2022 AMC 10A Problem 15

1. Spot the right triangles. Draw diagonal $AC$, which splits the quadrilateral into triangle $ABC$ (sides $7$, $24$) and triangle $ACD$ (sides $15$, $20$). Notice
$$7^2 + 24^2 = 49 + 576 = 625 = 25^2, \qquad 15^2 + 20^2 = 225 + 400 = 625 = 25^2.$$
So if the angles at $B$ and $D$ are right angles, both triangles have long side $AC = 25$.
2. Why the angles really are right angles. In a quadrilateral inside a circle, opposite angles add to $180^\circ$, so $\angle B + \angle D = 180^\circ$.
- If $\angle B$ were more than $90^\circ$, triangle $ABC$ would be "opened wider" than a right triangle, so $AC$ would be longer than $25$. But then $\angle D$ would be less than $90^\circ$, making $AC$ shorter than $25$. Both cannot happen.
- The same clash happens if $\angle B$ is less than $90^\circ$.

So both angles are exactly $90^\circ$ and $AC = 25$.
3. The circle. A right angle whose corner is on a circle stands on a diameter. So $AC = 25$ is a diameter, the radius is $\frac{25}{2}$, and the circle's area is
$$\pi \left(\tfrac{25}{2}\right)^2 = \frac{625\pi}{4}.$$
4. The quadrilateral. It is two right triangles:
$$\tfrac12 \cdot 7 \cdot 24 = 84, \qquad \tfrac12 \cdot 15 \cdot 20 = 150, \qquad 84 + 150 = 234.$$
5. The region outside the quadrilateral. Write $234 = \frac{936}{4}$ so both parts have the same bottom:
$$\frac{625\pi}{4} - \frac{936}{4} = \frac{625\pi - 936}{4}.$$
6. Read off the answer. $625 = 5^4$ and $4 = 2^2$ share no prime factor, so $a = 625$, $b = 936$, $c = 4$, and $a + b + c = 1565$.

The answer is $\boxed{\textbf{(D)}\ 1565}$.

When the sides of a quadrilateral fit two Pythagorean triples with the same long side, suspect that this long side is a diagonal with right angles on both sides of it. A right angle drawn inside a circle always stands on a diameter, so the circle's size follows at once, and the quadrilateral is just two right triangles.

Draw diagonal $AC$. The sides $7, 24$ and $15, 20$ are the legs of the Pythagorean triples $7$-$24$-$25$ and $15$-$20$-$25$, which share the hypotenuse $25$. This suggests $AC = 25$ with right angles at $B$ and $D$, and that is consistent with the cyclic condition, since opposite angles $B$ and $D$ would sum to $180^\circ$. It is in fact forced: $\angle B$ and $\angle D$ are supplementary, so $\cos D = -\cos B$, and computing $AC^2$ in both triangles gives $625 - 336\cos B = 625 + 600\cos B$, so $\cos B = 0$.

Right angles at $B$ and $D$ mean $AC$ is a diameter, so the radius is $\frac{25}{2}$ and the circle's area is $\pi \left(\frac{25}{2}\right)^2 = \frac{625\pi}{4}$.

The quadrilateral is two right triangles: $[ABC] = \frac12 \cdot 7 \cdot 24 = 84$ and $[ACD] = \frac12 \cdot 15 \cdot 20 = 150$, total $234 = \frac{936}{4}$.

The region inside the circle but outside the quadrilateral has area
$$\frac{625\pi}{4} - \frac{936}{4} = \frac{625\pi - 936}{4}.$$
Since $\gcd(625, 4) = 1$, $a + b + c = 625 + 936 + 4 = 1565$.

The answer is $\boxed{\textbf{(D)}\ 1565}$.