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| − | ==Problem==
| + | #REDIRECT [[2016_AMC_10A_Problems/Problem_24]] |
| − | A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}.</math> Three of the sides of this quadrilateral have length <math>200.</math> What is the length of its fourth side?
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| − | <math>\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500</math>
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| − | ==Solution 1==
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| − | <asy>
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| − | pathpen = black; pointpen = black;
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| − | size(6cm);
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| − | draw(unitcircle);
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| − | pair A = D("A", dir(50), dir(50));
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| − | pair B = D("B", dir(90), dir(90));
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| − | pair C = D("C", dir(130), dir(130));
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| − | pair D = D("D", dir(170), dir(170));
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| − | pair O = D("O", (0,0), dir(-90));
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| − | draw(A--C, red);
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| − | draw(B--D, blue+dashed);
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| − | draw(A--B--C--D--cycle);
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| − | draw(A--O--C);
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| − | draw(O--B);
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| − | </asy>
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| − | Let <math>s = 200</math>. Let <math>O</math> be the center of the circle. Then <math>AC</math> is twice the altitude of <math>\triangle OBC</math>. Since <math>\triangle OBC</math> is isosceles we can compute its area to be <math>s^2 \sqrt7/4</math>, hence <math>CA = 2 \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{7/2}</math>.
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| − | Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s = 500</math>.
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| − | ==Solution 2==
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| − | Using trig. Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>.
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| − | Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>.
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| − | First, <math>\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}</math> and <math>\cos 2\theta=\frac{3}{4}</math> by Pythagorean.
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| − | <cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</cmath>
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| − | <cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(C)}\text{ 500}}</cmath>
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