IMO 1986 · Problem 4

Let \(A, B\) be adjacent vertices of a regular \(n\)-gon \((n \geq 5)\) in the plane having center at \(O\). A triangle \(XYZ\), which is congruent to and initially conincides with \(OAB\), moves in the plane in such a way that \(Y\) and \(Z\) each trace out the whole boundary of the polygon, \(X\) remaining inside the polygon. Find the locus of \(X\).

Official text of the problem. Official problem paper (PDF)

Answer

Write \(R=OA\) and \(\theta=\pi/n\). For each vertex \(V\), let \(P_V\) lie beyond \(O\) on the ray from \(V\) through \(O\), with \[ OP_V=R\left(\sec\theta-1\right). \] The locus is the union of the \(n\) segments \(OP_V\).

Solution

Let \[ s=AB=2R\sin\theta,\qquad r=R\cos\theta. \] Thus \(XY=XZ=R\), \(YZ=s\), and the altitude of triangle \(XYZ\) from \(X\) is \(r\).

We first show that two boundary points at distance \(s\) can always be regarded as lying on two adjacent sides.

If they lie on the same side, they must be its endpoints, and the assertion follows. Now consider two nonadjacent sides. Their smaller cyclic separation is some integer \[ 2\leq k\leq \lfloor n/2\rfloor. \] Take \(O\) as origin and rotate the coordinates so that the radii through their midpoints have arguments \(-k\theta\) and \(k\theta\). The endpoints of the upper side have arguments \((k-1)\theta,(k+1)\theta\), and those of the lower side have the negatives of these arguments.

Since \[ \sin((k+1)\theta)-\sin((k-1)\theta) =2\cos(k\theta)\sin\theta\geq0, \] every point on the upper side has vertical coordinate at least \(R\sin((k-1)\theta)\), and every point on the lower side has vertical coordinate at most its negative. Hence the distance between any such two points is at least \[ 2R\sin((k-1)\theta)\geq2R\sin\theta=s. \] Here \(\theta\leq(k-1)\theta<\pi/2\), so equality requires \(k=2\). Moreover, \(\cos(2\theta)>0\), because \(n\geq5\). Thus equality requires the points to be the endpoints with arguments \(-\theta,\theta\), which are adjacent vertices. These too can be assigned to two adjacent sides. This proves the assertion, including all vertex cases.

Now let \(V\) be the common vertex of two sides containing \(Y,Z\). Choose local coordinates with \[ V=(0,0),\qquad O=(R,0), \] and, interchanging \(Y,Z\) for this calculation if necessary, write \[ Y=(u\sin\theta,u\cos\theta),\qquad Z=(v\sin\theta,-v\cos\theta), \qquad 0\leq u,v\leq s. \] Put \(m=u+v\) and \(d=u-v\). The condition \(YZ=s\) becomes \[ s^2=m^2\cos^2\theta+d^2\sin^2\theta. \tag{1} \]

The midpoint of \(YZ\) is \[ M=\left(\frac{m\sin\theta}{2},\frac{d\cos\theta}{2}\right), \] and a perpendicular vector to \(YZ\), of length \(s\), is \[ (m\cos\theta,-d\sin\theta). \] Since \(r/s=\tfrac12\cot\theta\), the two possible positions of the apex are \[ X_+=\left(\frac{m}{2\sin\theta},0\right), \qquad X_-=\left(-\frac{m\cos2\theta}{2\sin\theta},d\cos\theta\right). \] The whole polygon lies in the half-plane \(x\geq0\). Because \(m>0\) and \(\cos2\theta>0\), the point \(X_-\) is outside the polygon. Therefore \[ X=X_+=\left(\frac{m}{2\sin\theta},0\right). \tag{2} \]

As \(|d|\leq m\), equation (1) gives \[ m^2\cos^2\theta\leq s^2\leq m^2, \] so \[ s\leq m\leq\frac{s}{\cos\theta}. \] Using (2) and \(s=2R\sin\theta\), we obtain \[ R\leq VX\leq R\sec\theta. \] Thus \(X\) belongs to the claimed segment \(OP_V\).

All these segments lie strictly inside the polygon. Indeed, \(\theta<\pi/4\), so \[ OP_V=R(\sec\theta-1) <R(\sqrt2-1) <\frac{R}{\sqrt2} <R\cos\theta=r. \] The disk centered at \(O\) with radius \(r\) is the polygon’s incircle.

It remains to prove that the entire segments are attained in the prescribed motion. For a fixed vertex \(V\), let \(u\) decrease continuously from \(s\) to \(0\), and set \[ v=-u\cos2\theta+ \sqrt{s^2-u^2\sin^22\theta}. \] This is the unique nonnegative solution of \[ u^2+v^2+2uv\cos2\theta=s^2. \tag{3} \] It lies in \([0,s]\) and varies continuously from \(0\) to \(s\). Moreover, because the left-hand side of (3) is strictly increasing in either nonnegative variable, \(v\) increases as \(u\) decreases.

Consequently, \(Y\) travels along one side toward \(V\), while \(Z\) travels along the next side away from \(V\). At the beginning and end, \(m=s\), so \(X=O\). At \[ u=v=\frac{s}{2\cos\theta}, \] we have \(m=s/\cos\theta\), so \(X=P_V\). Continuity therefore ensures that \(X\) traverses the whole segment \(OP_V\).

Starting with \(V=B\), this motion begins at \((X,Y,Z)=(O,A,B)\). Repeating it at successive vertices produces a continuous motion in which both \(Y\) and \(Z\) trace the whole boundary and every claimed segment is traversed.

Finally, the locus is the same for any permitted motion covering the boundary. The orientation of the nondegenerate triangle cannot change continuously. Orienting the polygon so that \(A,B\) occur consecutively in that order, this orientation forces \(Z\) onto the side following the side containing \(Y\). Equation (3) then determines \(Z\), and hence \(X\), uniquely from \(Y\). Therefore, when \(Y\) covers the entire boundary, all the configurations described above must occur. The locus is exactly the stated union.

The solution was checked against John Scholes' IMO solution archive (Kalva).