Difference between revisions of "2010 AIME II Problems/Problem 14"
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Let <math>O</math> be the [[circumcenter]] of <math>ABC</math> and let the intersection of <math>CP</math> with the [[circumcircle]] be <math>D</math>. It now follows that <math>\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>. | Let <math>O</math> be the [[circumcenter]] of <math>ABC</math> and let the intersection of <math>CP</math> with the [[circumcircle]] be <math>D</math>. It now follows that <math>\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>. | ||
− | Denote <math>E</math> the projection of <math>O</math> onto <math>CD</math>. Now <math>CD = CP + DP = 3</math>. By the [[ | + | Denote <math>E</math> the projection of <math>O</math> onto <math>CD</math>. Now <math>CD = CP + DP = 3</math>. By the [[Pythagorean Theorem]], <math>OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}</math>. Now note that <math>EP = \frac {1}{2}</math>. By the Pythagorean Theorem, <math>OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}</math>. Hence it now follows that, |
<cmath>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</cmath> | <cmath>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</cmath> | ||
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Let <math>\angle{ACP}</math> be equal to <math>x</math>. Then by Law of Sines, <math>PB = -\frac{\cos{x}}{\cos{3x}}</math> and <math>AP = \frac{\sin{x}}{\sin{3x}}</math>. We then obtain <math>\cos{3x} = 4\cos^3{x} - 3\cos{x}</math> and <math>\sin{3x} = 3\sin{x} - 4\sin^3{x}</math>. Solving, we determine that <math>\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}</math>. Plugging this in gives that <math>\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}</math>. The answer is <math>\boxed{007}</math>. | Let <math>\angle{ACP}</math> be equal to <math>x</math>. Then by Law of Sines, <math>PB = -\frac{\cos{x}}{\cos{3x}}</math> and <math>AP = \frac{\sin{x}}{\sin{3x}}</math>. We then obtain <math>\cos{3x} = 4\cos^3{x} - 3\cos{x}</math> and <math>\sin{3x} = 3\sin{x} - 4\sin^3{x}</math>. Solving, we determine that <math>\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}</math>. Plugging this in gives that <math>\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}</math>. The answer is <math>\boxed{007}</math>. | ||
+ | |||
+ | |||
+ | (You can derive that <math>\cos{3x} = 4\cos^3{x} - 3\cos{x},</math> and similarly for <math>\sin{3x},</math> by considering the expansion of <math>(\text{cis}(x))^3,</math> equating real parts to <math>\cos{x}</math> and imaginary parts to <math>\sin{x},</math> then substituting with <math>1-\sin^2{x}</math> to finish. ~happypi31415) | ||
==Solution 4 (The quickest and most elegant)== | ==Solution 4 (The quickest and most elegant)== | ||
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Let <math>\alpha=\angle{ACP}</math>, <math>\beta=\angle{ABC}</math>, and <math>x=BP</math>. By Law of Sines, | 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{x}{sin(90-\alpha)}\implies sin(\beta)=\frac{cos(\alpha)}{x}</math> (1), and | + | <math>\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}</math> (1), and |
− | <math>\frac{4-x}{sin(\alpha)}=\frac{ | + | <math>\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}</math>. (2) |
Then, substituting (1) into (2), we get | Then, substituting (1) into (2), we get | ||
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The answer is <math>\boxed{007}</math>. | The answer is <math>\boxed{007}</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | Let <math>O</math> be the circumcenter of <math>\triangle ABC</math>. Since <math>\triangle ABC</math> is a right triangle, <math>O</math> will be on <math>\overline{AB}</math> and <math>\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2</math>. Let <math>\overline{OP} = x</math>. | ||
+ | |||
+ | Next, let's do some angle chasing. Label <math>\angle ACP = \theta^{\circ}</math>, and <math>\angle APC = 2\theta^{\circ}</math>. Thus, <math>\angle PAC = (180-3\theta)^{\circ}</math>, and by isosceles triangles, <math>\angle ACO = (180-3\theta)^{\circ}</math>. Then, by angle subtraction, <math>\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}</math>. | ||
+ | |||
+ | Using the Law of Sines: <cmath>\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}</cmath>Using trigonometric identies, <math>\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}</math>. Plugging this back into the Law of Sines formula gives us: <cmath>\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}</cmath> | ||
+ | |||
+ | <cmath>-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}</cmath> | ||
+ | <cmath>-4\cos (2\theta)^{\circ}=x</cmath> | ||
+ | <cmath>\cos(2\theta)^{\circ}=\frac{-x}4</cmath> | ||
+ | |||
+ | Next, using the Law of Cosines: <cmath>2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}</cmath> | ||
+ | Substituting <math>\cos(2\theta)^{\circ}=\frac{-x}4</math> gives us: | ||
+ | <cmath>2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4</cmath> | ||
+ | <cmath>4=1+x^2+\frac{x^2}{2}</cmath> | ||
+ | |||
+ | Solving for x gives <math>x=\sqrt 2</math> | ||
+ | |||
+ | Finally: <math>\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2</math>, which gives us an answer of <math>3+2+2=\boxed{007}</math>. ~adyj | ||
== See also == | == See also == |
Latest revision as of 23:31, 18 November 2023
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.