Problem of the Week
Hilton Memorial Lecture
Summer Challenge (due August 31)
Fix a circle $\Gamma$. Let $T=\triangle ABC$ be a triangle inscribed in the circle $\Gamma$ which is not a right triangle and let $H$ be the orthocenter of $T$. The line $HA$ intersects the circle $\Gamma$ at $A$ and at a second point $A_1$ (which can be $A$ if the line is tangent to the circle). Likewise, the line $HB$ intersects $\Gamma$ at a second point $B_1$, and the line $HC$ intersects $\Gamma$ at a second point $C_1$. The triangle $\triangle A_1B_1C_1$ is again inscribed in $\Gamma$. We denote this triangle by $\Phi(T)$. Warning: $\Phi(T)$ can be a right triangle.
a) Show that triangles $T$ and $\Phi(T)$ are congruent if and only if either $T$ is equilateral or the angles of $T$ are $\pi/7$, $2\pi/7$, $4\pi/7$.
b) For every integer $k>0$ find the number $t_k$ of non-congruent triangles $T$ inscribed in $\Gamma$ such that $\Phi^k(T)$ and $T$ are congruent. Here $\Phi^k$ denotes the composition $\Phi\circ \Phi \circ \ldots \circ\Phi$ of $\Phi$ with itself $k$ times. Thus, according to a), we have $t_1=2$.
c) Is it true that if $\Phi^k(T)$ and $T$ are congruent then $\Phi^m(T)=T$ for some $m$?
d) Find and prove your own results about $\Phi$.