Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
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<Diagram Needed> | <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> | 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> | ||
Revision as of 20:52, 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 1
Solution 2 (Trigonometry Bash)
<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: ![\[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\]](http://latex.artofproblemsolving.com/6/7/b/67b9eea04634c2968312ac6b4ec924102450e2ae.png)
Substituting the values in, we get ![\[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\]](http://latex.artofproblemsolving.com/3/7/8/37849b9fb56a8b33685b1fd01cc630851704ba6e.png)
Canceling out, we get ![\[\cos\theta=\frac{3}{4}\]](http://latex.artofproblemsolving.com/5/7/4/5741c56a5c288f95f9313ff7967e8ac807e71103.png)
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 
: ![\[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)\]](http://latex.artofproblemsolving.com/c/0/6/c064b137b9092a480515455a50d70685e7f7babc.png)
![\[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)\]](http://latex.artofproblemsolving.com/3/0/9/309fef4bacf1cefe1093ffc9efc7338e8f3cc8d2.png)
![\[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\]](http://latex.artofproblemsolving.com/a/6/3/a638f69803f3be5eb47caa360f20a917e66715ac.png)
Plugging in 
, we get 
. Now, applying law of cosines on triangle , we get ![\[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\]](http://latex.artofproblemsolving.com/9/e/6/9e6cccb20fb8db36966d05bfe489a301f50133c0.png)
![\[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\]](http://latex.artofproblemsolving.com/7/2/e/72e07ba400228b3fb8c8b2843a8bc2372b1ee017.png)
![\[AD=200 \cdot \frac{5}{2}=\boxed{500}\]](http://latex.artofproblemsolving.com/8/f/1/8f152870a54a68baa34070334de2813fa5b88be6.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
