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Difference between revisions of "1996 AHSME Problems/Problem 30"

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==Problem==
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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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n</math>.
  
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<math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409 </math>
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==Solution 1==
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In hexagon <math>ABCDEF</math>, let <math>AB=BC=CD=3</math> and let <math>DE=EF=FA=5</math>. Since arc <math>BAF</math> is one third of the circumference of the circle, it follows that <math>\angle BCF = \angle BEF=60^{\circ}</math>. Similarly, <math>\angle CBE =\angle CFE=60^{\circ}</math>. Let <math>P</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CF}</math>, <math>Q</math> that of <math>\overline{BE}</math> and <math>\overline{AD}</math>, and <math>R</math> that of <math>\overline{CF}</math> and <math>\overline{AD}</math>. Triangles <math>EFP</math> and  <math>BCP</math> are equilateral, and by symmetry, triangle <math>PQR</math> is isosceles and thus also equilateral.
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<asy>
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import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
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real angleUnit = 15;
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draw(Circle(origin,1));
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pair D = dir(22.5);
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pair C = dir(3*angleUnit + degrees(D));
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pair B = dir(3*angleUnit + degrees(C));
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pair A = dir(3*angleUnit + degrees(B));
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pair F = dir(5*angleUnit + degrees(A));
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pair E = dir(5*angleUnit + degrees(F));
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draw(A--B--C--D--E--F--cycle);
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dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F);
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draw(A--D^^B--E^^C--F);
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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);
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</asy>
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Furthermore, <math>\angle BAD</math> and <math>\angle BED</math> subtend the same arc, as do <math>\angle ABE</math> and <math>\angle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</cmath> It follows that <cmath>\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad
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\mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.</cmath> Solving the two equations simultaneously yields <math>AD=360/49,</math> so <math>m+n=\boxed{409}. \blacksquare</math>
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==Solution 2==
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All angle measures are in degrees.
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Let the first trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=3</math>.  Then the second trapezoid is <math>AFED</math>, where <math>AF=FE=ED=5</math>.  We look for <math>AD</math>.
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Since <math>ABCD</math> is an isosceles trapezoid, we know that <math>\angle BAD=\angle CDA</math> and, since <math>AB=BC</math>, if we drew <math>AC</math>, we would see <math>\angle BCA=\angle BAC</math>.  Anyway, <math>\widehat{AB}=\widehat{BC}=\widehat{CD}</math> (<math>\widehat{AB}</math> means arc AB).  Using similar reasoning, <math>\widehat{AF}=\widehat{FE}=\widehat{ED}</math>.
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Let <math>\widehat{AB}=2\phi</math> and <math>\widehat{AF}=2\theta</math>.  Since <math>6\theta+6\phi=360</math> (add up the angles), <math>2\theta+2\phi=120</math> and thus <math>\widehat{AB}+\widehat{AF}=\widehat{BF}=120</math>.  Therefore, <math>\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120</math>.  <math>\angle CDE=120</math> as well.
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Now I focus on triangle <math>FAB</math>.  By the Law of Cosines, <math>BF^2=3^2+5^2-30\cos{120}=9+25+15=49</math>, so <math>BF=7</math>.  Seeing <math>\angle ABF=\theta</math> and <math>\angle AFB=\phi</math>, we can now use the Law of Sines to get:
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<cmath>\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.</cmath>
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Now I focus on triangle <math>AFD</math>.  <math>\angle AFD=3\phi</math> and <math>\angle ADF=\theta</math>, and we are given that <math>AF=5</math>, so
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<cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.</cmath>
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We know <math>\sin{\theta}=\frac{5\sqrt{3}}{14}</math>, but we need to find <math>\sin{3\phi}</math>.  Using various identities, we see
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<cmath>\begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\
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&=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\
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&=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\
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&=\sin{\phi}(3-4\sin^2{\phi})\\
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&=\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}
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\end{align*}</cmath>
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Returning to finding <math>AD</math>, we remember <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.</cmath>
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Plugging in and solving, we see <math>AD=\frac{360}{49}</math>. Thus, the answer is <math>360 + 49 = 409</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>.
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==Solution 3==
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Let <math>x</math> be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides <math>a, b, c, d</math> the circumradius <math>R</math> satisfies <cmath>R^2=\frac{1}{16}\cdot\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)},</cmath> where <math>s</math> is the semiperimeter. Applying this to the trapezoid with sides <math>3, 3, 3, x</math>, we see that many terms cancel and we are left with <cmath>R^2=\frac{27}{9-x}</cmath> Similar canceling occurs for the trapezoid with sides <math>5, 5, 5, x</math>, and since the two quadrilaterals share the same circumradius, we can equate: <cmath>\frac{27}{9-x}=\frac{125}{15-x}</cmath> Solving for <math>x</math> gives <math>x=\frac{360}{49}</math>, so the answer is <math>\fbox{(E) 409}</math>.
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== See also ==
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{{AHSME box | year = 1996 | num-b = 29 | after = Last Problem}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Revision as of 14:31, 18 June 2017

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

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 3

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)
Preceded by
Problem 29 		Followed by
Last Problem
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