# Solution to the Upside-Down Semi-Circles in a Semi-Circle Puzzle +-- {.image} [[UpsideDownSemiCirclesinaSemiCircle.png:pic]] > What proportion of each diagram is shaded? Give your answers as decimals. What comes next? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[UpsideDownSemiCirclesinaSemiCircleLabelled.png:pic]] =-- Fix a diagram with $n$ small semi-circles. Let $r$ be the radius of the small semi-circles and $R$ of the outer one. In that diagram, triangle $O B A$ is [[right-angled triangle|right-angled]]. The length of $O A$ is $r$ while the length of $O B$ is $n r$. Applying [[Pythagoras' theorem]] shows that: $$ R^2 = r^2 + n^2 r^2 = (n^2 + 1) r^2 $$ The shaded area is $n \times \frac{1}{2} \pi r^2$ while the area of the outer semi-circle is $\frac{1}{2} \pi R^2 = (n^2 + 1) \times \frac{1}{2} \pi r^2$. The proportion that is shaded is therefore $\frac{n}{n^2 + 1}$. The first few values of this are: $$ \begin{aligned} \frac{1}{2} &= 0.5 \\ \frac{2}{5} &= 0.4 \\ \frac{3}{10} &= 0.3 \\ \frac{4}{17} &\simeq 0.23529411764705882 \\ \frac{5}{26} &\simeq 0.19230769230769232 \\ \frac{6}{37} &\simeq 0.16216216216216217 \\ \frac{7}{50} &=0.14 \\ \frac{8}{65} &\simeq 0.12307692307692308 \\ \frac{9}{82} &\simeq 0.10975609756097561 \end{aligned} $$