Difference between revisions of "2010 AIME II Problems/Problem 14"
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</asy></center> | </asy></center> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>AC=b</math>, <math>BC=a</math> by convention. Also, Let <math>AP=x</math> and <math>BP=y</math>. Finally, let <math> \angle ACP=\theta</math> and <math> \angle APC=2\theta</math>. | ||
+ | |||
+ | We are then looking for <math> \frac{AP}{BP}=\frac{x}{y}</math> | ||
+ | |||
+ | Now, by arc interceptions and angle chasing we find that <math> \triangle BPD \sim \triangle CPA</math>, and that therefore <math> BD=yb.</math> Then, since <math> \angle ABD=\theta</math> (it intercepts the same arc as <math> \angle ACD</math>) and <math> ABD</math> is right, | ||
+ | |||
+ | <math> \cos\theta=\frac{DB}{AB}=\frac{by}{4}</math>. | ||
+ | |||
+ | |||
+ | Using law of sines on APC, we additionally find that <math> \frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.</math> Simplification by the double angle formula <math> \sin 2\theta=2\sin \theta\cos\theta</math> yields | ||
+ | |||
+ | <math> \cos \theta=\frac{b}{2x}</math>. | ||
+ | |||
+ | |||
+ | We equate these expressions for <math> \cos\theta</math> to find that <math> xy=2</math>. Since <math> x+y=AB=4</math>, we have enough information to solve for x and y. We obtain <math> x,y=2 \pm \sqrt{2}</math> | ||
+ | |||
+ | Since we know x>y, we use <math> \frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}</math> | ||
== See also == | == See also == |
Revision as of 23:56, 10 April 2010
Contents
[hide]Problem
Triangle with right angle at , and . Point on $\overbar{AB}$ (Error compiling LaTeX. Unknown error_msg) is chosen such that and . The ratio can be represented in the form , where , , are positive integers and is not divisible by the square of any prime. Find .
Solution
Let be the circumcenter of and let the intersection of with the circumcircle be . It now follows that . Hence is isosceles and .
Denote the projection of onto . Now . By the pythagorean theorem, . Now note that . By the pythagorean theorem, . Hence it now follows that,
This gives that the answer is .
Solution 2
Let , by convention. Also, Let and . Finally, let and .
We are then looking for
Now, by arc interceptions and angle chasing we find that , and that therefore Then, since (it intercepts the same arc as ) and is right,
.
Using law of sines on APC, we additionally find that Simplification by the double angle formula yields
.
We equate these expressions for to find that . Since , we have enough information to solve for x and y. We obtain
Since we know x>y, we use
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |