Regular pentagons inscribed in a triangle

Question

I know that inscribing a square into a triangle has been researched a lot. But has there been any research on the problem inscribing a regular pentagon into a triangle? Can anyone tell me more on the subject.

Answer 1

Suppose we have a triangle with a regular pentagon inscribed. Since no side of the triangle can contain three corners of the regular pentagon (because no three of its corners are collinear), it must be that there are two of the triangle's sides that each contain two corners of the pentagon.

This means that the triangle must have one corner angle equal to 36°.

(In the degenerate case where the triangle and the pentagon has a corner in common, the triangle must be a 36°-36°-108° triangle, which still satisfies this).

On the other hand, if we have a triangle with one of the angles being 36° and the two other angles $\ge$36° we can inscribe a regular pentagon in it. Clearly we can draw a tiny regular pengaton near the 36° corner which has four corners on the triangle and the fifth corner in the inside it. Now blow up this pentagon until the "free" corner hits the opposite side. This will happen before the two far pentagon corners on the neigboring sides run out of triangle side exactly if both of the opposite angles of the triangle is at least 36°.

So the triangles in which a regular pentagon can be inscribed are exactly those where (a) every angle is at least 36°, AND (b) at least one angle is 36° exactly.