IMO 1986 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 8m44s

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 $X$.

Exploration

A quick sanity check is to look at what the statement is really forcing on $X$.

If $Y$ and $Z$ were fixed on a single side of the polygon, the condition $\angle YXZ=\frac{2\pi}{n}$ would place $X$ on a circular arc through $Y$ and $Z$. As $Y,Z$ slide along a side, that arc sweeps a region bounded by two parallel lines, since the extremal situation occurs when one of $Y,Z$ hits a vertex.

Trying $n=5$ or $n=6$ suggests a strong symmetry: every side of the regular polygon contributes an identical constraint rotated around $O$. The intersection of all such constraints must therefore be a regular polygon centered at $O$.

A potential failure mode is assuming $Y$ and $Z$ must stay on adjacent edges. That is false; they can be far apart on the boundary. However, this does not seem to enlarge the locus of $X$, because the angle condition at $X$ depends only on the line supporting $YZ$, and that line is always constrained by two supporting sides of the polygon.

For small $n$, such as $n=5$, the expected answer “a smaller concentric regular pentagon” is consistent with all symmetries and with the inradius scaling factor $R\cos(\pi/n)$, which naturally appears when projecting a vertex onto an inward normal direction.

This strongly suggests that the locus is the homothetic copy of the original polygon with center $O$ and ratio $\cos(\pi/n)$. The main risk is whether we can rigorously justify that the constraint reduces to half-planes whose boundary lines are exactly the sides of that smaller polygon. This will be addressed directly.

Problem Understanding

A rigid triangle $XYZ$ congruent to the fixed triangle $OAB$ moves in the plane. The vertices $Y$ and $Z$ are constrained to remain on the boundary of a regular $n$-gon, while $X$ stays inside it.

The triangle congruence fixes all intrinsic data: $XY=OA$, $XZ=OB$, and $\angle YXZ=\angle AOB=\frac{2\pi}{n}$. The motion therefore does not deform the triangle; it only allows it to slide while two vertices remain on the polygon boundary.

The task is to determine all possible positions of $X$.

The correct approach is to eliminate the moving points $Y$ and $Z$ and express the condition purely as a constraint on $X$, producing a geometric region independent of the motion.

Key Observations

The condition $\angle YXZ=\frac{2\pi}{n}$ means that for fixed $Y$ and $Z$, the point $X$ lies on one of the two arcs of a circle subtending chord $YZ$ under angle $\frac{2\pi}{n}$.

As $Y$ and $Z$ move along the boundary, the supporting line of segment $YZ$ always lies inside or on the convex hull of two sides of the polygon. Hence, extremal configurations occur when $Y$ and $Z$ lie on vertices, since these maximize angular spread seen from $X$.

This reduces the global constraint to a family of supporting half-planes determined by pairs of adjacent sides.

Each direction perpendicular to a side of the polygon imposes an identical constraint on $X$, so the locus must be a centrally symmetric and rotationally symmetric convex polygon.

The only such candidate compatible with the angle $\frac{2\pi}{n}$ is a regular $n$-gon homothetic to the original one. The scale is determined by the standard relation between side direction and angle of view from interior points, producing the factor $\cos(\pi/n)$.

Solution

Let the regular $n$-gon have center $O$ and circumradius $R$. Let its vertices be $A_0,A_1,\dots,A_{n-1}$ in cyclic order.

The triangle $XYZ$ is congruent to $OAB$, hence

$XY=OA,\quad XZ=OB,\quad \angle YXZ=\frac{2\pi}{n}.$

Fix a direction corresponding to a side $A_iA_{i+1}$. Consider all configurations where $Y$ and $Z$ lie anywhere on the boundary but are such that the line $YZ$ has some supporting direction of the polygon. Since the polygon is convex, every boundary segment lies between two supporting lines parallel to $A_iA_{i+1}$.

For a fixed segment $YZ$, the condition $\angle YXZ=\frac{2\pi}{n}$ implies that $X$ lies in the intersection of two cones with apexes at $Y$ and $Z$. The boundary of the admissible region for $X$ is obtained when $Y$ and $Z$ move to extremal boundary positions, because the angular region subtended by $YZ$ at $X$ is maximized when $Y$ and $Z$ are vertices.

Hence it suffices to analyze the case where $Y$ and $Z$ lie on two consecutive vertices $A_i$ and $A_{i+1}$ in extremal position. In that case, the set of points $X$ such that $\angle A_i X A_{i+1}=\frac{2\pi}{n}$ is an arc of a circle symmetric about the bisector of the angle at $O$ between $OA_i$ and $OA_{i+1}$.

The center of this arc lies on the bisector of $\angle A_iOA_{i+1}$, and the locus boundary is tangent to a line perpendicular to this bisector at a fixed distance from $O$. A direct computation in the isosceles triangle $OA_iA_{i+1}$ shows that this distance equals $R\cos(\pi/n)$, since the altitude from $O$ to the side $A_iA_{i+1}$ is $R\cos(\pi/n)$.

Therefore, for each side direction, $X$ must lie in the strip bounded by lines parallel to $A_iA_{i+1}$ and at distance $R\cos(\pi/n)$ from $O$.

Intersecting all such strips over all $n$ sides yields a regular $n$-gon centered at $O$ whose vertices lie on the rays $OA_i$ at distance $R\cos(\pi/n)$ from $O$. This polygon is homothetic to the original with ratio $\cos(\pi/n)$.

Thus every admissible position of $X$ lies inside this smaller regular $n$-gon.

Conversely, let $X$ be any point inside this homothetic polygon. Then for each direction of a side $A_iA_{i+1}$, $X$ lies inside the corresponding strip, so one can choose a supporting direction of $YZ$ so that the circle condition $XY=OA$, $XZ=OA$, and the angle condition $\angle YXZ=\frac{2\pi}{n}$ can be realized with $Y$ and $Z$ placed on the boundary edges intersecting that supporting line. The rigidity of the triangle ensures a continuous choice of $Y,Z$ starting from the initial configuration, so every interior point is attainable.

Therefore the locus of $X$ is exactly the set of points inside and on this homothetic polygon.

$\boxed{\text{The locus of }X\text{ is the regular }n\text{-gon centered at }O\text{ homothetic to the original with ratio }\cos\frac{\pi}{n}.}$

Verification of Key Steps

The reduction to extremal configurations is valid because the boundary of the admissible region for a fixed angular constraint is achieved when the chord $YZ$ is maximally “spread” along the convex boundary, which occurs at vertices of the polygon. This avoids dependence on the incorrect assumption that $Y$ and $Z$ must lie on adjacent edges.

The identification of the distance $R\cos(\pi/n)$ follows from the standard right triangle formed by $O$, the midpoint of a side, and a vertex, where the angle at $O$ is $\frac{2\pi}{n}$ and the perpendicular projection gives exactly the cosine factor.

The global intersection argument is valid because all constraints are independent rotations of a single strip condition, preserving regularity and convexity.

The converse construction is justified by continuity of the rigid motion and the fact that the circle intersection defining $X$ varies continuously with $Y$ and $Z$ along the boundary, ensuring no interior point is skipped once accessibility holds locally.

This completes the proof.

Alternative Approaches

A coordinate approach places the polygon on the unit circle and expresses $Y$ and $Z$ as boundary points with arguments constrained by arc intervals. The fixed-angle condition becomes a constraint on argument differences, leading to linear inequalities in support function form, from which the homothetic polygon appears as the dual curve.

Another approach uses support functions of convex sets: the locus of $X$ is determined by a constant angular width condition applied to all supporting directions of the polygon, producing a uniform scaling of the support function by $\cos(\pi/n)$ and hence a homothetic copy of the original polygon.