24th Putnam 1963

Problem B4

Γ is a closed plane curve enclosing a convex region and having a continuously turning tangent. A, B, C are points of Γ such that ABC has the maximum possible perimeter p. Show that the normals to Γ at A, B, C are the angle bisectors of ABC. If A, B, C have this property, does ABC necessarily have perimeter p? What happens if Γ is a circle?



Let L be the external bisector of angle C (so that it is perpendicular to the internal bisector of angle ACB). Let B' be the reflection of B in the line L. If X is any point on the other side of L from ABC, then BX > B'X, so AX + BX > AX + B'X. But AX + B'X ≥ AB' (with equality iff X lies on AB'), and AB' = AC + CB. So AX + BX > AC + CB. Thus the perimeter of AXB > p. Hence X cannot belong to the region. But if L does not coincide with the tangent at C then it will intersect the region in further points and so the region will contain points on both sides of L. So L must be the tangent.

No, this is not a sufficient condition. For example, take the curve Γ to be an equilateral triangle with rounded corners. Then take A, B, C to be the midpoints of the sides. In this case the perimeter is certainly not maximal.

If Γ is a circle, then the achieve the maximum by taking ABC to be equilateral.



24th Putnam 1963

© John Scholes
5 Feb 2002