# Flex a square into a circle, and prove...

Let points $A$, $B$, $C$, and $D$ be the vertices on a square. Let $\overline{CD}$'s midpoint be $E$. Flex the square into a circle (so they'll have equal perimeter/circumference), and translate the circle so it touches point $E$ (i.e. if $\overline{CD}$ is the bottom line, then translate the circle up). Prove or disprove whether the points $A$ and $B$ will lie inside the circle or not. Using graphing tools I've determined that points $A$ and $B$ do lie inside the circle, but not by much. Need a nice, simple proof though. Thanks.

• By flex the square do you mean construct a circle tangent to the side at E with circumference equal to the perimeter of the square? You might as well let the square be a unit square. Can you compute the radius of the circle? Commented Apr 9, 2015 at 3:48

Suppose for computational convenience that we begin with the square centered at $(0,0)$. Since a circle is all points a certain distance (the radius, which I'll call $r$) from its center, you want to show that $(x_A-x_{newcenter})^2 + (y_A-y_{newcenter})^2 \le r^2$. Since the circle and square are symmetric, A and B really present the same problem and I'll only do the one in the first quadrant.
If the square has sides of length $s$, then the coordinates of A are $(s/2,s/2)$ and the perimeter is $4s$. The circumference of the square is $2\pi r$, so $r=\frac{2s}{\pi}$. Then we have to find the new center. The new center is at $(0,r+y_E)=(0,\frac{2s}{\pi}-\frac{s}{2})$ since it's a distance of $r$ directly above E. Now just plug everything back into the inequality and divide through by $s^2$ - we're trying to show that $1/4+(1-2/\pi)^2\le 4/\pi^2$. That's just a matter of calculation.
In fact, I end up with $5\pi \le 16$ at the very end. Yep, just barely.