IMO 1983 · Problem 4

Let \(ABC\) be an equilateral triangle and \(\mathcal E\) the set of all points contained in the three segments \(AB\), \(BC\) and \(CA\) (including \(A\), \(B\) and \(C\)). Determine whether, for every partition of \(\mathcal E\) into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle. Justify your answer.

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

Answer

Yes. Every such partition contains a right-angled triangle entirely within one of its two subsets.

Solution

Colour the points of the two subsets red and blue. Suppose, for a contradiction, that there is no monochromatic right-angled triangle.

Trisect each side, naming the points in the following orders: \[ A,P,Q,B;\qquad B,R,S,C;\qquad C,T,U,A. \] The six points \(P,Q,R,S,T,U\) form a regular hexagon. Indeed, the corner triangles \(AUP,BQR,CST\) are equilateral, so all six hexagon sides have length \(AB/3\), and all its interior angles are \(120^\circ\). Its opposite vertex pairs are \[ (P,S),\qquad (Q,T),\qquad (R,U), \] each forming a diameter of its circumcircle.

We first show that each opposite pair has different colours. Suppose an opposite pair has the same colour, say red. Any third red hexagon vertex would form a red right-angled triangle with this diameter, by Thales’ theorem. Thus the other four hexagon vertices must all be blue. But these four contain another opposite pair and a third vertex, giving a blue right-angled triangle. This contradiction proves the claim.

Among \(P,R,T\), two have the same colour. By cyclically relabelling \(A,B,C\), and interchanging the colour names if necessary, we may assume that \(P\) and \(R\) are red. Their opposite vertices \(S\) and \(U\) are therefore blue.

Now \[ QP=QB=QR, \] because the side is trisected and \(BQR\) is equilateral. Hence \(P,B,R\) lie on the circle with centre \(Q\) and diameter \(PB\). Thales’ theorem gives \[ PR\perp BC. \] The central symmetry of the regular hexagon sends \(P\) to \(S\) and \(R\) to \(U\), so \(SU\parallel PR\). Therefore \[ SU\perp BC. \]

If \(B\) is red, \(BPR\) is a red triangle right-angled at \(R\). If \(B\) is blue, \(BSU\) is a blue triangle right-angled at \(S\). Both triangles are nondegenerate: their vertices are distinct, and \(P\) and \(U\) lie off the line \(BC\).

Either colour of \(B\) gives a contradiction. Thus every partition has the required triangle.

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