Quadrilateral and Bisectors

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Quadrilateral and Bisectors

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Source: 2018 Korean Mathematical Olympiad Problem 5

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#1 h Nov 11, 2018, 8:37 AM • 5 Y

Y by anantmudgal09, Adventure10, Mango247, Mango247, Mango247

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 $ZM \cdot ZK + ZL \cdot ZN$ .

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#2 h Nov 11, 2018, 8:40 AM • 3 Y

Y by rkm0959, Adventure10, Mango247

Proving that $E,G,P,R$ and $F,H,Q,S$ are collinear and using Ptolemy theorem kills the problem

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#3 h Nov 11, 2018, 8:59 AM • 4 Y

Y by rkm0959, aopsuser305, Adventure10, Mango247

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.

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#4 h Nov 16, 2018, 10:19 AM • 2 Y

Y by Adventure10, Mango247

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.

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#5 h Nov 16, 2018, 10:36 AM • 4 Y

Y by falantrng, aopsuser305, Adventure10, Mango247

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.

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