1.Let $ABC$ be a triangle, $D$ the foot of the altitude from $A$ and $M$ the midpoint of the side $BC$. Let $S$ be a point on the closed segment $DM$ and let $P, Q$ the projections of $S$ on the lines $AB$ and $AC$ respectively. Prove that the length of the segment $PQ$ does not exceed one quarter the perimeter of the triangle $ABC$.
Problem
Source: Danube Contest 2016, Romania
Tags: geometric inequality, inequalities
GGPiku
01.11.2017 21:56
Since $APSQ$ is cyclic with diameter $AS$, we get that $PQ=AS*sin(\angle BAC)$. So $PQ$ is maximum when $AS$ is maximum, and that happens when $S$ is the midpoint of $BC$. In this case, let $E,F$ be the midpoints of $AB$,$AC$ and $G,H$ the midpoints of $ME$ and $MF$. Then $PQ\leq PG+GH+HQ=\frac{SE+EF+FS}{2}=\frac{P_{ABC}}{4}$, as desired.
leonardg
01.11.2017 22:09
What if either B or C is obtuse ? Parerea mea !
GGPiku
01.11.2017 22:27
leonardg wrote: What if either B or C is obtuse ? Parerea mea ! I don't know how you call it english, but doesn't matter if the angles are obtuse, the inequality of the line("Inegalitatea liniei frante") still holds. If that's what you meant by your observation, sir.
