Difference between revisions of "2010 AIME II Problems/Problem 14"
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== Problem == | == Problem == | ||
[[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>\overline{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>. | [[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>\overline{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>. |
Revision as of 16:03, 9 August 2018
Contents
[hide]Problem
Triangle with right angle at , and . Point on 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 .
An alternate finish for this problem would be to use Power of a Point on and . By Power of a Point Theorem, . Since , we can solve for and , giving the same values and answers as above.
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 , 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 and . We obtain
Since we know , we use
Solution 3
Let be equal to . Then by Law of Sines, and . We then obtain and . Solving, we determine that . Plugging this in gives that . The answer is .
Solution 4 (The quickest and most elegant)
Let , , and . By Law of Sines,
(1), and
. (2)
Then, substituting (1) into (2), we get
The answer is . ~Rowechen
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.