On the sides $AB{}$ and $AC{}$ of the acute-angled triangle $ABC{}$ the points $M{}$ and $N{}$ are chosen such that $MN$ passes through the circumcenter of $ABC.$ Let $P{}$ and $Q{}$ be the midpoints of the segments $CM{}$ and $BN{}.$ Prove that $\angle POQ=\angle BAC.$
Problem
Source: Russian TST 2014, Day 8 P3 (Groups A & B)
Tags: geometry, angles
NO_SQUARES
08.01.2024 20:49
Complex bash Let $(ABC)$ be a unit circle and $MN$ be a real axis. Since $B,M,A$ are collinear, we have $\frac{a-m}{b-m}=\frac{\frac{1}{a}-m}{\frac{1}{b}-m}$, so $m=\frac{a+b}{ab+1}$. Similarly, $n=\frac{a+c}{ac+1}$. So, $p = (c+m)/2=\frac{abc+a+b+c}{2(ab+1)}$ and $n = \frac{abc+a+b+c}{2(ac+1)}$. We need to check that $\frac{\frac{b-a}{c-a}}{\frac{q-0}{p-0}} \in \mathbb{R}$ which is $\frac{(b-a)(ac+1)}{(c-a)(ab+1)}$ and its conjugate is also $\frac{(b-a)(ac+1)}{(c-a)(ab+1)}$, so we are done!
math_comb01
08.01.2024 21:47
Angle BASH Chase! Introduce $F$ and $G$ to be the midpoints of $BM$ and $CN$. Claim: $\triangle OPQ \sim \triangle ANM $
$\measuredangle OQP = \measuredangle OQF+\measuredangle FQP = \measuredangle OQF + \measuredangle MQP - \measuredangle MQF =^{by parallel lines} \measuredangle AMN$, other follows similarly
Hence we're done.
LoloChen
10.01.2024 07:46
This is 2011 China Southeast P4.