Olympiad Maths Prep

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, 2007

Combinatorics Difficulty 8.5 Shortlist Prove it IMO

Given a convex nn-gon PP in the plane. For every three vertices of PP, consider the triangle determined by them. Call such a triangle good if all its sides are of unit length.
Prove that there are not more than 23n\frac{2}{3} n good triangles.

Solution

Consider all good triangles containing a certain vertex AA. The other two vertices of any such triangle lie on the circle ωA\omega_{A} with unit radius and center AA. Since PP is convex, all these vertices lie on an arc of angle less than 180180^{\circ}. Let LARAL_{A} R_{A} be the shortest such arc, oriented clockwise (see Figure 1). Each of segments ALAA L_{A} and ARAA R_{A} belongs to a unique good triangle. We say that the good triangle with side ALAA L_{A} is assigned counterclockwise to AA, and the second one, with side ARAA R_{A}, is assigned clockwise to AA. In those cases when there is a single good triangle containing vertex AA, this triangle is assigned to AA twice.
There are at most two assignments to each vertex of the polygon. (Vertices which do not belong to any good triangle have no assignment.) So the number of assignments is at most 2n2 n.

Consider an arbitrary good triangle ABCA B C, with vertices arranged clockwise. We prove that ABCA B C is assigned to its vertices at least three times. Then, denoting the number of good triangles by tt, we obtain that the number KK of all assignments is at most 2n2 n, while it is not less than 3t3 t. Then 3tK2n3 t \leq K \leq 2 n, as required.

Actually, we prove that triangle ABCA B C is assigned either counterclockwise to CC or clockwise to BB. Then, by the cyclic symmetry of the vertices, we obtain that triangle ABCA B C is assigned either counterclockwise to AA or clockwise to CC, and either counterclockwise to BB or clockwise to AA, providing the claim.

Figure 1
Figure 1
Figure 2
Figure 2

Assume, to the contrary, that LCAL_{C} \neq A and RBAR_{B} \neq A. Denote by AA', BB', CC' the intersection points of circles ωA\omega_{A}, ωB\omega_{B} and ωC\omega_{C}, distinct from A,B,CA, B, C (see Figure 2). Let CLCLCC L_{C} L_{C}' be the good triangle containing CLCC L_{C}. Observe that the angle of arcLCA\operatorname{arc} L_{C} A is less than 120120^{\circ}. Then one of the points LCL_{C} and LCL_{C}' belongs to arc BAB' A of ωC\omega_{C}; let this point be XX. In the case when LC=BL_{C}=B' and LC=AL_{C}'=A, choose X=BX=B'.

Analogously, considering the good triangle BRBRBB R_{B}' R_{B} which contains BRBB R_{B} as an edge, we see that one of the points RBR_{B} and RBR_{B}' lies on arc ACA C' of ωB\omega_{B}. Denote this point by YY, YAY \neq A. Then angles XAYX A Y, YABY A B, BACB A C and CAXC A X (oriented clockwise) are not greater than 180180^{\circ}. Hence, point AA lies in quadrilateral XYBCX Y B C (either in its interior or on segment XYX Y ). This is impossible, since all these five points are vertices of PP.

Hence, each good triangle has at least three assignments, and the statement is proved.

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