# 2018 AIME I Problems/Problem 15

## Problem 15

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$, which can each be inscribed in a circle with radius $1$. Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$, and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\frac{2}{3}$, $\sin\varphi_B=\frac{3}{5}$, and $\sin\varphi_C=\frac{6}{7}$. All three quadrilaterals have the same area $K$, which can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution 1

Suppose our four sides lengths cut out arc lengths of $2a$, $2b$, $2c$, and $2d$, where $a+b+c+d=180^\circ$. Then, we only have to consider which arc is opposite $2a$. These are our three cases, so $$\varphi_A=a+c$$ $$\varphi_B=a+b$$ $$\varphi_C=a+d$$ Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$, $\overarc{BC}=2b$, $\overarc{CD}=2c$, and $\overarc{DA}=2d$.

Then, by Law of Sines, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ and $BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)$. Therefore, $$K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},$$ so our answer is $24+35=\boxed{059}$.

By S.B.

## Solution 2

Let the four stick lengths be $a$, $b$, $c$, and $d$. WLOG, let’s say that quadrilateral $A$ has sides $a$ and $d$ opposite each other, quadrilateral $B$ has sides $b$ and $d$ opposite each other, and quadrilateral $C$ has sides $c$ and $d$ opposite each other. The area of a convex quadrilateral can be written as $\frac{1}{2} d_1 d_2 \sin{\theta}$, where $d_1$ and $d_2$ are the lengths of the diagonals of the quadrilateral and $\theta$ is the angle formed by the intersection of $d_1$ and $d_2$. By Ptolemy's theorem $d_1 d_2 = ad+bc$ for quadrilateral $A$, so, defining $K_A$ as the area of $A$, $$K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}$$ Similarly, for quadrilaterals $B$ and $C$, $$K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}$$ and $$K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}$$ Multiplying the three equations and rearranging, we see that $$K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}$$ $$K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)$$ $$\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)$$ The circumradius $R$ of a cyclic quadrilateral with side lengths $a$, $b$, $c$, and $d$ and area $K$ can be computed as $R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$. Inserting what we know, $$1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}$$ $$4K = \sqrt{\frac{70}{3}K^3}$$ $$16K^2 = \frac{70}{3}K^3$$ $$\frac{24}{35} = K$$ So our answer is $24 + 35 = \boxed{059}$.

~Solution by divij04

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