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.