problem 7 of Indian Mathematical Olympiad 1989

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problem 7 of Indian Mathematical Olympiad 1989

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perpendicular bisector

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Source: Plane Geometry (intersection of two circles)

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makar

#1 h Sep 1, 2009, 3:40 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Let $A$ be one of the two points of intersection of two circles with centers $X, Y$ respectively.The tangents at $A$ to the two circles meet the circles again at $B, C$ . Let a point $P$ be located so that $PXAY$ is a parallelogram. Show that $P$ is also the circumcenter of triangle $ABC$ .

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599 posts

#2 h Sep 1, 2009, 5:59 PM • 1 Y

Y by Adventure10

Excelent problem.
Please check my solution because i am not sure about it.
Consider the circumcircle of $ABC$ with circumcenter $O$ .Let $Q$ the second intersection point of $(X,Y)$ .In order to prove that $P$ is the center of the circle $ABC$ we have to prove that: $AB,AC$ is perpendicular to $PX$ and $PY$ respectively.(because AB,AC,AQ are the radical axes of $(X,O),(Y,O),(X,Y)$ ,so we will have that $O \equiv P$ .
$AQ$ is perpendicular at $XY$ ,so we only need to prove that $\angle BAQ(=x)=\angle YXP(=y)$ .
But $y=\angle XYA$ (because of the parallelogram).So it suffices to prove that $\angle XYA=\angle BAQ$ which is true because these angles have their lines perpendicular.
So we prove that $\angle BAQ=\angle YXP$
Analogously we can prove that $\angle CAQ=\angle PYX$ .

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#3 h Sep 2, 2009, 1:29 AM • 4 Y

Y by PME2018, Adventure10, Mango247, and 1 other user

$PY//AX, AC\perp AX$ then $PY\perp AC \Rightarrow P$ lies on perpendicular bisector of $[AC]$ , similar for $AB$ . We are done!

Happy the Independence Day of Vietnam!

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#4 h Dec 16, 2013, 10:34 AM • 1 Y

Y by Adventure10

Let $AB \cap XP=L,AC \cap PY=M$ , $\angle BAY=90$ so $\angle L=90$ ,hence $XP$ is the perpendicular bisector of $AB$ ,similarly $YP$ is also perpendicular bisector of $AC$ .hence done.

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lifeismathematics

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lifeismathematics

#5 h Nov 6, 2022, 11:14 AM

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400th post!

This post has been edited 4 times. Last edited by lifeismathematics, Nov 6, 2022, 11:20 AM

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