## sábado, 20 de junio de 2020

### Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

In Cut-the-knot's Relations between various elements of a triangle, the formulae

$$\sin^2{\frac{\gamma}{2}} = \frac{(s-a)(s-b)}{ab}\quad and\quad\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$

are derived using Heron's formula. Here we give an alternative proof without using Heron's Formula and we demonstrate several well-known theorems based on these formulae as a sample of its power. We will be using standard notation: $BC=a$, $AC=b$, $AB=c$, $\Delta$ for the area, $s$ for the semiperimeter, $R$ for the circumradius and $r$ for the inradius. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$.

Notice that $\frac{\cot{\frac{\gamma}{2}}}{s-c} = \frac{1}{r}$. Also, We know $Δ = rs$ and $Δ =\frac {ab\sin{\gamma}}{2}$, hence

$$\frac{\cot{\frac{\gamma}{2}}}{s-c}=\frac{1}{r}=\frac{s}{\Delta}=\frac{2s}{ab\sin{\gamma}}$$

But, $\sin{\gamma} = 2\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}$, so

$$\frac{\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}\cdot{\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}} = \frac{s(s-c)}{ab}$$

from which we get $\cos^2{\frac{\gamma}{2}} = \frac{s(s-c)}{ab}$.

The other formula can be obtained replacing $\cos^2{\frac{\gamma}{2}}$ by $1 - \sin^2{\frac{\gamma}{2}}$. Indeed,

\begin{aligned} 1-\sin^2{\frac{\gamma}{2}} &= \frac{s(s-c)}{ab} \\ \sin^2{\frac{\gamma}{2}} &= 1-\frac{s(s-c)}{ab} \\ &= \frac{\left(ab-s(s-c)\right)}{ab} \\ &= \frac{\left((y+z)(x+z)-(x+y+z)(z)\right)}{ab}\\ &= \frac{xy}{ab} \\ &=\frac{(s-a)(s-b)}{ab} \end{aligned}

$\square$

A proof of Heron's Formula
Making use of the formulae proven above and the double angle identity for sine we have

$$\sin{\gamma}=2\sqrt{\frac{s(s-c)}{ab}}\sqrt{\frac{(s-a)(s-b)}{ab}}=2\frac{\sqrt{s(s-a)(s-b)(s-c)}}{ab}$$

Since $\Delta=\frac{ab\sin{\gamma}}{2}$, it follows

$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

$\square$

A proof of the Law of Cosines
Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=s(s-c)-(s-a)(s-b)$$
Substituting and multiplying by 4,
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

Proofs for some trigonometric identities associated to a triangle
a) $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$.

As a consequence of the formulae proven at the beginning of the note,

$$\tan{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}, \quad\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\quad and \quad \tan{\frac{\gamma}{2}}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

So, by canceling and simplifying you get

\begin{aligned}\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}} &=\frac{s-c}{s}+\frac{s-b}{s}+\frac{s-a}{s}\\ &=\frac{z+y+x}{s}\\ &=\frac{s}{s}=1\end{aligned}

$\square$

b) $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$.

We make use of the well-known relationship $abc=4R\Delta$ (see here for a proof) and Heron's Formula.

\begin{aligned}r&=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\\&=4R\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{16R^2\Delta^2}}\\&=\frac{\Delta}{s}\\&=\frac{rs}{s}\\&=r\end{aligned}

$\square$

c) $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$.

\begin{aligned}s&=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-b)}{ac}}\sqrt{\frac{s(s-c)}{ab}}\\&=4R\sqrt{\frac{s^2\Delta^2}{a^2b^2c^2}}\\&=4R\frac{s\Delta}{abc}\\&=4R\frac{s\Delta}{4R\Delta}\\&=s\end{aligned}

$\square$

Consequently, the following relationship also holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{r}{s}$$

or

$$\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\frac{s}{r}$$

d) $\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}$.

To prove the above identity we will show that the right hand side equals $\frac{s}{r}$.

\begin{aligned}\frac{s}{r}&= \cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}\\&=\frac{\sqrt{s(s-a)}}{\sqrt{(s-b)(s-c)}}+\frac{\sqrt{s(s-b)}}{\sqrt{(s-a)(s-c)}}+\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\\&=\frac{\Delta(s-a)+\Delta(s-b)+\Delta(s-c)}{(s-a)(s-b)(s-c)}\\&=\frac{\Delta(x+y+z)}{xyz}\\&=\Delta\frac{s^2}{\Delta^2}\\&=\frac{s^2}{rs}\\&=\frac{s}{r}\end{aligned}

$\square$

We invite the reader to prove the following identity (possibly new) on their own.
$$\frac{\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}+\cot{\frac{\alpha}{2}}\cot{\frac{\gamma}{2}}+\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}}{(s-a)(s-b)+(s-a)(s-c)+(s-b)(s-c)}=\frac{1}{r^2}$$

Two similar formulae (possibly new) in a cyclic quadrilateral
Similar formulae can be derived from a cyclic quadrilateral. Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\frac{a+b+c+d}{2}$. If $\angle{BAD}=\alpha$, then

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc}$$

First we will find an expression for $\cos{\alpha}$ in terms of $a$, $b$, $c$ and $d$. Let $\angle{BCD}=\gamma$. By the Law of Cosines and keeping in mind that $\alpha$ and $\gamma$ are supplementary, we have

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{(180^\circ-\alpha)}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$. Now, making use of the half angle formula for cosine,

\begin{aligned} \cos^2{\frac{\alpha}{2}}&=\frac{2ad+2bc+a^2+d^2-b^2-c^2}{4(ad+bc)}\\ &=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\\&=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\\&=\frac{1}{ad+bc}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)\\&=\frac{(s-b)(s-c)}{ad+bc}\end{aligned}

$\square$

A similar procedure gives us the formula $\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}$.

Remark. Notice that we can get the analogous formulae for a triangle by making $b$ or $c$ equal zero.

No explicit formulae are given in Durell, C. V.; Robson, A. (2003) [1930], Advanced Trigonometry for $\sin^2{\frac{\alpha}{2}}$ and $\cos^2{\frac{\alpha}{2}}$ in terms of the semiperimeter and sides lengths of a cyclic quadrilateral. Wikipedia neither provides such formulae.

A Proof of the Brahmagupta's formula
Denote $\Delta_c$ the area of the cyclic quadrilateral, $ABCD$. Then

\begin{aligned}\Delta_c&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\\&=\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-b)(s-c)}{ad+bc}}(ad+bc)\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\end{aligned}

$\square$

## sábado, 13 de junio de 2020

### Proof Without Words for the Addition Formula for Sine and Subtraction Formula for Cosine

I just hope this is new. What better gift than that on my birthday? :)

$$\sin{(\alpha+\beta)}=\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}\qquad\left(\alpha+\beta<\frac{\pi}{2}\right)$$

$$\cos{(\alpha-\beta)}=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\qquad\left(\alpha-\beta<\frac{\pi}{2}\right)$$

Related material.

### Geometric Proof of the Sum Angle Formula (Sine, Cosine)

I wouldn't say the simpler one but at least $\sin(α+β)$ can be obtained almost without words.

$\frac{\cos{\alpha}}{\cos{\beta}}-\frac{\sin{\beta}(\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta})}{\cos{\beta}}=\frac{\cos{\alpha}-\sin^2{\beta}\cos{\alpha}-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=\frac{\cos{\alpha}(1-\sin^2{\beta})-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=$

$\frac{\cos{\alpha}\cos^2{\beta}-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=\cos{\alpha}\cos{\beta}-\sin{\beta}\sin{\alpha}=\cos{(\alpha+\beta)}.$

## viernes, 5 de junio de 2020

### Yet Another Proof of the Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using standard notation, the law of cosines states

$$c^2=a^2+b^2-2ab\cos{\gamma},$$

where $\gamma$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$. For the same figure, the other two relations are analogous:

$$a^2=b^2+c^2-2ac\cos{\alpha},$$
$$b^2=a^2+c^2-2ac\cos{\beta}.$$

Proof. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$. We start from two well-known relationships of a triangle: $$\sin^2{\frac{\gamma}{2}}=\frac{(s-a)(s-b)}{ab} \qquad\text{and}\qquad \cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}$$
(See Cut-the-knot's Relations between various elements of a triangle for proofs), where $s$ denotes the semiperimeter of $\triangle{ABC}$. Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=sz-xy$$
Substituting and multiplying by 4,
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

AcknowledgementMy sincerest thanks to Angina Seng for giving helpful comments which allowed me to simplify the proof.

Related material.

## miércoles, 20 de mayo de 2020

### Another Simple Proof of Johnson's Theorem

Other proofs can be found in cut-the-knot.org. See also Johnson's Three Circles Theorem Revisited and Johnson's theorem proof.

Johnson's theorem: Let three equal circles with centers $J_a$, $J_b$, and $J_c$ intersect in a single point $H$ and intersect pairwise in the points $A$, $B$, and $C$. Then the circumcircle of the triangle $\triangle{ABC}$ is congruent to the original three.

Proof. Since $\odot{ABH}$ and $\odot{BCH}$ are congruent and as $\angle{BAH}$ and $\angle{BCH}$ are angles subtended by the same arc, it follows that $\angle{BAH}=\angle{BCH}$. Analogously, $\angle{CAH}=\angle{CBH}$. Let $O$, $J_a$ be the centers of $\odot{ABC}$ and $\odot{BCH}$, respectively, then

$$\angle{COB}=2\angle{CAB}=2\angle{CAH}+2\angle{BAH}=2\angle{CBH}+2\angle{BCH}=\angle{BJ_aC}.$$

It follows that $\triangle{BCO}$ and $\triangle{BCJ_a}$ are similar isosceles triangles sharing a common side, $BC$, so by $ASA$ we deduce that $\triangle{BCO}\cong{\triangle{BCJ_a}}$. Consequently, $\odot{ABC}$ is congruent to the original three.

$\square$

Note: The point $H$ may cross the side lines of the triangle $\triangle{ABC}$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.

## sábado, 9 de mayo de 2020

### The Pythagorean Theorem by Reductio ad Absurdum

Other proofs by contradiction can be found in cut-the-knot.org (see proof #122) and in Loomis' collection (see proofs # 16/2 and # 32).

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$, respectively. Also, let $AE=AF=x$; $BD=BF=y$; $CD=CE=r$, where $r$ is the inradius of $\triangle{ABC}$.

Assume to the contrary that $a^2+b^2>c^2$. Then,

$$(r+y)^2+(x+r)^2>(x+y)^2.$$
Expanding, collecting like terms and simplifying we get
$$ry+rx+r^2>xy.$$

Notice that $ry+rx+r^2=r(y+x+r)=rs=\Delta$, where $\Delta$ denotes the area of $\triangle{ABC}$ and $s$ its semiperimeter. Moreover, it is well-known that $xy=\Delta$. So $ry+rx+r^2>xy$ is equivalent to write $\Delta>\Delta$, which is a contradiction. A similar situation arise if you assume $a^2+b^2<c^2$.

There is a simple direct proof starting from $\Delta=rs$ and $\Delta=xy$, can you find it? My sincerest thanks to Andrius Navas and José Hernández for pointing me out this.

## jueves, 7 de mayo de 2020

### Yet Another Proof of the Pythagorean Theorem - Simpler version

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here I give another proof which I hope to be new.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Below I will be using standard notations: $a$, $b$, $c$, for the side length, $s$ for the semiperimeter, $\Delta$ for the area and $r$ for the inradius.

$$s=c+r$$
$$s-\frac{c}{2}=\frac{c}{2}+r$$
$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$
$$a+b=c+2r$$

Squaring both sides,

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

It is well-known $\Delta=rs$, which implies $2ab=4r(c+r)$. So the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

Below is the comment by John Molokach, whom I consider an expert on this topic (he has proved the theorem in more than ten different ways).