Problem 6

In a plane there are

100

points, no three collinear. Consider all triangles whose vertices are among these points. Prove that no more than

70\%

of these triangles are acute-angled.

Step 1 of 4: Bound four-point configurations

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\#\{\text{acute triangles in any 4-point set}\}\le3

Detailed analysis

If the convex hull of four points is a quadrilateral, its four interior angles sum to $360^\circ$ , so at least one is at least $90^\circ$ ; the corresponding triangle is non-acute. If the convex hull is a triangle with one point inside, the three angles around the interior point sum to $360^\circ$ , so at least two are at least $90^\circ$ . Thus every four-point set has at most three acute triangles.