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.

$\blacksquare$

Sources

• 1991: David Wells: Curious and Interesting Geometry ... (previous) ... (next): geometrical illusions

• Weisstein, Eric W. "Bullseye Illusion." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/BullseyeIllusion.html