Difference between revisions of "2010 AIME II Problems/Problem 14"
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~Rowechen | ~Rowechen | ||
+ | ==Solution 5== | ||
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+ | Let <math>\angle{ACP}=x</math>. Then, <math>\angle{APC}=2x</math> and <math>\angle{A}=180-3x</math>. Let the foot of the angle bisector of <math>\angle{APC}</math> on side <math>AC</math> be <math>D</math>. Then, | ||
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+ | <math>CD=DP</math> and <math>\triangle{DAP}\sim{\triangle{APC}}</math> due to the angles of these triangles. | ||
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+ | Let <math>CD=a</math>. By the Angle Bisector Theorem, <math>\frac{1}{a}=\frac{AP}{AD}</math>, so <math>AD=a\cdot{AP}</math>. Moreover, since <math>CD=DP=a</math>, by similar triangle ratios, <math>\frac{AP}{a+a\cdot{AP}}=a</math>. Therefore, <math>AP = \frac{a^2}{1-a^2}</math>. | ||
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+ | Construct the perpendicular from <math>D</math> to <math>AP</math> and denote it as <math>F</math>. Denote the midpoint of <math>CP</math> as <math>M</math>. Since <math>PD</math> is an angle bisector, <math>PF</math> is congruent to <math>PM</math>, so <math>PF=\frac{1}{2}</math>. | ||
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+ | Also, <math>\triangle{DFA}\sim{\triangle{BCA}}</math>. Thus, <math>\frac{FA}{AC}=\frac{AD}{AB}\Longrightarrow\frac{\frac{a^2}{1-a^2}-\frac{1}{2}}{a+\frac{a^3}{1-a^2}}=\frac{\frac{a^3}{1-a^3}}{4}</math>. After some major cancellation, we have <math>7a^4-8a^2+2=0</math>, which is a quadratic in <math>a^2</math>. Thus, <math>a^2 = \frac{4\pm\sqrt{2}}{7}</math>. | ||
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+ | Taking the negative root implies <math>AP<BP</math>, contradiction. Thus, we take the positive root to find that <math>AP=2+\sqrt{2}</math>. Thus, <math>BP=2-\sqrt{2}</math>, and our desired ratio is <math>\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies</math>3+2\sqrt{2}<math>. The answer is </math>3+2+2=\boxed{7}$. | ||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=13|num-a=15|n=II}} | {{AIME box|year=2010|num-b=13|num-a=15|n=II}} |
Revision as of 19:01, 19 August 2018
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 . ~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 3+2\sqrt{2}3+2+2=\boxed{7}$.
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.