2010 AIME II Problems/Problem 14
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 1
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 .
(You can derive that and similarly for by considering the expansion of equating real parts to and imaginary parts to then substituting with to finish. ~happypi31415)
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
Solution 5
Let . Then, and . Let the foot of the angle bisector of on side be . Then,
and due to the angles of these triangles.
Let . By the Angle Bisector Theorem, , so . Moreover, since , by similar triangle ratios, . Therefore, .
Construct the perpendicular from to and denote it as . Denote the midpoint of as . Since is an angle bisector, is congruent to , so .
Also, . Thus, . After some major cancellation, we have , which is a quadratic in . Thus, .
Taking the negative root implies , contradiction. Thus, we take the positive root to find that . Thus, , and our desired ratio is .
The answer is .
Solution 6
Let be the circumcenter of . Since is a right triangle, will be on and . Let .
Next, let's do some angle chasing. Label , and . Thus, , and by isosceles triangles, . Then, by angle subtraction, .
Using the Law of Sines: Using trigonometric identies, . Plugging this back into the Law of Sines formula gives us:
Next, using the Law of Cosines: Substituting gives us:
Solving for x gives
Finally: , which gives us an answer of . ~adyj
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 |
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