IMO 1970 SL 12

We are given 100 points in the plane, no three of which are

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IMO 1970 SL 12

Origin: USS

Problem

We are given 100 points in the plane, no three of which are on the same line. Consider all triangles that have all vertices chosen from the 100 given points. Prove that at most 70% of these triangles are acute angled.

Solution

Lemma. Five points are given in the plane such that no three of them are collinear. Then there are at least three triangles with vertices at these points that are not acute-angled. Proof. We consider three cases, according to whether the convex hull of these points is a triangle, quadrilateral, or pentagon. (i) Let a triangle ABC be the convex hull and two other points D and E lie inside the triangle. At least two of the triangles ADB, BDC and CDA have obtuse angles at the point D. Similarly, at least two of the triangles AEB, BEC and CEA are obtuse-angled. Thus there are at least four non-acute-angled triangles. (ii) Suppose that ABCD is the convex hull and that E is a point of its interior. At least one angle of the quadrilateral is not acute, determining one non-acute-angled triangle. Also, the point E lies in the interior of either \triangleABC or \triangleCDA hence, as in the previous case, it determines another two obtuse-angled triangles. (iii) It is easy to see that at least two of the angles of the pentagon are not acute. We may assume that these two angles are among the angles corresponding to vertices A, B, and C. Now consider the quadrilateral ACDE. At least one its angles is not acute. Hence, there are at least three triangles that are not acute-angled.

Now we consider all combinations of 5 points chosen from the given 100. There are 100  such combinations, and for each of them there are at least three non-acute-angled triangles with vertices in it. On the other hand, vertices of each of the triangles are counted 97  times. Hence there are at least 3 100  / 97  non-acute-angled triangles with vertices in the given 100 points. Since the number of all triangles with vertices in the given points is 100  , the ratio between the number of acute-angled triangles and the number of all triangles cannot be greater than 1 − 100  97 100  = 0.7.