# Occurrence of $e$ in intersecting circles.

Consider two identical circles that share a radius such that they intersect. The radii of the circles are $\pi\over 2$. If this new shape sits such that its major axis is horizontal and the shortest geometric diameter is $\sqrt3$ and is vertical. Now the center top intersection can be labelled $A$. The furthest point to the left can be labelled $B$ and the geometric center of the shape can be labelled $C$.

There should be a shape that looks like a typical venn diagram. The question comes to matter when the sides of the triangle constructed form points $A,B$ and $\space C$ are taken into account.

$BC = \pi$ as stated before.

how ever is it coincidence that $AB$ $\approx e$

• Would you mind drawing it? Your description is a little unclear for me. Commented Jul 23, 2013 at 23:04
• @CameronWilliams There you go. Commented Jul 23, 2013 at 23:18
• Pure coincidence. Commented Jul 23, 2013 at 23:27

In fact, the number I get is not really all that close to $e$, but is really close to $3$.

Consider the right triangle formed by the center of the left circle $O$, the point $A$, and the center of the football $D$. Let the distance from $F$ to the rightmost point on the left circle be $x$; then $|OD|=(\pi/2)-x$, and

$$\frac{\pi^2}{4}=\left ( \frac{\pi}{2}-x\right)^2+\frac{3}{4}$$

or

$$x=\frac{\pi-\sqrt{\pi^2-3}}{2}$$

Let $E$ be the leftmost point of the left circle; then $|ED|=\pi-x=\frac{\pi}{2}+\frac12 \sqrt{\pi^2-3}$, and the length in question is

$$\sqrt{(\pi-x)^2+\frac{3}{4}} = \sqrt{\frac{\pi}{2} (\pi+\sqrt{\pi^3-3})} \approx 3.00863$$

• Indeed. The triangle $ABC$ also happens to be right angled at angle $BAC$. This shortens length $AB$ to $\sqrt {\pi^2 - \frac 34}$. Which is also around 3. Commented Jul 24, 2013 at 10:26