Are Not More Than 70 Percent of Triangles Acute-Angled?

[Ackbach]

Here is this week's POTW:

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Seventy-five coplanar points are given, no three collinear. Prove that, of all the triangles which can be drawn with these points as vertices, not more than seventy percent are acute-angled.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

I'm going to give another week on this one, as this is a particularly nice problem.

 

[Ackbach]

Congratulations to castor28 for his correct solution to this POTW, which I note with dismay was THREE WEEKS ago. My excuse is I had company. Anyway, here is his solution to this POTW, which was Problem 453 in the MAA challenges.

[sp]
We will call an angle "large" if it is right or obtuse, and a triangle "bad" if it contains a large angle. Note that a triangle can contain at most one large angle; distinct large angles define distinct bad triangles. We must show that there are at least 30 percent of bad triangles.

We recall that, in a convex polygon, the sum of the exterior angles is 360°; this implies that there are at most three acute interior angles, and a convex polygon of $n$ sides has at least $(n-3)$ large interior angles.

We will first prove the proposition with 5 points, i.e., we will prove that at least 3 of the $\binom{5}{2} = 10$ triangles are bad.

Case 1 : the convex hull is a triangle ABC. Let P be an interior point. As we have 3 triangles APB, BPC, and CPA, the three angles at P are smaller than 180°; as the sum of these angles is 360°, at most one of these angles is acute, and we have at least two bad triangles containing P. As the same argument applies to the other interior point Q, we have at least 4 bad triangles (these are distinct, because none of these triangles contains both P and Q).

Case 2: the convex hull is a quadrilateral ABCD, and P is the interior point. By the remark above, at least one of the interior angles of the quadrilateral is large, and this defines a bad triangle, for example ABD.

We draw a diagonal, for example AC. Because no three points are collinear, P lies within one of the two triangles thus defined, say ABC. The argument of case 1 shows that there are two bad triangles containing P. Together with the triangle ABD (which does not contain P), this gives three bad triangles.

Case 3: the five points are the vertices of a convex pentagon ABCDE. By the remark above, at least two of the interior angles are large, giving at least two bad triangles.

If there are three large interior angles, we have three bad triangles and we are done.

Otherwise, the two bad triangles have one or two vertices in common (depending on the positions of the two large angles). If A is a common point of these triangles, the convex quadrilateral BCDE contains a large interior angle that defines a bad triangle. That triangle is distinct from the first two bad triangles, because it does not contain the vertex A. We have therefore at least three bad triangles in this case also.

We have shown that each 5-clique contains at least three bad triangles. There are $\binom{75}{5} = 17259390$ 5-cliques, and each triangle belongs to $\binom{72}{2}=2556$ cliques. We have therefore at least:

$$\left\lceil\frac{3\times17259390}{2556}\right\rceil = 20258$$
bad triangles, and this is more than 30 percent of the $\binom{75}{3}=67525$ triangles.
[/sp]