Let $P$ and $Q$ be points inside triangle $ABC$ satisfying $\angle PAC=\angle QAB$ and $\angle PBC=\angle QBA$. a) Prove that feet of perpendiculars from $P$ and $Q$ on the sides of triangle $ABC$ are concyclic. b) Let $D$ and $E$ be feet of perpendiculars from $P$ on the lines $BC$ and $AC$ and $F$ foot of perpendicular from $Q$ on $AB$. Let $M$ be intersection point of $DE$ and $AB$. Prove that $MP\bot CF$.
Problem
Source: Serbia Additional TST 2012, Problem 3
Tags: geometry, circumcircle, radical axis, geometry proposed
Luis González
19.05.2012 18:04
a) See Nine-point circle generalization, isogonal, pedal circle theorem and elsewhere. b) Let $K$ be the orthogonal projection of $P$ on $AB.$ Let the circumcircles of $\triangle PKF$ and $PDCE$ meet again at $N.$ $DE,PN,FK$ are pairwise radical axes of $\odot(PDCE),$ $\odot(PKF)$ and $\odot(DEFK)$ concurring at their radical center $M,$ i.e. $N \in PM.$ $\angle PNC=\angle PEC=90^{\circ},$ i.e. $MP \perp CF.$
crazyfehmy
25.05.2012 01:02
The part b) was the last question in BMO2012 Shortlist.
Swistak
26.05.2012 18:53
Here's solution to b) using vectors. Let $G$ be the foot of the perpendicular to $AB$ from $P$ and $H$ foot of the perpendicular to $AC$ from $Q$. $CQ \perp DE$ since $P$ ang $Q$ are isogonal conjugate. Let me write $XY$ instead of $\vec{XY}$, because writing second ones many times is exhausting . $MP \cdot CF=MP \cdot PF - MP \cdot PC$ $MP \cdot PF =MP \cdot (PQ + QF)=MP \cdot PQ + (MG + GP) \cdot QF=MP \cdot PQ + GP \cdot QF$ $- MP \cdot PC=-(ME+EP) \cdot PC =MY(CQ+QP)-EP \cdot PC=ME \cdot QP - EP \cdot (PE+EC)=ME \cdot QP + PE^2$ $MP \cdot CF=MP \cdot PF - MP \cdot PC=MP \cdot PQ + GP \cdot QF + ME \cdot QP + PE^2=(MP-ME) \cdot PQ+ PE^2+GP \cdot QF = EP(PQ+EP)+GP \cdot QF=EP\cdot EQ-PG \cdot QF= EP \cdot HQ-PG \cdot QF=0$, since $AEPG$ and $AFQH$ are similar.
KOSNITA
16.07.2012 18:26
The thing is to notice that P and Q are isogonal conjugates of each other. Then the proofs are corollaries of the property.