geometry hard olympiad problem

Question

Prove that $\left[BC\right]\:$ cross that $[AP]'s$ midpoint

Answer 1

Let $a=AB'=AC'$, $b=BC'=BA'$ and $c=CA'=CB'$. Let $B'C'$ meet $AA'$ at $X$ and $BC$ at $Y$. There is a unique rectangle $AA'ZW$ such that $W$ lies on $BC$. Let $U$ be the center of this rectangle, so $UA=UA'=UZ=UW$.

We have $$\begin{eqnarray*} \frac{|\triangle AB'C'|}{|\triangle ABC|}&=&\frac{a^2}{(a+b)(a+c)},\\ \frac{|\triangle A'B'C'|}{|\triangle ABC|}&=&1 -\frac{|\triangle AB'C'|}{|\triangle ABC|} -\frac{|\triangle A'BC'|}{|\triangle ABC|} -\frac{|\triangle A'B'C|}{|\triangle ABC|}\\ &=&1-\frac{a^2}{(a+b)(a+c)}-\frac{b^2}{(a+b)(b+c)}-\frac{c^2}{(a+c)(b+c)}\\ &=&\frac{2abc}{(a+b)(a+c)(b+c)}. \end{eqnarray*}$$ Hence $$ \frac{A'X}{AX}=\frac{|\triangle A'B'C'|}{|\triangle AB'C'|} =\frac{2bc}{a(b+c)}. $$ By Ceva's theorem, $$ 1=\frac{CY}{YB}\cdot\frac{BC'}{C'A}\cdot\frac{AB'}{B'C} =\frac{YA'+c}{YA'-b}\cdot\frac{b}{a}\cdot\frac{a}{c}. $$ Solving, $$ YA'=\frac{2bc}{c-b}. $$ Let $\alpha=\angle AA'B$ and $p=AA'$. By the law of cosines, $$ a^2+2ab=p^2-2bp\cos\alpha, $$ $$ a^2+2ac=p^2+2cp\cos\alpha. $$ Thus $$ p^2=\frac{(a^2+2ab)c+(a^2+2ac)b}{b+c}=a^2+\frac{4abc}{b+c}, $$ $$ p\cos\alpha=\frac{a(c-b)}{b+c}. $$ Hence $$ A'W=\frac{p}{\cos\alpha}=\frac{a(b+c)+4bc}{c-b}. $$ This gives $$ UY=UA'-YA'=\frac12 A'W-YA'=\frac{a(b+c)}{2(c-b)}. $$ Finally we can compute $$ \frac{AZ}{ZU}\cdot\frac{UY}{YA'}\cdot\frac{A'X}{XA} =2\cdot\frac{a(b+c)}{4bc}\cdot\frac{2bc}{a(b+c)}=1. $$ By Ceva's theorem, $X$, $Y$ and $Z$ are collinear. But $B'$ and $C'$ also lie on the line $XY$. That is, $Z$ is the intersection of $B'C'$ and $A'Z$. Since $A'Z$ is perpendicular to $AA'$, $Z=P$. The midpoint of $AP$ is $U$ which lies on $BC$.