Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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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== | ||
+ | <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 . Three of the sides of this quadrilateral have length . What is the length of the fourth side?
Solution
<Diagram Needed>
Construct quadrilateral 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 to and . Apply law of cosines on triangle with angle as . We get the following equation: Substituting the values in, we get Canceling out, we get To find the remaining side (), we simply have to apply law of cosines with angle on since the other triangles , , and are congruent. Now, to find , we can derive a formula that only uses : Plugging in , we get . Now, applying law of cosines on triangle , we get
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
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 |
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