HMMT-Nov 2022 gen, Problem 3

Let $A B C D$ be a rectangle with $A B=8$ and $A D=20$. Two circles of radius 5 are drawn with centers in the interior of the rectangle - one tangent to $A B$ and $A D$, and the other passing through both $C$ and $D$. What is the area inside the rectangle and outside of both circles?

Solution: Solution: Let $O_{1}$ and $O_{2}$ be the centers of the circles, and let $M$ be the midpoint of $\overline{C D}$. We can see that $\triangle O_{2} M C$ and $\triangle O_{2} M D$ are both 3-4-5 right triangles. Now let $C^{\prime}$ be the intersection of circle $O_{2}$ and $\overline{B C}$ (that isn't $C$ ), and let $D^{\prime}$ be the intersection of circle $O_{2}$ and $\overline{A D}$ (that isn't $D$ ). We know that $A D^{\prime}=B C^{\prime}=14$ because $B C^{\prime}=2 O_{2} M=6$. All of the area of $A B C D$ that lies outside circle $O_{2}$ must lie within rectangle $A B C^{\prime} D^{\prime}$ because $C^{\prime} C D D^{\prime}$ is completely covered by circle $O_{2}$. Now, notice that the area of circle $O_{2}$ that lies inside $A B C^{\prime} D^{\prime}$ is the same as the area of circle $O_{1}$ that lies outside $A B C^{\prime} D^{\prime}$. Thus, the total area of $A B C^{\prime} D^{\prime}$ that is covered by either of the two circles is exactly the area of one of the circles, $25 \pi$. The remaining area is $8 \cdot 14-25 \pi$, which is our answer. $\fbox{112-25 \pi}$.

Answer: 112-25 \pi