# Bullseye Illusion

## Classic Problem

The circle of radius $3 a$ at the center has the same area as the outer annulus whose width is $1$.

However, it appears to be larger.

## Proof

From the diagram, it is seen that the circles are evenly spaced.

Let $C$ denote the circle of radius $3 a$ at the center, colored orange.

Let $A$ denote the outer annulus of width $a$ and outer radius $5 a$, colored blue.

We have that:

 $\ds \map \Area C$ $=$ $\ds \pi \paren {3 a}^2$ Area of Circle $\ds$ $=$ $\ds 9 \pi a^2$

 $\ds \map \Area A$ $=$ $\ds \pi \paren {\paren {5 a}^2 - \paren {4 a}^2}$ Area of Annulus $\ds$ $=$ $\ds \pi a^2 \paren {25 - 16}$ $\ds$ $=$ $\ds 9 \pi a^2$

Hence the result.

$\blacksquare$