Let $A,B,C$ be the midpoints of the three sides $B'C', C'A', A'B'$ of the triangle $A'B'C'$ respectively. Let $P$ be a point inside $\Delta ABC$, and $AP,BP,CP$ intersect with $BC, CA, AB$ at $P_a,P_b,P_c$, respectively. Lines $P_aP_b, P_aP_c$ intersect with $B'C'$ at $R_b, R_c$ respectively, lines $P_bP_c, P_bP_a$ intersect with $C'A'$ at $S_c, S_a$ respectively. and lines $P_cP_a, P_cP_b$ intersect with $A'B'$ at $T_a, T_b$, respectively. Given that $S_c,S_a, T_a, T_b$ are all on a circle centered at $O$. Show that $OR_b=OR_c$.
Problem
Source: 2018 Taiwan TST Round 2
Tags: geometry, Taiwan
math_pi_rate
11.06.2020 22:30
Troll problem! Here's the solution: We show that $A$ is also the midpoint of $R_bR_c$, and similar results hold for $B,C$. This would suffice since then the perpendicular bisectors of $R_bR_c,S_cS_a$ and $T_aT_b$ are always concurrent at the circumcenter of $\triangle A'B'C'$ (regardless of whether $S_bS_cT_aT_b$ is cyclic or not). To show this result, let $P_bP_c \cap BC=K$. Then $$-1=(B,C;K,P_a) \stackrel{A}{=} (P_c,P_b;AK \cap P_bP_c,P_a) \stackrel{P_a}{=} (R_c,R_b;A,\infty_{BC})$$and so $A$ is the midpoint of $R_bR_c$, as desired. $\blacksquare$
HamstPan38825
29.06.2022 06:00
The key claim is that $A$ is the midpoint of $\overline{R_aR_c}$, and similar properties hold for $B, C$. There are many proofs of this: we present a simple length chasing one here. Notice that $$\frac{R_bA}{CP_a} \cdot \frac{BP_a}{R_cA} = \frac{AP_b}{P_bC} \cdot \frac{BP_c}{P_cA} = \frac{BP_a}{CP_a}$$by Ceva on $\triangle ABC$, so $R_bA = R_cA$; similar results hold for $B$ and $C$. As a result, $\overline{OC} \perp \overline{A'B'}$ and $\overline{OB} \perp \overline{A'C'}$. It follows $O$ is the circumcenter of $A'B'C'$, so $\overline{OA} \perp \overline{B'C'}$ as well, from where the result follows.