# Solution to the Four Parallel Semi-Circles Puzzle +-- {.image} [[FourParallelSemiCircles.png:pic]] > The bases of all $4$ semicircles are parallel, and the largest one has diameter $8$. What’s the total shaded area? =-- ## Solution by Properties of [[Chords]] and [[Pythagoras' Theorem]] +-- {.image} [[FourParallelSemiCirclesLabelled.png:pic]] =-- In the above diagram, the points labelled $G$ and $M$ are the centres of their semi-circles. Since $A B$ is a [[chord]] on the largest semi-circle, and $G$ is its [[midpoint]], the line $G M$ is [[perpendicular]] to $A B$ and so also perpendicular to $C F$. Triangles $G M D$ and $M G A$ are therefore a [[right-angled triangles]]. Let $a$, $b$, and $c$ be the radii of the semi-circles other than the largest, with the length of $C D$ as $2a$, of $D F$ as $2 b$, and of $A B$ as $2 c$. Since $C F$ has length $8$, $2 a + 2 b = 8$ so $a + b = 4$. As $M$ is the midpoint of $C F$, $C M$ has length $a + b$, so $D M$ has length $b - a$. Let $d$ be the length of $G M$. Applying [[Pythagoras' theorem]] to triangles $G M D$ and $M G A$ shows that: $$ \begin{aligned} 16 &= c^2 + d^2 \\ c^2 &= d^2 + (b - a)^2 \\ \end{aligned} $$ Putting these together and eliminating $d^2$ shows that: $$ 16 - c^2 = c^2 - (b - a)^2 = c^2 - b^2 - a^2 + 2 a b $$ Since $a + b = 4$, $4^2 = (a + b)^2 = a^2 + 2 a b + b^2$, so $2 a b = 16 - a^2 - b^2$. Substituting this in shows that: $$ 16 = 2 c^2 - b^2 - a^2 + 16 - a^2 - b^2 = 16 + 2 c^2 - 2 a^2 - 2 b^2 $$ So $c^2 = a^2 + b^2$. From the [[area of a circle]], the shaded area is given by: $$ \frac{1}{2} \pi 16 - \frac{1}{2} \pi c^2 + \frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 = \frac{1}{2} \pi 16 = 8 \pi $$ ## Solution by [[Agg Invariance Principle]] The point labelled $D$ in the above diagram can move along the [[diameter]] $C F$. +-- {.image} [[FourParallelSemiCirclesSpecialA.png:pic]] =-- With $D$ at the [[midpoint]] the radii of the circles are in the ratio $1:\sqrt{2}:2$ so their areas are in the ratio $1 : 2 : 4$. The two smaller semi-circles therefore have the same area as the middle one, leaving the equivalent of the largest semi-circle in total area. +-- {.image} [[FourParallelSemiCirclesSpecialB.png:pic]] =-- Placing $D$ at the end of the diameter, the white semi-circle coincides with one of the upper semi-circles, and the other has zero size. The shaded area is therefore just the lower semi-circle.