quadratic equations

 This note presents alternative trigonometric formulas for finding the roots of quadratic equations where $a$, $b$, and $c$ are non-zero real numbers.

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 identities that appear useful for integrating irrational functions. Thus, the sole purpose of this note is to document the origins of these identities.

In the formulas presented, each inverse trigonometric or hyperbolic function assumes real values only when its argument lies in the corresponding real domain. Specifically:

$\alpha = \sin^{-1}\left(\frac{b}{2\sqrt{ac}}\right)$ and $\gamma = \cos^{-1}\left(\frac{b}{2\sqrt{ac}}\right)$ are real-valued if and only if 

$$\left|\frac{b}{2\sqrt{ac}}\right|\;\le\;1.$$

$\beta = \csc^{-1}\left(\frac{b}{2\sqrt{ac}}\right)$ and $\delta = \sec^{-1}\left(\tfrac{b}{2\sqrt{ac}}\right)$ are real-valued if and only if 

$$\left|\frac{b}{2\sqrt{ac}}\right|\;\ge\;1.$$

$\eta = \tanh^{-1}\left(\frac{b}{2\sqrt{ac}}\right)$ is real-valued if and only if 

$$\left|\frac{b}{2\sqrt{ac}}\right|\;<\;1.$$

$\theta = \coth^{-1}\left(\frac{b}{2\sqrt{ac}}\right)$ is real-valued if and only if 

$$\left|\frac{b}{2\sqrt{ac}}\right|\;>\;1.$$

Consequently, each theorem in this note implicitly presumes that \(b/(2\sqrt{ac})\) lies within the necessary interval so that the stated inverse trigonometric or hyperbolic expression remains real-valued. If one allows complex-valued functions, these restrictions need not apply, but the resulting roots may lie in the complex plane.

Theorem 1. Let $a$, $b$, and $c$ be non-zero real numbers. For the quadratic equation $ax^2+bx+c=0$, the roots are given by:

$$x_{1}=  ie^{i\alpha}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad  x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}},\tag{1}$$

where $\alpha=\sin^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. 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

$$p^2+bp+q^2=0.$$

Let's make the substitution $\sin{\alpha} = \frac{b}{2q}$. Consequently,

$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sin{\alpha}+q^2\\&=p^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)+4qp\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+q^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 + \left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2  -i^2\left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2 .\end{aligned}$$

Factorizing the difference of squares and then factorizing again, one of the factor of the quadratic equation can be express as follows

$$p\left(\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}\right).$$

Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain 

$$\begin{aligned}x_1&=-\left(\frac{\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}}{\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\sin{\alpha}-i\cos{\alpha})\sqrt{\frac{c}{a}}\\&=ie^{i\alpha}\sqrt{\frac{c}{a}}. \end{aligned}$$

The other factor is given by

$$p\left(\sin{\frac{\alpha}{2}}-i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}-i\sin{\frac{\alpha}{2}}\right).$$

And similarly,

$$x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}}.$$ 

Theorem 2. For the quadratic equation $ax^2+bx+c=0$ with non-zero real numbers $a$, $b$, and $c$, the roots are given by:

$$x_{1}=-\tan{\frac{\beta}{2}}\sqrt{\frac{c}{a}}\qquad \text{and} \qquad x_{2}=-\cot{\frac{\beta}{2}}\sqrt{\frac{c}{a}},\tag{2}$$

where $\beta=\csc^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Multiplying by $a$ the quadratic equation $ax^2+bx+c=0$ and then substituting $ax=p$ and $ac=q^2$, we have

$$p^2+bp+q^2=0.$$

Use the substitution $\csc{\beta}=\frac{b}{2q}$, then

$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\csc{\beta}+q^2\\&=\csc{\beta}\left(p^2\sin{\beta}+2qp+q^2\sin{\beta}\right)\\&=2p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+2qp\left(\sin^2{\frac{\beta}{2}}+\cos^2{\frac{\beta}{2}}\right)+2q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&=p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+qp\sin^2{\frac{\beta}{2}}+ qp\cos^2{\frac{\beta}{2}}+q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&= p\sin{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)+q\cos{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\\&= \left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\left(p\sin{\frac{\beta}{2}}+q\cos{\frac{\beta}{2}}\right).\end{aligned}$$

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 $(2)$.

Remark. We acknowledge that formulas similar to those presented in Theorem 2 were previously obtained in the work titled Lehrbuch der Gleichungen des II. Grades (Quadratische Gleichungen).

Theorem 3. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:

$$x_{1}=  -e^{i\gamma}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad  x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}},\tag{3}$$

where $\gamma=\cos^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields

$$p^2+bp+q^2=0.$$

By making the substitution $\cos{\gamma} = \frac{b}{2q}$, it follows that

$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\cos{\gamma}+q^2\\&=p^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)+4qp\left(\cos^2{\frac{\gamma}{2}}-\sin^2{\frac{\gamma}{2}}\right)+q^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2+\cos^2{\frac{\gamma}{2}}(p+q)^2\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2-i^2\cos^2{\frac{\gamma}{2}}(p+q)^2.\end{aligned}$$

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

$$p\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right)+q\left(-\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$

Setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain

$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\cos{\gamma}+i\sin{\gamma})\sqrt{\frac{c}{a}}\\&=-e^{i\gamma}\sqrt{\frac{c}{a}}.\end{aligned}$$

The other factor is given by

$$p\left(\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}\right)-q\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$

And similarly,

$$x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}}.$$

Remark. Similar formulas to those in Theorem 3 were obtained in the paper: Simons, Stuart, “Alternative approach to complex roots of real quadratic equations,” Mathematical Gazette 93, March 2009, 91–92.

Theorem 4. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by: 

$$x_{1,2}=\frac{\tan{\frac{\delta}{2}\pm1}}{\tan{\frac{\delta}{2}\mp1}}\sqrt{\frac{c}{a}},\tag{4}$$

where $\delta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields

$$p^2+bp+q^2=0.$$

Use the substitution $\sec{\delta}=\frac{b}{2q}$, then

$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sec{\delta}+q^2\\&=\sec{\delta}\left(p^2\cos{\delta}+2qp+q^2\cos{\delta}\right)\\&=\cos{\delta}(p^2+q^2)+2qp\\&= \cos{\delta}(p+q)^2-2qp\cos{\delta}+2qp\\&= \left(\cos^2{\frac{\delta}{2}}-\sin^2{\frac{\delta}{2}}\right)(p+q)^2-2qp\left(1-2\sin^2{\frac{\delta}{2}}\right)+2qp\\&=  \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p+q)^2 +4qp\sin^2{\frac{\delta}{2}}\\&=   \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}\left((p+q)^2 -4qp\right)\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p-q)^2\\&= \left(\cos{\frac{\delta}{2}}(p+q) +\sin{\frac{\delta}{2}}(p-q)\right) \left(\cos{\frac{\delta}{2}}(p+q) -\sin{\frac{\delta}{2}}(p-q)\right).\end{aligned}$$

Expanding and then factorizing again, one of the factors of the quadratic equation is given by

$$p\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right).$$ 

By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain

$$\begin{aligned}x_1&=\frac{\sin{\frac{\delta}{2}}+\cos{\frac{\delta}{2}}}{\sin{\frac{\delta}{2}}-\cos{\frac{\delta}{2}}}\sqrt{\frac{c}{a}}\\&= \frac{\tan{\frac{\delta}{2}+1}}{\tan{\frac{\delta}{2}-1}}\sqrt{\frac{c}{a}}. \end{aligned}$$

The other factor is given by

$$p\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right).$$

Similarly, 

$$x_2=\frac{\tan{\frac{\delta}{2}-1}}{\tan{\frac{\delta}{2}+1}}\sqrt{\frac{c}{a}}.$$

Theorem 5. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:

$$x_{1,2}=\frac{i\pm e^{\eta}}{i\mp e^{\eta}}\sqrt{\frac{c}{a}},\tag{5}$$

where $\eta=\tanh^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields

$$p^2+bp+q^2=0.$$

Use the substitution $\tanh{\eta}=\frac{b}{2q}$, then

$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\tanh{\eta}+q^2\\&=(e^{2\eta} + 1) p^2 + 2 (e^{2\eta} - 1) p q + (e^{2\eta} + 1) q^2\\&=e^{2\eta}(p + q)^2 + (p - q)^2\\&=e^{2\eta}(p + q)^2 - i^2(p - q)^2 \\&=(e^{\eta}p + i p + e^{\eta}q - i q)(e^{\eta}p - i p + e^{\eta}q + i q)\\&= \left(p(e^{\eta} + i) + q(e^{\eta} - i)\right)\left(p(e^{\eta} - i) + q(e^{\eta} + i)\right)\end{aligned}$$

Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain 

$$x_{1}=\frac{i- e^{\eta}}{i+e^{\eta}}\sqrt{\frac{c}{a}}.$$

And similarly,

$$x_{2}=\frac{i+ e^{\eta}}{i-e^{\eta}}\sqrt{\frac{c}{a}}.$$

Theorem 6. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:

$$x_{1,2}=\frac{1\pm e^{\theta}}{1\mp e^{\theta}}\sqrt{\frac{c}{a}},\tag{6}$$

where $\theta=\coth^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields

$$p^2+bp+q^2=0.$$

Use the substitution $\coth{\theta}=\frac{b}{2q}$, Then, similar to how we proceeded previously,

$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\coth{\theta}+q^2\\&=(e^{2\theta} - 1) p^2 + 2 (e^{2\theta} + 1) p q + (e^{2\theta} - 1) q^2\\&=e^{2\theta}(p + q)^2 - (p - q)^2 \\&=\left( p e^{\theta} + p + q e^{\theta} - q \right) \left( p e^{\theta} - p + q e^{\theta} + q \right)\\&=\left(p(e^x + 1) + q(e^x - 1)\right)\left(p (e^x - 1) + q (e^x + 1)\right)\end{aligned}$$

Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain 

$$x_{1}=\frac{1- e^{\theta}}{1+e^{\theta}}\sqrt{\frac{c}{a}}.$$

And similarly,

$$x_{2}=\frac{1+ e^{\theta}}{1-e^{\theta}}\sqrt{\frac{c}{a}}.$$

Integrals yielding $e^{\pi}$ or $e^{-\pi}$

 Lately, I've been playing a lot with integrals, and coincidentally (with a bit of algebraic manipulation), I've come across these two beauties:

$$\int_{0}^{1} \left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right) - 1\right) \, dx = e^{\pi},\tag{1}$$

$$\int_{0}^{1} \left( -\frac{5}{2\left( x - \sqrt{x^2 - 1}\right)^{2i}} - \frac{5x^4}{2} + 1 \right) \, dx = e^{-\pi}.\tag{2}$$

$e^{\pi}$ is known as Gelfond's constant.

I have provided these integrals as a response to a question on MathSE. The proofs are left as exercises for the reader.

A generalization of Burlet's theorem to cyclic quadrilaterals

 The Burlet's theorem is a result in Euclidean geometry, which can be formulated as follows:

Theorem 1. 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

$$\Delta_0=mn\cot{\frac{\gamma}{2}},\tag{1}$$

A triangle $ABC$ with $AP=m$ and $BP=n$.

where $\Delta_0$ denotes the area of $ABC$. Note that when $\gamma=\frac{\pi}{2}$, then $\Delta_0=mn$.

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.

Proof. Let $AB=a$, $BC=b$ and $AC=c$. By Heron's formula, we have

$$\begin{aligned}\Delta_0&=\sqrt{s(s-a)(s-b)(s-c)}\\&=\sqrt{s(s-a)(s-b)(s-c)}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}.\end{aligned}$$

However, using half-angle formulas for a triangle, 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 $(1)$.

$\square$

Theorem 2 (Generalization). 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:

$$\Delta_1=(s-a)(s-d)\cot{\frac{\alpha}{2}},\tag{2}$$

A cyclic quadrilateral $ABCD$.

where $\Delta_1$ denotes the area of $ABCD$.

Proof. Similar to how we proceeded in theorem 1, using the Brahmagupta's formula, we have

$$\begin{aligned}\Delta_1&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\color{red}{\sqrt{\frac{(s-a)(s-d)}{(s-a)(s-d)}}}\\&=(s-a)(s-d)\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}.\end{aligned}$$

Now, invoking the half-angle formulas for cyclic quadrilaterals, we have that $\cot{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}$, from which the relation $(2)$ is deduced.

$\square$

Let $\angle{BCD}=\gamma$ and assume that $d=0$. We have that $\alpha=\pi-\gamma$, so substituting in $(2)$, 

When $d=0$, by the alternate segment theorem, $\alpha=\pi-\gamma$.

$$\begin{aligned}\Delta_1&=(s-a)(s-0)\cot{\left(\frac{\pi-\gamma}{2}\right)}\\&=s(s-a)\tan{\frac{\gamma}{2}}\\&=s(s-a)\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\\&=(s-b)(s-c)\cot{\frac{\gamma}{2}},\end{aligned}$$

which is Burlet's theorem for a triangle since $m=(s-b)$ and $n=(s-c)$.

Remark. A. Burlet (Dublin) presented this theorem as a problem in the Nouvelles annales de mathématiques, Vol. 15, 1856, p. 290. Consequently, the theorem might have derived its name from him.

The half-angle formulas are central!

 As the title suggests, the half-angle formulas are central. 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 "The Theoretical Importance of Half-Angle Formulas" where you can see the details of the proofs, as well as the new generalizations I have managed to derive from this novel approach.

Click on the image for a better view

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.

Solution to Problem 1315 of Gogeometry

 The following problem was proposed by my esteemed friend, Ángel Mejía. However, this is problem 1315 from Gogeometry. Here I give a trigonometric proof.

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 $DG = MN$.

Proof. 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

$$GM = FG \cdot \sin{\delta},\tag{1}$$

$$DG = \frac{FG \cdot \sin{\epsilon}}{\sin{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}.\tag{2}$$

Dividing $(1)$ by $(2)$ and then squaring, we get

$$\left(\frac{GM}{DG}\right)^2=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\cdot{\sin^2{\frac12\left(\frac{\pi}{2}-\frac{\beta}{2}\right)}}=\frac{\sin^2{\delta}}{\sin^2{\epsilon}}\left(\frac{1-\sin{\frac12\beta}}{2}\right).\tag{3}$$

Now, applying the sine law once more, this time in triangles $AFO$ and $ADF$, we obtain

$$\sin{\delta}=\frac{AO\cdot{\sin{\frac12(\alpha+\beta)}}}{AF},\tag{4}$$

$$\sin{\epsilon}=\frac{AD\cdot{\sin{\frac14(3\pi+\beta})}}{AF}.\tag{5}$$

Dividing $(4)$ by $(5)$ and then squaring, we have

$$\frac{\sin^2{\delta}}{\sin^2{\epsilon}}=\frac{\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}}{\sin^2{\frac14(3\pi+\beta})}=\sec^2{\frac12\alpha}\cos^2{\frac12\gamma}\left(\frac{2}{1-\sin{\frac12\beta}}\right).\tag{6}$$

Substituting $(6)$ in $(3)$ and taking square roots,

$$\frac{GM}{DG}=\frac{\cos{\frac12\gamma}}{\cos{\frac12\alpha}}.$$

Analogously,

$$\frac{MN}{EN}=\frac{\cos{\frac12\alpha}}{\cos{\frac12\gamma}}.$$

This implies $\frac{GM}{DG}=\frac{EN}{MN}$ and, since $DM=ME$, we conclude that $DG=MN$.

$\square$

Another Identity on Triangle Areas Arising from Reflection

In a previous post, we proved an identity about triangle areas arising from reflection. In this occasion, we prove another interesting identity associated with triangle areas.

Theorem. 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 

$$[A'B'C']-[A''B''C''] = 3[ABC].$$

The brackets $[\, ]$ represent the area of the enclosed figure.

Lemma 1. $[A'B'C]=[ABA''B'']$, $[A'C'B]=[ACA''C'']$ and $[B'C'A]=[BCB''C'']$.

Proof. 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 $\theta$.

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

$$[AB'C']=[BCB'C']-3[ABC]=4[BCEF]-3[ABC]=4[ABC]-4[AEF]-3[ABC]=[ABC]-4[AEF].\tag{1}$$

Let $H$ and $I$ be the midpoints of $AC$ and $AB$, respectively. Then

$$[ABC]=4[AHI]=4[AEF]+4[HEFI].\tag{2}$$

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,

$$[AB'C']=[BCC''B''].$$

$\square$

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.

$\color{blue}{[A'B'C]=[ABA''B'']}$, $\color{red}{[A'C'B]=[ACA''C'']}$ and $\color{green}{[B'C'A]=[BCB''C'']}$.

Remark. 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.

Back to the main problem

Notice that 

$$[A'B'C']=[A'C'B'C]-[A'B'C]=4[ABC]+[AB'C']+[BA'C']-[A'B'C].$$

But $[AB'C']=[ABC]-4[AEF]$ and similarly $[BA'C']=[ABC]-4[BDF]$ and $[A'B'C]=4[CDE]-[ABC]$, so

$$[A'B'C']= 7[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{3}$$

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

$$[A''B''C'']=4[DEF]=4[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{4}$$

Substracting $(4)$ from $(3)$ we get 

$$[A'B'C']-[A''B''C'']=3[ABC].$$

$\square$

A curious family of integrals that give rational multiples of $\pi$

 I have noticed experimentally that:

$$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$

$$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$

$$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$

So slight variations in the numerator always seem to produce something like $n\pi$, where $n$ is a rational number.

At MathSE 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:

$$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx=\frac\pi8\color{green}{(c+b-a)}+\frac\pi{3\sqrt3}\color{green}{(a+c)}.\tag{4}$$

However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula $(34)$ in this list of $\pi$ formulas:

$$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$

Notice that the denominator is different. But again, a slight variation in the numerator and it still produces something like $n\pi$:

$$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$

More generally, 

$$\int_{0}^{1} \frac{ax^2+bx+c}{x^4-2x^3+4x-4}\,dx=\frac{\pi}{16}\color{green}{(2a-c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt32}\color{green}{(b+c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt8}\color{green}{a}.\tag{7}$$

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.

A more intimidating integral

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:

$$\int_{0}^{1} \frac{2(1-x)(x^5-5x^4-10x^3-4x^2+8x-8)}{x^8-2x^6-2x^5+9x^4-2x^3-16x^2+12x-4}=\pi.$$

Addendum. 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 $n\pi$?