Difference between revisions of "2010 AIME II Problems/Problem 14"
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+ | == Problem 14 == | ||
+ | Triangle <math>ABC</math> with right angle at <math>C</math>, <math>\angle BAC < 45^\circ</math> and <math>AB = 4</math>. Point <math>P</math> on <math>\overbar{AB}</math> is chosen such that <math>\angle APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> can be represented in the form <math>p + q\sqrt{r}</math>, where <math>p</math>, <math>q</math>, <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>. | ||
+ | |||
+ | == Solution == | ||
+ | |||
Label the center of the circumcircle of <math>ABC</math> as <math>O</math> and the intersection of <math>CP</math> with the circumcircle as <math>D</math>. It now follows that <math>\angle{DOA} = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>. | Label the center of the circumcircle of <math>ABC</math> as <math>O</math> and the intersection of <math>CP</math> with the circumcircle as <math>D</math>. It now follows that <math>\angle{DOA} = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>. | ||
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<math>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</math> | <math>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</math> | ||
− | This gives that the answer is <math>007</math>. | + | This gives that the answer is <math>\boxed{007}</math>. |
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2010|num-b=14|num-a=15|n=I}} |
Revision as of 17:13, 3 April 2010
Problem 14
Triangle with right angle at , and . Point on $\overbar{AB}$ (Error compiling LaTeX. ! Undefined control sequence.) 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
Label the center of the circumcircle of as and the intersection of with the circumcircle as . 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 .
See also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |