2022 AMC 10A Problem 15

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 $\boxed{\textbf{(D)}\ 1565}$.

Side lengths that complete two Pythagorean triples with a common hypotenuse are a strong hint that the shared hypotenuse is a diagonal and both opposite angles are right, and a right inscribed angle always marks a diameter. Then the whole configuration is known: the circle from the diameter, the quadrilateral as two right triangles.

• Not converting 234 to 936/4 before reading off a, b, c, which gives 625 + 234 + 4 = 863, not a choice, or using radius 25 instead of 25/2.
• Missing the Pythagorean triples and attempting Brahmagupta's formula or Ptolemy's theorem, which is slow though it still yields area $234$.