Let there be a convex quadrilateral $ABCD$. The angle bisector of $\angle A$ meets the angle bisector of $\angle B$, the angle bisector of $\angle D$ at $P, Q$ respectively. The angle bisector of $\angle C$ meets the angle bisector of $\angle D$, the angle bisector of $\angle B$ at $R, S$ respectively. $P, Q, R, S$ are all distinct points. $PR$ and $QS$ meets perpendicularly at point $Z$. Denote $l_A, l_B, l_C, l_D$ as the exterior angle bisectors of $\angle A, \angle B, \angle C, \angle D$. Denote $E = l_A \cap l_B$, $F= l_B \cap l_C$, $G = l_C \cap l_D$, and $H= l_D \cap l_A$. Let $K, L, M, N$ be the midpoints of $FG, GH, HE, EF$ respectively. Prove that the area of quadrilateral $KLMN$ is equal to $ZM \cdot ZK + ZL \cdot ZN$.
Problem
Source: 2018 Korean Mathematical Olympiad Problem 5
Tags: geometry, bisectors, angle bisector, exterior angle
lminsl
11.11.2018 11:40
Proving that $E,G,P,R$ and $F,H,Q,S$ are collinear and using Ptolemy theorem kills the problem
62861
11.11.2018 11:59
First of all note that $P$, $R$, $E$, $G$ are all collinear on the internal bisector of $\angle (\overline{AD}, \overline{BC})$ and likewise $Q$, $S$, $F$, $H$ are collinear on the internal bisector of $\angle (\overline{AB}, \overline{CD})$. These bisectors are perpendicular, so $KLMN$ is a rectangle. Finally note that $\angle KZL + \angle MZN = 180^{\circ}$, so translating $\triangle KLZ \to \triangle NMZ'$ and applying Ptolemy to $ZMZ'N$ gives the desired claim.
falantrng
16.11.2018 13:19
CantonMathGuy wrote: First of all note that $P$, $R$, $E$, $G$ are all collinear on the internal bisector of $\angle (\overline{AD}, \overline{BC})$ and likewise $Q$, $S$, $F$, $H$ are collinear on the internal bisector of $\angle (\overline{AB}, \overline{CD})$. These bisectors are perpendicular, so $KLMN$ is a rectangle. Finally note that $\angle KZL + \angle MZN = 180^{\circ}$, so translating $\triangle KLZ \to \triangle NMZ'$ and applying Ptolemy to $ZMZ'N$ gives the desired claim. I have one question.Where do you know these bisectors are perpendicular? I think your claim is true when $ABCD$ is concyclic.
62861
16.11.2018 13:36
falantrng wrote: Where do you know these bisectors are perpendicular? rkm0959 wrote: $PR$ and $QS$ meets perpendicularly at point $Z$. Indeed, this does imply $ABCD$ is cyclic.