AoPS - Problem

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Problem

Source: 2015 Saudi Arabia JBMO TST 1.3

Tags: geometry, circumcircle, right triangle, right angle

parmenides51

21.07.2020 11:14

A right triangle $ABC$ with $\angle C=90^o$ is inscribed in a circle. Suppose that $K$ is the midpoint of the arc $BC$ that does not contain $A$ . Let $N$ be the midpoint of the segment $AC$ , and $M$ be the intersection point of the ray $KN$ and the circle.The tangents to the circle drawn at $A$ and $C$ meet at $E$ . prove that $\angle EMK = 90^o$

Gryphos

21.07.2020 12:53

Let

$L$ be the point diametrically opposite to

$K$ on

$(ABC)$ . We need to prove that

$E,M,L$ are collinear, i.e. that

$ML$ is the

$M$ -symmedian in

$\triangle ACM$ . Since

$AC \perp BC \perp KL$ , we have

$KL \parallel AC$ or

$|AL|=|CK|$ . This means that

$\angle AML = \angle KMC = \angle NMC$ , hence

$ML$ is indeed the symmedian.

dgreenb801

22.07.2020 06:21

Let $O$ be the center of the circle. By symmetry, $E$ , $N$ , and $O$ are collinear. Also, $\angle ECO = \angle EAO = 90$ , so $ECOA$ is cyclic. Then $MN \cdot NK = AN \cdot NC = EN \cdot NO$ , so $EMOK$ is cyclic. Then $\angle EMK = \angle EOK = \angle MOK = 90$ (since $M$ and $K$ are both midpoints of arcs $\widehat{AC}$ and $\widehat{BC}$ , respectively, and arc $\widehat{ACB}=180$ ).