Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 3: Line 3:
  
 
<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 1==
  
 
<asy>
 
<asy>
Line 24: Line 24:
  
 
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>.
 +
 +
==Solution 2==
 +
Using trig. Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triange 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}}</math>=\frac{\sqrt{2}}{4}<math>. Similarly, the cosine is </math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}<math>.
 +
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>.
 +
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}$ by Pythagorean.
 +
<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>.
 +
<cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac5\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>

Revision as of 16:02, 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 1

[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 $CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s = 500$.

Solution 2

Using trig. Since all three sides equal $200$, they subtend three equal angles from the center. The right triange between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}$=\frac{\sqrt{2}}{4}$. Similarly, the cosine is$\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$. Since there are three sides, and since$\sin\theta=\sin\left(180-\theta\right)$,we seek to find$2r\sin 3\theta$. First,$\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$and$\cos 2\theta=\frac{3}{4}$ by Pythagorean. 

\[\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}\] (Error compiling LaTeX. Unknown error_msg)

. 

\[2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac5\sqrt{2}{}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(C)}\text{ 500}}\] (Error compiling LaTeX. Unknown error_msg)
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_21&oldid=75467"