Difference between revisions of "2010 AIME II Problems/Problem 14"
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==Solution 4 (The quickest and most elegant)== | ==Solution 4 (The quickest and most elegant)== | ||
− | Let <math>\alpha=\angle{ACP}</math>, | + | Let <math>\alpha=\angle{ACP}</math>, <math>\beta=\angle{ABC}</math>, and <math>x=BP</math>. By Law of Sines, |
− | <math>\frac{1}{sin(\beta)}=\frac{ | + | <math>\frac{1}{sin(\beta)}=\frac{x}{sin(90-\alpha)}\implies sin(\beta)=\frac{cos(\alpha)}{x}</math> (1), and |
− | <math>\frac{ | + | <math>\frac{4-x}{sin(\alpha)}=\frac{4sin(\beta)}{sin(2\alpha)} \implies 4-x=\frac{2sin(\beta)}{cos(\alpha)}</math>. (2) |
Then, substituting (1) into (2), we get | Then, substituting (1) into (2), we get | ||
− | <math>4-x=\frac{2}{ | + | <math>4-x=\frac{2}{x} \implies x^2-4x+2=0 \implies x=2-\sqrt{2} \implies \frac{4-x}{x}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}</math> |
The answer is <math>\boxed{007}</math>. | The answer is <math>\boxed{007}</math>. | ||
− | |||
== See also == | == See also == |
Revision as of 16:10, 26 November 2017
Contents
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 .
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.