Difference between revisions of "2016 AMC 10A Problems/Problem 24"

(Problem)
Line 4: Line 4:
 
<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==
 +
<Diagram Needed>
 +
Construct quadrilateral <math>ABCD</math> on the circle with AD being the missing side (Notice that since the side length is less than the radius, it will be very small on the bottom of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on triangle <math>OBC</math> with angle <math>BOC</math> as <math>\theta</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath>
 +
To find the remaining side (<math>AD</math>), we simply have to apply law of cosines with angle <math>3\theta</math> on <math>OAD</math> since the other triangles <math>OAB</math>, <math>OBC</math>, and <math>OCD</math> are congruent. Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:46, 3 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 the 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

<Diagram Needed>

Construct quadrilateral $ABCD$ on the circle with AD being the missing side (Notice that since the side length is less than the radius, it will be very small on the bottom of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply law of cosines on triangle $OBC$ with angle $BOC$ as $\theta$. We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\] To find the remaining side ($AD$), we simply have to apply law of cosines with angle $3\theta$ on $OAD$ since the other triangles $OAB$, $OBC$, and $OCD$ are congruent. Now, to find $\cos 3\theta$, we can derive a formula that only uses $\cos\theta$: \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] Plugging in $\cos\theta=\frac{3}{4}$, we get $\cos 3\theta= -\frac{9}{16}$. Now, applying law of cosines on triangle $OAD$, we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\]

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png