Notes
four quarter circles solution

Solution to the Four Quarter Circles Puzzle

Solution by Area of a Circle and Pythagoras' Theorem

Let $a$, $b$, $c$, $d$ be the radii of the quarter circles in increasing length. By considering the width of the rectangle $O A C E$ then $a + b = c$. The diagonals $A E$ and $O C$ have the same length, so $d = b + c = a + 2 b$.

The equations are slightly simpler with everything expressed in terms of $b$ and $c$, so $a = c - b$ and $d = b + c$. Applying Pythagoras' theorem to triangle $O A C$ yields:

But also $d^2 = (b + c)^2 = b^2 + 2 b c + c^2$ so $4c^2 = 6 c b$ and so $2 c = 3 b$. Then $a = c - b = \frac{1}{2} b$ so $b = 2 a$, $c = 3 a$, and $d = 5 a$.

If using $a$ and $b$ the resulting equation is $4 a^2 + 2 a b - 2 b^2 = 0$ which leads to $(2 a - b)(a + b) = 0$ and so, as $a$ and $b$ are both lengths, $b = 2 a$ then $c = 3 a$, and $d = 5 a$.

The area of the outer quarter circle is $\frac{1}{4} \pi d^2 = \frac{25}{4} \pi a^2$ and the sum of the areas of the shaded quarter circles is $\frac{1}{4} \pi c^2 + \frac{1}{4} \pi b^2 + \frac{1}{4} \pi a^2 = \frac{1}{4} \pi 14 a^2$. Hence the fraction that is shaded is $\frac{14}{25}$.