33 Concurso Nacional de la OMM - Problema 2

Sea $H$ el ortocentro de un triángulo acutángulo $ABC$ y $M$ el punto medio de $AH$. La recta $BH$ corta a $AC$ en $D$. Considera un punto $E$ de manera que $BC$ sea mediatriz del segmento $DE$. Los segmentos $CM$ y $AE$ se cortan en $F$. Muestra que $BF$ es perpendicular a $CM$.

Demostración. Claramente $\triangle{ADH}\sim{\triangle{BCD}}$ y $\angle{AHD}=\angle{ACB}$. Como $M$ es el punto medio de la hipotenusa $AH$, resulta que $AM=HM=DM$, implicando que $\triangle{MDH}$ es isósceles con $\angle{MDH}=\angle{MHD}=\angle{ACB}$. Note que $\angle{MDC}=\angle{ADE}=90^\circ+\angle{ACB}$. También 

$$\frac{MD}{AD}=\frac{\frac{AH}{2}}{AD}=\frac{AH}{2AD}.$$

Llamemos $G$ el punto donde $DE$ corta $BC$, entonces, 

$$\frac{DC}{DE}=\frac{DC}{2DG}.$$

Como $\triangle{ADH}\sim{\triangle{DCG}}$, entonces, $\frac{AH}{AD}=\frac{DC}{DG}$, implicando $\frac{MD}{AD}=\frac{DC}{DE}$. Por el criterio de semejanza $LAL$, los triángulos $\triangle{MDC}$ y $\triangle{ADE}$ son semejantes con $\angle{DCM}=\angle{DCF}=\angle{AED}=\angle{FED}$. De lo anterior se deduce que $F$ yace en la circunferencia circunscrita del cuadrilátero $BDCE$ (claramente cíclico), significando que $\angle{BDC}=\angle{BFC}$. Pero $\angle{BDC}=90^\circ$, por lo tanto, $BF\perp{CM}$. 

Conics intersecting the sides of a triangle

Problem 1. Consider a triangle $ABC$ and a parabola, $P_a$, whose focus is $A$ and directrix, $BC$. Call $C_a$, $B_a$ the intersections of $P_a$ with the sides $AB$, $AC$, respectively. Define $A_b$, $C_b$, $A_c$ and $B_c$ cyclically. Prove that $C_a$, $B_a$, $A_b$, $C_b$, $A_c$ and $B_c$ lie on a conic. 

My proof can be found here and a generalization (by Barry Wolk) here.

The external version.

Problem 2. Consider a triangle, $ABC$, and its Incenter, $I$. A perpendicular line to $AI$ in $I$, cut the sides $AB$, in $A_c$, and $AC$, in $A_b$. Define $B_c$, $B_a$, $C_a$ and $C_b$ cyclically. Prove that $A_c$, $A_b$, $B_c$, $B_a$, $C_a$ and $C_b$ lie on a conic.

My proof (in Spanish) can be found here.

Problem 3. Consider a triangle $ABC$ and its $A$-mixtilinear incircle, $\tau_a$. Call $A_b$ the intersection of $\tau_a$ with the side $BC$ closer to $B$. Define $A_c$ similarly. Construct $B_a$, $B_c$, $C_a$ and $C_b$ cyclically. Prove that $A_b$,  $A_c$, $B_a$, $B_c$, $C_a$ and $C_b$ lie on a conic. 

A proof by Ivan Zelich can be found here.

Problem 4. Let $ABC$ be a triangle and $DEF$ its orthic triangle. Construct a parabola, $P_a$, being $F$ and line $DE$ its focus and directrix, respectively. Prove that this parabola is tagential to sides $AB$, $AC$ and to the altitudes $BD$, $CE$. 

My proof can be found here.

Problem 4-a. Consider the parabola, $P_a$, described in problem 4. Let $A_b$ be the intersection of $P_a$ with the side $BC$ closer to $B$. Define $A_c$ similarly. Construct $B_c$, $B_a$, $C_a$ and $C_b$ cyclically. Prove that  $A_b$, $A_c$, $B_c$, $B_a$, $C_a$ and $C_b$ lie on a conic. 

A proof by Ivan Zelich can be found here.

Problem 4-b. Consider again the parabola, $P_a$ described in problem 4. Call $A'_b$ and $A'_c$ the points of tangency of $P_a$ with the sides $AB$ and $AC$, respectively. Construct $B'_c$, $B'_a$, $C'_a$ and $C'_b$ cyclically. Prove that $A'_b$, $A'_c$, $B'_c$, $B'_a$, $C'_a$ and $C'_b$ lie on a conic.  

Problem 5. Let $O_a$, $O_b$ and $O_c$ be the centers of three congruent circles. Let the line $AB$ be a common tangent line to the circles $O_a$ and $O_b$ farther from $O_c$. Construct lines $BC$ and $AC$ similarly. Let $CO_a$ meet $AB$ in $C_a$. Similarly construct $C_b$. Define $A_b$, $A_c$, $B_c$ and $B_a$ cyclically. Prove that $C_a$, $C_b$, $A_b$, $A_c$, $B_c$ and $B_a$ lie on a circle. (Not proven yet.)

Problem 6. Consider two points, $P$ and $Q$, in the interior of a triangle, $ABC$. Let $\triangle{P_aP_bP_c}$ and $\triangle{Q_aQ_bQ_c}$ be the cevian triangles of $P$ and $Q$, respectively. Construct outwardly semicircles with diameters $BP_a$ and $CQ_a$. Let $A_1$ be the second intersection of semicircle $(BP_a)$ with the circumcircle of $ABC$. Define $A_2$ similarly. Denote $A_b$ the intersection of $AA_1$ and $BC$. Define $A_c$ similarly. Construct $B_c$, $B_a$, $C_a$ and $C_b$ cyclically. Prove that $A_b$, $A_c$, $B_c$, $B_a$, $C_a$ and $C_b$ lie on a conic. (Not proven yet.)

Related material.
Carnot's theorem (conics)
Conics Related To In- and Excircles

Solución alternativa a un problema de admisión

Considera un cuadrado $ABCD$. $E$ y $F$ son puntos en los lados $DC$ y $AD$, respectivamente. Si $\angle{EFB}=90^\circ$, $BF=4$, $EF=3$ y $BE=5$, determina la longitud del segmento $CE=x$.

Solución. Note que $\angle{EFB}+\angle{ECB}=180^\circ$, por lo que el cuadrilátero $BFEC$ es cíclico. Por propiedad de ángulos inscritos en una misma circunferencia, los ángulos $\angle{FCE}$ y $\angle{FBE}$ son congruentes (intersecan un mismo arco). Pero $\angle{FCE}=\angle{FCD}$, implicando que $\triangle{BFE}\sim\triangle{CDF}$, de donde resulta la siguiente proporción:

$$\frac{CF}{5}=\frac{a}{4},$$

donde $a$ es la longitud de los lados del cuadrado.

Podemos expresar el segmento $CF$ en términos de $x$ y $a$ por medio del teorema de Ptolomeo, es decir, 

$$3a+4x=5CF$$

De vuelta a nuestra proporción, 

$$\frac{\frac{3a+4x}{5}}{5}=\frac{a}{4}$$

Despejando para $x$, resulta

$$x=\frac{13a}{16}$$

Aplicando Pitágoras en el triángulo $\triangle{BCE}$, 

$$\left(\frac{13a}{16}\right)^2+a^2=25$$

Resolviendo para $a$ y descartando valores negativos, obtenemos $a=\frac{16\sqrt{17}}{17}$. Finalmente, $x=\frac{13a}{16}=\frac{13}{16}\cdot{\frac{16\sqrt{17}}{17}}=\frac{13\sqrt{17}}{17}\approx3.15296312...$

Perpendicularity in a right-triangle

This problem was proposed by Anthony Becerra in the Facebook group "Romantics of Geometry".

Consider a right-triangle $ABC$ with $\angle{ABC}=90^\circ$. Let $P$, $Q$ and $R$ be on sides $AB$, $BC$ and $AC$, respectively, such that $BPRQ$ is a square. Call $X$ the intersection of $AQ$ and $CP$. Prove that $BX\perp{AB}$.

Proof. By Ceva's theorem, $\frac{AH}{CH}\cdot{\frac{CQ}{BQ}}\cdot{\frac{BP}{AP}}=1$. By similarity, $\frac{CR}{AC}=\frac{CQ}{BC}$; $\frac{BQ}{BC}=\frac{AR}{AC}$, then, $\frac{CQ}{BQ}=\frac{CR}{AR}$. But, because of the Angle Bisector theorem, $\frac{CR}{AR}=\frac{BC}{AB}$. Now, as $\triangle{APR}\sim\triangle{CQR}$, it follows $\frac{BP}{AP}=\frac{CR}{AR}=\frac{BC}{AB}$. Back to Ceva: $\frac{AH}{CH}\cdot{\frac{BC^2}{AB^2}}=1$, which implies the perpendicularity by the converse of Euclid's theorem.

Meme de la tarjeta de crédito y la integral - Solución

Uno de mis estudiantes me desafió a resolver la integral que aparece en un meme que se ha hecho viral en las redes sociales. Me refiero a este: 

Omitiré los límites de la integral de momento. Note que la integral puede ser reescrita de la siguiente manera:

$$\int\frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=\int\frac{(x-1)(3x^2+2x+4)}{\sqrt{(x-1)(x-2)}}dx=\int\frac{\sqrt{x-1}(3x^2+2x+4)}{\sqrt{x-2}}dx$$

Si sustituimos $\sqrt{x-1}$ por $u$, entonces, $dx=2udu$ y $x=u^2+1$. Reescribiendo la integral en términos de $u$ y simplificando, 

$$\int\frac{\sqrt{x-1}(3x^2+2x+4)}{\sqrt{x-2}}dx=6\int\frac{u^6}{\sqrt{u^2-1}}du+16\int\frac{u^4}{\sqrt{u^2-1}}du+18\int\frac{u^2}{\sqrt{u^2-1}}du$$

Por sustitución trigonométrica, $u=\sec{\theta}$, $du=\sec{\theta}\tan{\theta}d\theta$ y $\tan{\theta}=\sqrt{u^2-1}$. Reescribiendo nuevamente, 

$$6\int\frac{u^6}{\sqrt{u^2-1}}du+16\int\frac{u^4}{\sqrt{u^2-1}}du+18\int\frac{u^2}{\sqrt{u^2-1}}du$$

$$=6\int\sec^7{\theta}d\theta+16\int\sec^5{\theta}d\theta+18\int\sec^3{\theta}d\theta$$

Haciendo uso de la Fórmula de Reducción de la Secante donde $n\neq{1}$,

$$\int\sec^n{\theta}d\theta=\frac{\sec^{n-2}{\theta}\tan{\theta}}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}{\theta}d\theta$$

y teniendo en cuenta la muy conocida integral $\int\sec{\theta}d\theta=\ln{\lvert\tan{\theta}+\sec{\theta}\rvert}+C$, tenemos:

$$6\int\sec^7{\theta}d\theta+16\int\sec^5{\theta}d\theta+18\int\sec^3{\theta}d\theta$$

$$=\sec^5{\theta}\tan{\theta}+\frac{21}{4}\sec^3{\theta}\tan{\theta}+\frac{135}{8}\sec{\theta}\tan{\theta}+\frac{135}{8}\ln{\lvert\tan{\theta}+\sec{\theta}\rvert}+C$$

Reemplazando $\sec{\theta}$ por $u$ y a $\tan{\theta}$ por $\sqrt{u^2-1}$, 

$$6\int\frac{u^6}{\sqrt{u^2-1}}du+16\int\frac{u^4}{\sqrt{u^2-1}}du+18\int\frac{u^2}{\sqrt{u^2-1}}du$$

$$=u^5\sqrt{u^2-1}+\frac{21}{4}u^3\sqrt{u^2-1}+\frac{135}{8}u\sqrt{u^2-1}+\frac{135}{8}\ln{\lvert\sqrt{u^2-1}+u\rvert}+C$$

Reemplazando $u$ por $\sqrt{x-1}$ y simplificando, 

$$\int\frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=(x-1)^2\sqrt{x^2-3x+2}+\frac{21}{4}(x-1)\sqrt{x^2-3x+2}$$

$$+\frac{135}{8}\sqrt{x^2-3x+2}+\frac{135}{8}\ln{\lvert\sqrt{x-2}+\sqrt{x-1}\rvert}+C$$

Pero 

$$\frac{(135)(2)}{16}\ln{\lvert\sqrt{x-2}+\sqrt{x-1}\rvert}=\frac{135}{16}\ln{\lvert(\sqrt{x-2}+\sqrt{x-1})^2\rvert}$$

$$=\frac{135}{16}\ln{\lvert2\sqrt{x^2-3x+2}+2x-3\rvert}$$

Finalmente, 

$$\int\frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=(x-1)^2\sqrt{x^2-3x+2}+\frac{21}{4}(x-1)\sqrt{x^2-3x+2}$$

$$+\frac{135}{8}\sqrt{x^2-3x+2}+\frac{135}{16}\ln{\lvert2\sqrt{x^2-3x+2}+2x-3\rvert}+C$$

Para $x=1$ la función anterior se reduce a $0$. Así, evaluando,

$$\int_{0}^{1}\frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}=\frac{21\sqrt{2}}{4}-\frac{135\sqrt{2}}{8}-\sqrt{2}-\frac{135\ln{\lvert2\sqrt{2}-3\rvert}}{16}$$ 

$$=-2.981266944005536$$

Como la clave de una tarjeta de crédito solo admite cuatro dígitos (es decir, solo números), omitimos el signo negativo, el punto y  voilà, la clave es 2981.

Un problema propuesto por Ángel Mejía Brito

En el problema se pide encontrar $\angle{AED}$, donde $E$ y $D$ son puntos de tangencias. 

Solución 1. Llamemos a la circunferencia de diámetro $AB$, $\omega$, y al círculo que pasa por $E$ y $D$, $\tau$. Considera la circunferencia de inversión, $\rho$, de radio $\frac{AB}{2}$ y centro en $A$. Denotemos con $X$ e $Y$ los puntos de intersección de $\omega$ y $\rho$. La imagen invertida de $\omega$ es la línea $XY$. La imagen de $\tau$ es otra circunferencia tangente a $AB$, en $D'$ (la imagen de $D$) y a $XY$ en $E'$ (la imagen de $E$). Si llamamos $Z$ al punto donde $XY$ corta a $AB$, al ser $XY$ el eje radical de $\omega$ y $\rho$, las rectas $AB$ y $XY$ son perpendiculares, de donde inferimos que $\triangle{D'ZC'}$ es un triángulo recto isósceles, y por consiguiente $\angle{AD'E'} = 45^\circ$. Como en la inversión los ángulos se preservan, $\angle{AED} = 45^\circ$.

Solución 2. Considera un punto $F$ sobre $\omega$ tal que $AF$ es tangente a $\tau$. Note que $\tau$ es el incírculo mixtilíneo de $\triangle{ABF}$. Si $M$ es el punto medio de $\widehat{AB}$ (que no contiene a $F$), entonces, $E$, $D$ y $M$ están alineados (ver demostración aquí). De lo anterior se deduce que al ser $\angle{AFB}=90^\circ$, por propiedades de ángulos, $\angle{AFM}=\angle{BFM}=\angle{AEM}=\angle{AED}=45^\circ$.

Solución 3. Ver problema 2 aquí para una demostración usando cuaternas armónicas. 

How likely is it that a mathematics student can't solve IMO problems? Is there a fear of embarrassment in being a math Ph.D. who can't solve problems that high-school students can?

Response by Cornelius Goh on Quora:

Math Research and IMO (International Math Olympiad) Math are different 'things'.

A good analogy given by a famous Chinese Math professor, who criticized the "IMO Craze" in China, since 1985 as China wins the World IMO Championship and hundreds of IMO Gold medals for continuously 3 decades :--

Math Research = Martial Art (aka Kungfu 武术功夫);
IMO Math = Acrobatics (杂技, not real kungfu).

Prof S.S. Chern (陈省身 Wolf Prize 1983) ［Note *］and his PhD student Prof S.T. Yau (邱成桐, Fields Medal 1982) were always surrounded by keen IMO Math students for tough IMO questions, to whom the 2 professors just squarely replied "I don't know how to do it!". It was reported that some Chinese IMO Gold medalists entered PhD class at Harvard and failed, because PhD Math research problem has no quick solution with known techniques, it usually takes many years to see result, unlike IMO Math questions with known solution by astute tricks.

Many years ago in a Singapore seminar, I asked Prof. Pierre-Louis Lions (1956 - , Fields Medalist 1994) for his opinion on IMO. He told us when he represented France in the IMO competition, he spent the few days there staring at the ceiling, not knowing how to solve the problems.

Looking at the past winner list of Fields Medalists, not 100％ were IMO Medalists -- except Terence Tao (陶轩哲, Australia), Grigori Perelman (Russia), Timothy Gowers (UK), Ngô Báo Châu (吴宝珠, Vietnam), etc.

Interesting to note that the "IMO Champion" China has yet to produce a single Fields Medalist todate.

In parallel, the "Fields Medal Champion" France (wins 1/3 of the medals so far) has never been the "IMO (team nor individual) Champion".
However, a small country Singapore has produced an individual "IMO Champion 2012" (Lim Jeck 林捷) , the only contestant in the world that year with perfect score.

In conclusion, IMO Math is like acrobatics doing fantastic Math "stunts" (特技), it is impressive but not the real Math which requires deep thinking, perseverance in finding the Universe's truth, creating new mathematical tools (Category 范畴, Quantum group 量子群, Homology 同调, Homotopy 同伦, Sheaf 束, Motif 动力, Fiber Bundle 光纤丛, ...) to explore the vast scientific frontier...

P.S. Fields Medal is the equivalent of Nobel Prize (Math) but tougher to get, only awarded every 4 years and for age below 40. (unlike the yearly Nobel Prize for any living person of any age).

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