# 2021 AIME I Problems/Problem 11

## Problem

Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7$. Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C$, respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC$. The perimeter of $A_1B_1C_1D_1$ is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

Let $O$ be the intersection of $AC$ and $BD$. Let $\theta = \angle AOB$.

Firstly, since $\angle AA_1D = \angle AD_1D = 90^\circ$, we deduce that $AA_1D_1D$ is cyclic. This implies that $\triangle A_1OD_1 \sim \triangle AOD$, with a ratio of $\frac{A_1O}{AO} = \cos \angle A_1OA = \cos \theta$. This means that $\frac{A_1D_1}{AD} = \cos \theta$. Similarly, $\frac{A_1B_1}{AB} = \frac{B_1C_1}{BC} = \frac{C_1D_1}{CD} = \cos \theta$. Hence $$A_1B_1 + B_1C_1 + C_1D_1 + D_1A_1 = (AB + BC + CD + DA)\cos \theta$$ It therefore only remains to find $\cos \theta$.

From Ptolemy's theorem, we have that $(BD)(AC) = 4\times6+5\times7 = 59$. From Brahmagupta's Formula, $[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}$. But the area is also $\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta$, so $\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}$. Then the desired fraction is $(4+5+6+7)\cos\theta = \frac{242}{59}$ for an answer of $\boxed{301}$.