## domingo, 17 de octubre de 2021

### PAGMO 2021 - Problemas 2, 6

Estas son mis soluciones a los problemas $2$ y $6$ de la Olimpiada Panamericana Femenil de Matemáticas.

Problema 2. Considere el triángulo rectángulo isósceles $\triangle{ABC}$ con $\angle{BAC}=90^\circ$. Sea $l$ la recta que pasa por $B$ y el punto medio del lado $AC$. Sea $\tau$ la circunferencia con diámetro $AB$. La recta $l$ y la circunferencia $\tau$ se intersecan en el punto $P$, diferente de $B$. Muestre que la circunferencia que pasa por los puntos $A$, $C$ y $P$ es tangente a la recta $BC$ en $C$.

Demostración. Considera una línea, $m$, paralela a $AB$ que pase por $C$. Denota con $D$ la intersección de $l$ y $m$. Por el Teorema de Tales, $\angle{APD}=90^\circ$. Como $\angle{ACD}$ es recto, tenemos que $APCD$ es cíclico. Llamemos $M$ al punto medio de $AC$. Claramente, $\triangle{ABM}\cong{\triangle{CDM}}$ (por el criterio $ALA$) con $CD=AB=AC$, por lo que $\triangle{ACD}$ es también un triangulo rectángulo isósceles con $\angle{ADC}=\angle{ACB}$. Por la conversa del Teorema del Segmento Alterno, la circunferencia que pasa por los puntos $A$, $P$ y $C$ debe ser tangente a $BC$ en $C$.

Problema 6. Sea $ABC$ un triángulo con incentro $I$ y sea $\tau$ el excírculo opuesto al vértice $A$. Suponga que $\tau$ es tangente a las rectas $BC$, $AC$ y $AB$ en los puntos $A_1$, $B_1$ y $C_1$, respectivamente. Suponga que las rectas $IA_1$, $IB_1$ e $IC_1$ intersectan nuevamente a $\tau$ en los puntos $A_2$, $B_2$ y $C_2$, respectivamente. Sea $M$ el punto medio del segmento $AA_1$. Si las rectas $A_1B_1$ y $A_2B_2$ se intersectan en $X$ y las rectas $A_1C_1$ y $A_2C_2$ se intersectan en $Y$, demuestre que $MX=MY$.

Demostración. Claramente, $\angle{ABI}=\angle{BC_1A_1}$, por lo que  $BI\parallel{A_1C_1}$. Por propiedad de ángulos entre paralelas, $\angle{BIC_1}=\angle{IC_1A_1}$ y por propiedad del ángulo semiinscrito, $\angle{BC_1I}=\angle{C_1A_1C_2}$. Esto implica que $\triangle{IBC_1}\sim{\triangle{A_1C_1C_2}}$. Luego sigue que $\angle{IC_2A_1}=180^\circ-\angle{C_1C_2A_1}=180^\circ-\angle{C_1BI}=\angle{ABI}=\angle{A_1BI}=\frac{\angle{ABC}}{2}$. De donde se deduce que $IBC_2A_1$ es cíclico. Denotemos con $N$ la segunda intersección de $BC_2$ con $\tau$. Por el Teorema de Reim, $BI\parallel{A_2N}$. Note que $A_1NC_1C_2$ es un cuadrilátero armónico, por lo que si proyectamos desde $A_2$ sobre la recta $A_1C_1$, también paralela a $A_2N$ (recuerde que $BI\parallel{A_1C_1}$), tenemos

$$-1=(N, C_2; C_1, A_1)\stackrel{A_2}{=}(P_\infty, Y; C_1, A_1).$$

Esto implica que $Y$ es el punto medio de $A_1C_1$. Análogamente, $X$ es el punto medio de $A_1B_1$.  Por el Teorema del Segmento Medio, $MY=\frac{AC_1}{2}$ y $MX=\frac{AB_1}{2}$. Pero $AC_1=AB_1$ por ser tangentes comunes, por lo tanto $MY=MX$.

## lunes, 14 de junio de 2021

### Generalized half-angle formulas - Hyperbolic version

In Mathoverflow I wonder about the possibility of finding non-Euclidean versions of the generalized half-angle formulas $[1]$. The Russian mathematician Alexander Mednykh has kindly answered my question. Specifically, he has derived the hyperbolic version. Look at the following link:

Note: Generalized half-angle formula - Hyperbolic version

Reference

$[1]$ E. A. José García, Two Identities and their Consequences, MATINF, 6 (2020) 5-11.

## martes, 1 de junio de 2021

### Using the half-angle formula for cosine to derive Zelich's lemma on mixtilinear incircles

Lemma (Ivan Zelich). Let $w$ be the $A$-mixtilinear incircle in $\triangle{ABC}$ touching side $AB$ at $E$, side $AC$ at $F$. Then

$$AE=AF=\frac{bc}{s}\tag{1}$$

where $s$ is the semiperimeter.

Proof 1. The radius of $w$ inscribed in $\angle{CAB}=\alpha$ is given by

$$\rho_a=r\sec^2{\frac{\alpha}{2}}\tag{2}$$

where $r$ is the inradius of the reference triangle and $\rho_a$ is the radius of $w$ (Durell and Robson 1935).

A proof of $(2)$ can be found here (see pp. 13).

The half-angle formula for cosine states that

$$\cos^2{\frac{\alpha}{2}}=\frac{s(s-a)}{bc}\tag{3}$$

See here for a proof of $(3)$.

Call $I$ the Incenter of $\triangle{ABC}$ and $D$ the touchpoint between the incircle and $AC$. Denote $K$ the center of $w$. Notice that $\triangle{AID}\sim\triangle{AKF}$. Now, by similarity of triangles we have

$$\frac{r}{s-a}=\frac{\rho_a}{AF}\tag{4}$$

Combining $(2)$ and $(3)$ in $(4)$ we get $(1)$.

$\square$
Proof 2. We can also derive $(1)$ using the relationships $\Delta=rs$ and $\Delta=\frac{bc\sin{\alpha}}{2}$. Indeed, since

$$r=\frac{\Delta}{s}=\frac{\frac{bc\sin{\alpha}}{2}}{s}=\frac{bc\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}{s}.$$

Substituting in $(2)$ and simplifying we get

$$\rho_a=r\sec^2{\frac{\alpha}{2}}=\frac{bc}{s}\tan{\frac{\alpha}{2}}=\frac{bc\rho_a}{s\cdot{AF}}$$

from which the result holds.

$\square$

A proof using inversion can be found here

Related material

## viernes, 7 de mayo de 2021

### Using the half-angle formulas to derive Mahavira's identities

In a cyclic quadrilateral $ABCD$, let $a$, $b$, $c$, $d$ denote the lengths of sides $AB$, $BC$, $CD$, $DA$, and $m$, $n$ the lengths of the diagonals $BD$ and $BC$. Then Mahavira's result is expressed as
$$m^2=\frac{(ab+cd)(ac+bd)}{ad+bc}\tag{1}$$
$$n^2=\frac{(ac+bd)(ad+bc)}{ab+cd}\tag{2}.$$

Proof. By the Law of Cosines,

\begin{align*}m^2&=a^2+d^2-2ad\cos{A}\\&=a^2+d^2-2ad(2\cos^2{\frac{A}{2}}-1)\\&=(a+d)^2-4ad\cos^2{\frac{A}{2}}\end{align*}

Substituting from the half-angle formula (see formula $(5)$ in this page) we get

\begin{align*}m^2&=(a+d)^2-\frac{ad[(a+d)^2-(b-c)^2]}{ad+bc}\\&=\frac{bc(a+d)^2+ad(b-c)^2}{ad+bc}\\&=\frac{a^2bc+bcd^2+ab^2d+ac^2d}{ad+bc}\\&=\frac{(ab+cd)(ac+bd)}{ad+bc}.\end{align*}
$\square$

Similarly we can get $(2)$.

Related material

## martes, 30 de marzo de 2021

### Length of angle bisector: yet another application of the half-angle formulas

Let $\triangle{ABC}$ be a triangle. Let $AD$ be the angle bisector of $\angle{BAC}$ in $\triangle{ABC}$. Let $d$ be the length of $AD$. Then $d$ is given by:

$$d^2=\frac{bc}{(b+c)^2}\left[(b+c)^2-a^2\right]\tag{1}$$

where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

Lemma 1 (Tehebycheff). Let $\triangle{ABC}$ be a triangle. Let $AD$ be the angle bisector of $\angle{BAC}$ in $\triangle{ABC}$. Let $d$ be the length of $AD$ and $\angle{BAD}=\angle{CAD}=\frac{\alpha}{2}$. Then $d$ is given by:

$$d=\frac{2bc\cos{\frac{\alpha}{2}}}{b+c}\tag{2}$$

Proof. Let $BD=m$ and $CD=n$. We make use of the Angle Bisector Theorem and the Law of Cosines. Indeed, by the Angle Bisector Theorem we have

$$\frac{b}{c}=\frac{n}{m}$$

which can also be written as

$$\frac{b^2}{c^2}=\frac{n^2}{m^2}\tag{3}$$

Now, by the Law of Cosines,

$$n^2=b^2+d^2-2bd\cos{\frac{\alpha}{2}}\tag{4}$$

$$m^2=c^2+d^2-2cd\cos{\frac{\alpha}{2}}\tag{5}$$

Dividing $(4)$ by $(5)$ and substituting $\frac{n^2}{m^2}$ by $\frac{b^2}{c^2}$ we have

$$\frac{b^2}{c^2}=\frac{b^2+d^2-2bd\cos{\frac{\alpha}{2}}}{c^2+d^2-2cd\cos{\frac{\alpha}{2}}}$$

From which we get

$$d^2(b^2-c^2)=2bcd\cos{\frac{\alpha}{2}}(b-c)$$

Isolating $d$, factoring and simplifying we obtain $(2)$.

$\square$

Another proof of Lemma 1 can be found here.

Lemma 2. Let $\triangle{ABC}$ be a triangle and let $a$, $b$ and $c$ be the lengths of sides $BC$, $AC$ and $BC$, respectively. Then the following identity holds

$$2\cos{\frac{\alpha}{2}}=\sqrt{\frac{(b+c)^2-a^2}{bc}}\tag{6}$$

A proof of Lemma 2 can be found here.

$\square$

Main proof

Combining $(2)$ and $(6)$ we obtain

$$d=\frac{bc}{(b+c)}\cdot{\sqrt{\frac{(b+c)^2-a^2}{bc}}}$$

Raising both sides to the power of 2 we get $(1)$.

$\square$

Related material

Proving the length of angle bisector

## sábado, 5 de diciembre de 2020

### The Pythagorean Theorem from the Half-Angle Formula for Cosine

I've been trying to rescue the Half-Angle Formulas from oblivion and give them the status they deserve by showing its many applications and equivalence with other fundamental theorems (i.e., the Law of Cosines or the Pythagorean Theorem). In this page the Pythagorean Theorem is presented as a mere corollary of the aforementioned formulas.

Let $\triangle{ABC}$ be a right-triangle with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Also, let $s=\frac{a+b+c}{2}$. The proof is based on the formula

$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}\tag{1}$$

This is true for any triangle and can be proven independently of the Pythagorean theorem or the Pythagorean Identity. The proof of $(1)$ can be found here.

There has been controversy about whether it is possible to prove the Pythagorean Theorem using trigonometry. Elisha Loomis and many others believed and still believe that no trigonometric proof of the Pythagorean theorem is possible. This belief stems from the assumption that any such proof would rely on the most fundamental of trigonometric identities $\sin^2{\theta} + \cos^2{\theta} = 1$ which is nothing but a reformulation of the Pythagorean theorem proper. However, Jason Zimba showed it is possible to prove the Pythagoren Identity without a recourse to the Pythagorean theorem (see $[1]$). He then has used the Pythagorean Identity to prove the Pythagorean theorem. Another trigonometric proof is given by Luc Gheysens (see here).

Pythagorean Theorem. Given a right-triangle $\triangle{ABC}$ with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Then

$$a^2+b^2=c^2$$

Proof. From the Double-Angle Formula for cosine we know

$$\cos{2\theta}=2\cos^2{\theta}-1\tag{2}$$

The proof of $(2)$ does not rely on the Pythagorean Theorem (see this video). Using the unit circle, the proof of $\cos{90^\circ}=0$ only depends on the definition of Cosine (see here). So applying $(2)$ we can calculate $\cos{45^\circ}=\frac{\sqrt{2}}{2}$ without risk of Petitio Principii. Hence

$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{1}{2}$$

from which we have

\begin{align*}(a+b)^2-c^2&=2ab\\a^2+2ab+b^2-c^2&=2ab\\a^2+b^2-c^2&=0\end{align*}

I have to admit that this proof is not as satisfying as I would have hoped. Discussion of this proof along with geometric proofs (provided by Blue) of the Half-Angle Formulas can be seen here (pay no attention to the votedown, some intolerant people still believe in the impossibility of trigonometric proofs of the Pythagorean Theorem).

References
$[1]$ A. Bogomolny, More Trigonometric Proofs of the Pythagorean Theorem, from Cut-the-Knot.org