|
|
| Line 1: |
Line 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 <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
| |
| − | ==Solution==
| |
| − | 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>.
| |
| | | | |
| − | 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>.
| |
| − |
| |
| − | 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.
| |
| − |
| |
| − | 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:
| |
| − | <cmath>\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.</cmath>
| |
| − |
| |
| − | 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
| |
| − | <cmath>\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.</cmath>
| |
| − | 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
| |
| − | <cmath>\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*}</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>.
| |
