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>\ | + | [[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>. |
− | == Solution == | + | == Solution 1== |
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> | ||
This gives that the answer is <math>\boxed{007}</math>. | This gives that the answer is <math>\boxed{007}</math>. | ||
+ | |||
+ | An alternate finish for this problem would be to use Power of a Point on <math>BA</math> and <math>CD</math>. By Power of a Point Theorem, <math>CP\cdot PD=1\cdot 2=BP\cdot PA</math>. Since <math>BP+PA=4</math>, we can solve for <math>BP</math> and <math>PA</math>, giving the same values and answers as above. | ||
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ | <center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ | ||
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draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); | draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); | ||
</asy></center> | </asy></center> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>AC=b</math>, <math>BC=a</math> by convention. Also, Let <math>AP=x</math> and <math>BP=y</math>. Finally, let <math> \angle ACP=\theta</math> and <math> \angle APC=2\theta</math>. | ||
+ | |||
+ | We are then looking for <math> \frac{AP}{BP}=\frac{x}{y}</math> | ||
+ | |||
+ | Now, by arc interceptions and angle chasing we find that <math> \triangle BPD \sim \triangle CPA</math>, and that therefore <math> BD=yb.</math> Then, since <math> \angle ABD=\theta</math> (it intercepts the same arc as <math> \angle ACD</math>) and <math> ADB</math> is right, | ||
+ | |||
+ | <math> \cos\theta=\frac{DB}{AB}=\frac{by}{4}</math>. | ||
+ | |||
+ | |||
+ | Using law of sines on <math>APC</math>, we additionally find that <math> \frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.</math> Simplification by the double angle formula <math> \sin 2\theta=2\sin \theta\cos\theta</math> yields | ||
+ | |||
+ | <math> \cos \theta=\frac{b}{2x}</math>. | ||
+ | |||
+ | |||
+ | We equate these expressions for <math> \cos\theta</math> to find that <math> xy=2</math>. Since <math> x+y=AB=4</math>, we have enough information to solve for <math>x</math> and <math>y</math>. We obtain <math> x,y=2 \pm \sqrt{2}</math> | ||
+ | |||
+ | Since we know <math>x>y</math>, we use <math> \frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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)== | ||
+ | |||
+ | 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{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 | ||
+ | |||
+ | <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>. | ||
+ | ~Rowechen | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | 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, | ||
+ | |||
+ | <math>CD=DP</math> and <math>\triangle{DAP}\sim{\triangle{APC}}</math> due to the angles of these triangles. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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{3+2\sqrt{2}}</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 == | ||
− | {{AIME box|year=2010|num-b= | + | {{AIME box|year=2010|num-b=13|num-a=15|n=II}} |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
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.