Mathy posters, solve a problem for me!

[Chad Sexington]

How do I figure out if one polygon will fit inside of another?

Here's the problem: Regular pentagon, each side is 5'9.5" (5.792'), all the angles are equal, etc.

I have a 6' square. Can I fit the square inside the pentagon? Inherently I think no, but I can't really mathematically reason out how to solve the problem. This is actually relevant to a thing, and not just arbitrary if that motivates you.

I really just need a yes or no but if you want to expound at length as to how to solve the problem, I am listening.

 

[Papillon]

No.

Imagine drawing a 5 pointed star inside your pentagon. You can calculate the length of each side of the 5-pointed star by using the cosine rule, since two sides of the pentagon, and one side of the star will form a triangle, and you know the angle between the two sides of the pentagon (360/5 = 72 degrees = 2pi/5 radians). Thus, each side of the star will have a length
l_s = 2* l_p^2 - 2*l_p^2*cos(2*pi/5)
where l_p is the side length of the pentagon. In your case l_p = 69.5". This gives l_s = 81.70". This is the length of the longest straight line that can fit inside the pentagon without crossing a side.

The length of the diagonal of the square is 72*sqrt(2)" = 101.82", which is longer than l_s. Therefore, your square does not fit into your pentagon.

 

[Chad Sexington]

Okay thanks so much. I'm wondering, why is it sqrt(2)? Isn't the hypotenuse of the square 6^2+6^2=72, so you just take the sqrt of 72?

I'm just confused, I worked in inches and if I go 36^2 + 36^2 I get 2592, the sqrt of which is half the number you gave me. 50.91

I definitely trust your math over mine though.

Edit: As it turns out I am an idiot who thinks there are 36 inches in six feet, not 72.

Thanks so much again!