Ken's POTW 981113

Square in a Pentagon

What is the size of the largest square which can fit inside a regular pentagon with a side-length of 1?

Source: Original.

Solutions were received from Nick Baxter, Joseph W. DeVincentis, Carlos Rivera, Philippe Fondanaiche, Denis Borris, Robert T McQuaid. The solutions below are my own.

There are basically two different ways the square could be oriented. Either one side is parallel to a side of the pentagon, and the four corners are on the other four sides, OR one vertex is coincident with a vertex of the pentagon and two neighboring vertexes lie on the other sides. There are other orientations, but trying to find the maximum square will lead to one of these two.

A _/ \_ W---------X _/| s |\_ F | | B \ | | / \| |/ Z---------Y D-------C
If one side is parallel, let s=WX and t=AX.
Using the Sine Law in triangle AWX, s/sin(108)=t/sin(36).
Using the Sine Law in triangle BXY, s/sin(108)=(1-t)/sin(18).
Setting these equal, we find t=sin(36)/(sin(18)+sin(36)).
Solving for s we find s=sin(108)/(sin(18)+sin(36)) = 1.060497.
If we instead let one corner be coincident, using the figure above, assume Z and D are coincident.

If X actually lies on AB, the shortest the diagonal XZ could be is the height of the pentagon, or sin(36)+sin(72)=1.53884. In this case, the side of the square would be 1.08. This can only be the right answer if W and Y lie inside or on the pentagon. If we check the size of such a square and find a smaller value, that value will be the largest square possible.
Now we see how large the square can be if W and Y are on the pentagon. Angle CYZ (=CYD) is 63 degrees. So, again using the Sine Law on triangle CYZ, we find s/sin(108)=1/sin(63), so s=sin(108)/sin(63)=1.067395.

The final value of 1.067395 is the side-length of the largest square which can fit inside the pentagon.

Thanks to Nick Baxter for sending the following diagram to show the various squares discussed in this solution:

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