In triangle $ABC$ with incentre $I$, let $M_A,M_B$ and $M_C$ by the midpoints of $BC, CA$ and $AB$ respectively, and $H_A,H_B$ and $H_C$ be the feet of the altitudes from $A,B$ and $C$ to the respective sides. Denote by $\ell_b$ the line being tangent tot he circumcircle of triangle $ABC$ and passing through $B$, and denote by $\ell_b'$ the reflection of $\ell_b$ in $BI$. Let $P_B$ by the intersection of $M_AM_C$ and $\ell_b$, and let $Q_B$ be the intersection of $H_AH_C$ and $\ell_b'$. Defined $\ell_c,\ell_c',P_C,Q_C$ analogously. If $R$ is the intersection of $P_BQ_B$ and $P_CQ_C$, prove that $RB=RC$.
Problem
Source: 2018HKTST2P4
Tags: geometry, incenter, circumcircle
jlammy
21.10.2017 17:44
Here's the boring solution. First, note that $\ell_b'$ is the line through $B$ parallel to $AC$, and similarly for $\ell_c'$. Now we use barycentric coordinates with respect to $\triangle ABC$. Note that $P_B$ is the midpoint of $BP_B\cap CA=(a^2:0:-c^2)$, so using Conway's notation, $$P_B=(a^2:a^2-c^2:-c^2)=(S_B+S_C:S_C-S_A:-S_A-S_B),$$and likewise $P_C=(S_B+S_C:-S_A-S_B:S_B-S_A)$. Set $Q_B=(S_B:y:-S_B)$; then $$0=\begin{vmatrix}0&S_C&S_B \\ S_B&S_A&0 \\ S_B&y&-S_B\end{vmatrix}=-S_AS_B^2+S_B^2y+S_CS_B^2\implies y=S_A-S_C,$$so $Q_B=(S_B:S_A-S_C:-S_B)$ and analogously $Q_C=(S_C:-S_C:S_A-S_B)$. Now the equation of $P_BQ_B$ is \begin{align*}0&=\begin{vmatrix}S_B+S_C&S_C-S_A&-S_A-S_B \\ S_B&S_A-S_C&-S_B \\ x&y&z\end{vmatrix} \\&= \begin{vmatrix}2S_B+S_C&0&-S_A-2S_B \\ S_B&S_A-S_C&-S_B\\x&y&z\end{vmatrix} \\&= (S_A+2S_B)(S_A-S_C)x - S_B(S_A-S_C)y+(2S_B+S_C)(S_A-S_C)z,\end{align*}so $P_BQ_B$ has equation $$(S_A+2S_B)x-S_By+(2S_B+S_C)z=0,$$and similarly the equation for $P_CQ_C$ is $$(S_A+2S_C)x+(S_B+2S_C)y-S_Cz=0.$$Also, the perpendicular bisector of $\overline{BC}$ has equation $$(S_B-S_C)x+(S_B+S_C)(z-y)=0.$$Now using row/column operations, compute \begin{align*}\begin{vmatrix}S_A+2S_B&-S_B&2S_B+S_C \\ S_A+2S_C&S_B+2S_C&-S_C \\ S_B-S_C&-S_B-S_C&S_B+S_C\end{vmatrix} &= \begin{vmatrix}S_A+2S_B&-S_B&2S_B+S_C \\ S_A+S_B+S_C&S_C&S_B \\ S_B-S_C&-S_B-S_C&S_B+S_C\end{vmatrix} \\&= \begin{vmatrix}S_A+2S_B&-S_B&S_B+S_C \\ S_A+S_B+S_C&S_C&S_B+S_C \\ S_B-S_C&-S_B-S_C&0\end{vmatrix} \\&= \begin{vmatrix}S_A+2S_B&-S_B&S_B+S_C \\ S_A+2S_B&-S_B&S_B+S_C \\ S_B-S_C&-S_B-S_C&0\end{vmatrix} \\&=0. \end{align*}Thus $R=P_BQ_B\cap P_CQ_C$ lies on the perpendicular bisector of $\overline{BC}$, and the result follows.
polya78
23.10.2017 20:21
Let $w$ be the nine-point circle of $\triangle ABC$. We quickly see that $Q_b$ is tangent to $(BH_aH_c)$, so that the power of $Q_b$ with respect to $w$ is $Q_bB^2$. Similarly $P_bB$ is tangent to $(BM_aM_c)$, so the power of $P_b$ with respect to $w$ is $P_bB^2$. Thus if $B'$ is the inverse of $B$ with respect to $w$, $P_bQ_b$ is the perpendicular bisector of $BB'$. So $R$ is the center of $(BB'CC')$, which is clearly on the perpedicular bisector of $BC$.
Mosquitall
23.10.2017 21:41
Simplification of polya78's proof : $P_BQ_B$ is a radical line of $B$ and the nine-point circle, same for $P_CQ_C$, so they meet on the radical line of $B, C$ i.e on the perpendicular bisector to $BC$. Done