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1996 AHSME Problems/Problem 30

Revision as of 20:07, 27 September 2016 by Spg2500 (talk | contribs) (→‎Solution 1)

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

In hexagon $ABCDEF$, let $AB=BC=CD=3$ and let $DE=EF=FA=5$. Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$. Similarly, $\angle CBE =\angle CFE=60^{\circ}$. Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$, $Q$ that of $\overline{BE}$ and $\overline{AD}$, and $R$ that of $\overline{CF}$ and $\overline{AD}$. Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. [asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); real angleUnit = 15; draw(Circle(origin,1)); pair D = dir(22.5); pair C = dir(3*angleUnit + degrees(D)); pair B = dir(3*angleUnit + degrees(C)); pair A = dir(3*angleUnit + degrees(B)); pair F = dir(5*angleUnit + degrees(A)); pair E = dir(5*angleUnit + degrees(F)); draw(A--B--C--D--E--F--cycle); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F); draw(A--D^^B--E^^C--F); label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW); [/asy]

Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$. Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, \[\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.\] It follows that \[\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.\] Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}. \blacksquare$

Solution 2

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 $\fbox{(E) 409}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

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