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Difference between revisions of "1996 AHSME Problems/Problem 30"
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==Problem==
 
==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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n</math>.
 
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>.
−
<math>\textbf{(A)}</math> 309 <math>\textbf{(B)}</math> 349 <math>\textbf{(C)}</math> 369 <math>\textbf{(D)}</math> 389 <math>\textbf{(E)}</math> 409
+
<cmath>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409 </cmath>
 
==Solution==
 
==Solution==
 
All angle measures are in degrees.
 
All angle measures are in degrees.
 
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>.
 
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>.
  
−
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> (I couldn't find a better symbol; <math>\widehat{AB}</math> means arc AB).  Using similar reasoning, <math>\widehat{AF}=\widehat{FE}=\widehat{ED}</math>.
+
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>.
  
 
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.
 
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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\end{align*}</cmath>
 
\end{align*}</cmath>
 
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>
 
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>
−
Plugging in and solving, we see <math>AD=\frac{360}{49}</math>; <math>\boxed{49}</math>; <math>\boxed{\mathbb E}</math>.
+
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>.
 +
 
 +
== See also ==
 +
{{AHSME box | year = 1996 | num-b = 29 | after = Last Problem}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 10:10, 5 August 2011

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

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 $\boxed{\textbf{(E)}}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
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All AHSME Problems and Solutions
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