1986 IMO Problems/Problem 4

Problem

Let $A,B$ be adjacent vertices of a regular $n$ -gon ( $n\ge5$ ) with center $O$ . A triangle $XYZ$ , which is congruent to and initially coincides with $OAB$ , moves in the plane in such a way that $Y$ and $Z$ each trace out the whole boundary of the polygon, with $X$ remaining inside the polygon. Find the locus of $X$ .

Solution

Let $C \not = A$ the vertex which is adjacent to $B$ . While $XYZ$ moves from $OAB$ to $OBC$ , it is easy to see $XYBZ$ is cyclic. Thus $X$ lies on the bisector of $\angle YBZ = \angle ABC$ . Moreover, $X$ is the intersection of a circle passing through $B$ (the circumcircle of $XYBZ$ ) and with a fixed radius (the radius is a function of $\triangle XYZ$ ). Therefore $X$ varies in a line segment ended in $O$ . When $Y$ and $Z$ pass through the other sides, we get as locus $n$ distinct line segments, each passing throught $O$ and contained in $OV$ (but not in $\vec{OV}$ ) for some vertex $V$ of the polygon. Each two of these lines are obtained one from another by a rotation with center $O$ .

This solution was posted and copyrighted by feliz. The original thread for this problem can be found here: [1]

1986 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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1986 IMO Problems/Problem 4

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