Let $\triangle ABC$ be a acute triangle. Let $H$ the foot of the C-altitude in $AB$ such that $AH=3BH$, let $M$ and $N$ the midpoints of $AB$ and $AC$ and let $P$ be a point such that $NP=NC$ and $CP=CB$ and $B$, $P$ are located on different sides of the line $AC$. Prove that $\measuredangle APM=\measuredangle PBA$.
Problem
Source: Netherlands Team Selection Test 2016 Day 1-Problem 1
Tags: geometry
22.09.2016 10:59
23.09.2016 02:39
24.09.2016 15:31
Since $NP=NC=NA$ we have that $\angle APC=90$. Since $\angle APM=\angle PBA$ is equivalent to $\triangle APM \sim \triangle APB$ which is equivalent to $\frac{AP}{AM}=\frac{AB}{AP}$ which is equivalent to $AP^2=AB\cdot AM=\frac{AB^2}{2}$. From $\angle APC=90$ we have that $AP^2=AC^2-CP^2=AC^2-BC^2$. From $AH=3BH$ we have $AC^2=AH^2+HC^2$ $BC^2=BH^2+HC^2$ $\implies$ $AC^2-BC^@=AH^2-BH^2=\frac{AB^2}{2}$ so it's done.
17.03.2020 12:46
FabrizioFelen wrote: Let $\triangle ABC$ be a acute triangle. Let $H$ the foot of the C-altitude in $AB$ such that $AH=3BH$, let $M$ and $N$ the midpoints of $AB$ and $AC$ and let $P$ be a point such that $NP=NC$ and $CP=CB$ and $B$, $P$ are located on different sides of the line $AC$. Prove that $\measuredangle APM=\measuredangle PBA$. İt is also Azerbaijan BMO TST 2017 p1