Macedonian JBMO TST 2014, Problem 2

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Macedonian JBMO TST 2014, Problem 2

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#1 h Mar 30, 2015, 5:32 PM • 1 Y

Y by Adventure10

Point $M$ is an arbitrary point in the plane and let points $G$ and $H$ be the intersection points of the tangents from point M and the circle $k$ . Let $O$ be the center of the circle $k$ and let $K$ be the orthocenter of the triangle $MGH$ . Prove that ${\angle}GMH={\angle}OGK$ .

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#2 h Mar 30, 2015, 6:31 PM • 2 Y

Y by Adventure10, Mango247

Because $MG$ and $MH$ are tangents to $k$ we have that $OG$ is perpendicular to $MG$ and $OH$ is perpendicular to $MH$ so $MGOH$ is cyclic. Hence ${\angle}MGH = {\angle}MOH$ and ${\angle}MOH = {\angle}MOG$ because of symmetry. Also $GK$ is perpendicular to $MH$ , and because $MO$ is perpendicular to $GH$ , and $OH$ to is perpendicualr to $MH$ , we have that ${\angle}OKG = {\angle}KOH = {\angle}MOH = {\angle}MHG$ , hence ${\angle}OGK = {\angle}GMH$ .

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#3 h Mar 30, 2015, 7:03 PM • 1 Y

Y by Adventure10

$OG\bot GM, GK\bot MH\implies\angle OGK=\angle GMH$ , as having perpendicluar sides.

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