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Difference between revisions of "2016 AMC 12A Problems/Problem 21"

(Created page with "==Problem== 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 len...")
 
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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>
 
<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>
 
==Solution==
 
==Solution==
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pathpen = black; pointpen = black;
 
pathpen = black; pointpen = black;
 
size(6cm);
 
size(6cm);
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draw(A--O--C);
 
draw(A--O--C);
 
draw(O--B);
 
draw(O--B);
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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>.
 
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>.
  
 
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s = 500</math>.
 
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s = 500</math>.

Revision as of 15:00, 4 February 2016

Problem

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}.$ Three of the sides of this quadrilateral have length $200.$ What is the length of its fourth side?

$\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500$

Solution

[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]

Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$. Since $\triangle OBC$ is isosceles we can compute its area to be $s^2 \sqrt7/4$, hence $CA = 2 \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{7/2}$.

Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s = 500$.

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