# Solution to the Four Quarter Circles in a Circle Puzzle +-- {.image} [[FourQuarterCirclesinaCircle.png:pic]] > What fraction of the circle do these four quarter circles cover? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[FourQuarterCirclesinaCircleLabelled.png:pic]] =-- In the above diagram then $O$ is the centre of the circle, $M$ is the midpoint of $A B$, and $N$ is the midpoint of $C D$. The distances $a$, $b$, $c$, and $d$ are measured from the intersection point $P$, and $r$ is the radius of the outer circle. As $N$ is the midpoint of $C D$, $O N$ is [[perpendicular]] to $C D$ and similarly $O M$ is perpendicular to $A B$. Since angle $B \hat{P} D$ is a [[right-angle]], the [[quadrilateral]] $O M P N$ is a [[rectangle]]. In particular, $O M$ and $N P$ have the same length as each other, and $O N$ and $M P$ have the same length as each other. Since $N$ is the midpoint of $C D$, $N D$ has length $\frac{c + d}{2}$ and $N P$ has length $\frac{d - c}{2}$. Similarly, $B M$ has length $\frac{a + b}{2}$ and $M P$ has length $\frac{b - a}{2}$. Triangle $O N D$ is a [[right-angled triangle]] so applying [[Pythagoras' theorem]] results in: $$ r^2 = \left( \frac{c + d}{2} \right)^2 + \left( \frac{b - a}{2} \right)^2 $$ This simplifies to $$ 4 r^2 = c^2 + 2 c d + d^2 + b^2 - 2 a b + a^2 $$ A similar argument applied to triangle $O M B$ leads to $$ 4 r^2 = c^2 - 2 c d + d^2 + b^2 + 2 a b + a^2 $$ Taken together, these imply that $c d = a b$ (this could also be seen through use of the [[intersecting chords theorem]]) and so the original equation becomes $$ 4r^2 = c^2 + d^2 + b^2 + a^2 $$ The length $a$ is the hypotenuse of an [[isosceles]] [[right-angled triangle]] with radii of the yellow quarter circle forming the other two sides. This radius is therefore of length $\frac{a}{\sqrt{2}}$ so the area of that quarter circle is $$ \frac{1}{4} \pi \left( \frac{a}{\sqrt{2}} \right)^2 = \frac{1}{8} \pi a^2 $$ The total area of the quarter circles is therefore $$ \frac{1}{8} \pi a^2 + \frac{1}{8} \pi b^2 + \frac{1}{8} \pi c^2 + \frac{1}{8} \pi d^2 = \frac{1}{8} \pi (a^2 + b^2 + c^2 + d^2) = \frac{1}{2} \pi r^2 $$ Since the area of the outer circle is $\pi r^2$, the shaded area is thus half of the area of the full circle.