1986 IMO Problems/Problem 4


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 $X$.


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]

See Also

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