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.
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