Bullseye Illusion

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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.


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.