1996 AHSME Problems/Problem 30

Problem

A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$

Solution 1

All angle measures are in degrees. Let the first trapezoid be $ABCD$ , where $AB=BC=CD=3$ . Then the second trapezoid is $AFED$ , where $AF=FE=ED=5$ . We look for $AD$ .

Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$ , if we drew $AC$ , we would see $\angle BCA=\angle BAC$ . Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ ( $\widehat{AB}$ means arc AB). Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$ .

Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$ . Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$ . Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$ . $\angle CDE=120$ as well.

Now I focus on triangle $FAB$ . By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$ , so $BF=7$ . Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$ , we can now use the Law of Sines to get: $\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.$

Now I focus on triangle $AFD$ . $\angle AFD=3\phi$ and $\angle ADF=\theta$ , and we are given that $AF=5$ , so $\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.$ We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$ , but we need to find $\sin{3\phi}$ . Using various identities, we see $\begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\ &=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\ &=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\ &=\sin{\phi}(3-4\sin^2{\phi})\\ &=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343} \end{align*}$ Returning to finding $AD$ , we remember $\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.$ Plugging in and solving, we see $AD=\frac{360}{49}$ . Thus, the answer is $360 + 49 = 409$ , which is answer choice $\boxed{\textbf{(E)}}$ .

Solution 2

Let $x$ be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides $a, b, c, d$ the circumradius $R$ satisfies $R^2=\frac{1}{16}\cdot\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)},$ where $s$ is the semiperimeter. Applying this to the trapezoid with sides $3, 3, 3, x$ , we see that many terms cancel and we are left with $R^2=\frac{27}{9-x}$ Similar canceling occurs for the trapezoid with sides $5, 5, 5, x$ , and since the two quadrilaterals share the same circumradius, we can equate: $\frac{27}{9-x}=\frac{125}{15-x}$ Solving for $x$ gives $x=\frac{360}{49}$ , so the answer is $\fbox{(E) 409}$ .

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1996 AHSME Problems/Problem 30

Contents

Problem

Solution 1

Solution 2

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