# Acute triangles in “obtuse” polygons?

Let $$P$$ be a convex polygon. Suppose every interior angle of $$P$$ is obtuse. Is it always the case that there exist three vertices $$p, q, r$$ of $$P$$ such that $$\triangle pqr$$ is acute?

I conjecture that the answer is yes. I have tried different types of triangulations of $$P$$ (e.g. fan triangulation, triangulation by repeatedly connecting every other vertex, etc.) However, I haven't come up with a proof yet.

Edit: And how about if $$P$$ has at most one acute angle (others stay obtuse)? I think the result should be the same.

• Why the close vote? – Timothy Chow Apr 25 at 13:44
• Incidentally, I think adding extra acute angles to the original polygon should make it easier, not harder, for the answer to be yes. – Robin Saunders Apr 26 at 10:12

## 1 Answer

Take a very obtuse isosceles triangle and chop its acute angles.

• This also answers the variant with one acute angle (leave one of the acute angles unchopped). – Jukka Kohonen Apr 25 at 10:26
• @RobinSaunders I think that the point is that if the shape is wide enough, we can ensure that one of the angles in your proposed triangle is still obtuse. But I agree with you that it would be good to spell out the argument. – Timothy Chow Apr 25 at 13:43
• If the shape is wide enough, then each angle at the chopped corners must already be close to acute, which gives little wiggle room. I can imagine there's a balancing point where it works out, but it's not immediately obvious. – Robin Saunders Apr 25 at 13:48
• If you fix the obtuse isosceles triangle, chop off the corners at such an angle that the four angles at the bottom are equal, but do the chopping very, very, very close to the corners, then I think it's clear that it works. – Jeremy Rickard Apr 25 at 18:02
• Thanks. I had to draw a picture, but now agree. The key point is that by making the chopped edges short, the angle under the diagonal can be (much) smaller than half the angle at the corners that were chopped off. – Robin Saunders Apr 25 at 20:51