# 2022 AMC 10A Problems/Problem 15

## Problem

Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$ $\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$

## Diagram $[asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C, D; O = origin; A = (-25/2,0); C = (25/2,0); B = intersectionpoints(Circle(A,7),Circle(C,24)); D = intersectionpoints(Circle(A,15),Circle(C,20)); fill(Circle(O,25/2),yellow); fill(A--B--C--D--cycle,white); dot("A",A,1.5*W,linewidth(4)); dot("B",B,1.5*dir(B),linewidth(4)); dot("C",C,1.5*E,linewidth(4)); dot("D",D,1.5*dir(D),linewidth(4)); dot(O,linewidth(4)); draw(Circle(O,25/2)); draw(A--B--C--D--cycle); label("7",midpoint(A--B),rotate(90)*dir(midpoint(A--B)--A)); label("24",midpoint(B--C),rotate(-90)*dir(midpoint(B--C)--B)); label("20",midpoint(C--D),rotate(-90)*dir(midpoint(C--D)--C)); label("15",midpoint(D--A),rotate(90)*dir(midpoint(D--A)--D)); [/asy]$ ~MRENTHUSIASM

## Solution 1 (Inscribed Angle Theorem)

Opposite angles of every cyclic quadrilateral are supplementary, so $$\angle B + \angle D = 180^{\circ}.$$ We claim that $AC=25.$ We can prove it by contradiction:

• If $AC<25,$ then $\angle B$ and $\angle D$ are both acute angles. This arrives at a contradiction.
• If $AC>25,$ then $\angle B$ and $\angle D$ are both obtuse angles. This arrives at a contradiction.

By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$

The area of the requested region is $$\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.$$ Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

~MRENTHUSIASM

## Solution 2 (Brahmagupta‘s Formula)

When we look at the side lengths of the quadrilateral we see $7$ and $24,$ which screams out $25$ because of Pythagorean triplets. As a result, we can draw a line through points $A$ and $C$ to make a diameter of $25.$ See Solution 1 for a rigorous proof.

This can also be shown using the Law of Cosines: Since $7^2+24^2-2\cdot7\cdot24\cdot\cos B=15^2+20^2-2\cdot15\cdot20\cdot\cos D$ and $\cos B + \cos D = 0,$ it follows that $\cos B = \cos D = 0.$

Since the diameter is $25,$ we can see the area of the circle is just $\frac{625\pi}{4}$ from the formula of the area of the circle with just a diameter.

Then we can use Brahmagupta Formula $\sqrt{(s - a)(s - b)(s - c)(s - d)}$ where $a,b,c,d$ are side lengths, and $s$ is semi-perimeter to find the area of the quadrilateral.

If we just plug the values in, we get $\sqrt{54756}=234.$ So now the area of the region we are trying to find is $\frac{625\pi}{4} - 234 = \frac{625\pi-936}{4}.$

Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

~Gdking ~Oinava

We can guess that this quadrilateral is actually made of two right triangles: $\triangle CDA$ has a $3 \text{-} 4 \text{-} 5$ ratio in the side lengths, and $\triangle ABC$ is a $7 \text{-} 24 \text{-} 25$ triangle. (See Solution 1 for a proof.)

Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the $15-20-25$ triangle. The area of the triangle is equal to the product of the side lengths divided by $4$ times the circumradius. Therefore, $150 = \frac{15\cdot20\cdot25}{4r}$. Solving this simple algebraic equation gives us $r = \frac{25}{2}$.

Plugging in the values, we have $\frac{25}{2}^2\cdot\pi - \left(\frac{15\cdot20}{2}+\frac{7\cdot24}{2}\right) = \frac{625\cdot\pi}{4} - 234$. Rewriting this gives us $\frac{625\pi-936}{4}$.

Therefore, adding these values gets us $\boxed{\textbf{(D) } 1565}.$

## Video Solution 1

~Education, the Study of Everything

## Video Solution 2

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 