Difference between revisions of "2019 AIME I Problems/Problem 15"
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− | The | + | ==Problem 15== |
+ | |||
+ | Let <math>\overline{AB}</math> be a chord of a circle <math>\omega</math>, and let <math>P</math> be a point on the chord <math>\overline{AB}</math>. Circle <math>\omega_1</math> passes through <math>A</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circle <math>\omega_2</math> passes through <math>B</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>P</math> and <math>Q</math>. Line <math>PQ</math> intersects <math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ^2 = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair O, A, B, P, O1, O2, Q, X, Y; | ||
+ | O=(0, 0); | ||
+ | A=dir(140); B=dir(40); | ||
+ | P=(3A+5B)/8; | ||
+ | O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); | ||
+ | O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); | ||
+ | Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; | ||
+ | X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); | ||
+ | Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); | ||
+ | |||
+ | draw(circle(O, 1)); | ||
+ | draw(circle(O1, length(A-O1))); | ||
+ | draw(circle(O2, length(B-O2))); | ||
+ | draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); | ||
+ | |||
+ | dot("$O$", O, S); | ||
+ | dot("$A$", A, A); | ||
+ | dot("$B$", B, B); | ||
+ | dot("$P$", P, dir(70)); | ||
+ | dot("$Q$", Q, dir(200)); | ||
+ | dot("$O_1$", O1, SW); | ||
+ | dot("$O_2$", O2, SE); | ||
+ | dot("$X$", X, X); | ||
+ | dot("$Y$", Y, Y); | ||
+ | </asy> | ||
+ | Let <math>O_1</math> and <math>O_2</math> be the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. There is a homothety at <math>A</math> sending <math>\omega</math> to <math>\omega_1</math> that sends <math>B</math> to <math>P</math> and <math>O</math> to <math>O_1</math>, so <math>\overline{OO_2}\parallel\overline{O_1P}</math>. Similarly, <math>\overline{OO_1}\parallel\overline{O_2P}</math>, so <math>OO_1PO_2</math> is a parallelogram. Moreover, <cmath>\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,</cmath>whence <math>OO_1O_2Q</math> is cyclic. However, <cmath>OO_1=O_2P=O_2Q,</cmath>so <math>OO_1O_2Q</math> is an isosceles trapezoid. Since <math>\overline{O_1O_2}\perp\overline{XY}</math>, <math>\overline{OQ}\perp\overline{XY}</math>, so <math>Q</math> is the midpoint of <math>\overline{XY}</math>. | ||
+ | |||
+ | By Power of a Point, <math>PX\cdot PY=PA\cdot PB=15</math>. Since <math>PX+PY=XY=11</math> and <math>XQ=11/2</math>, <cmath>XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,</cmath> | ||
+ | and the requested sum is <math>61+4=\boxed{065}</math>. | ||
+ | |||
+ | (Solution by TheUltimate123) | ||
+ | |||
+ | ===Note=== | ||
+ | One may solve for <math>PX</math> first using PoAP, <math>PX = \frac{11}{2} - \frac{\sqrt{61}}{2}</math>. Then, notice that <math>PQ^2</math> is rational but <math>PX^2</math> is not, also <math>PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}</math>. The most likely explanation for this is that <math>Q</math> is the midpoint of <math>XY</math>, so that <math>XQ = \frac{11}{2}</math> and <math>PQ=\frac{\sqrt{61}}{2}</math>. Then our answer is <math>m+n=61+4=\boxed{065}</math>. One can rigorously prove this using the methods above | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let the tangents to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at <math>R</math>. Then, since <math>RA^2=RB^2</math>, <math>R</math> lies on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, which is <math>\overline{PQ}</math>. It follows that <cmath>-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).</cmath> | ||
+ | Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma, <cmath>RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,</cmath> | ||
+ | whence <math>Q=Q'</math>. Like above, <math>XP=\tfrac{11-\sqrt{61}}2</math>. Since <math>XQ=\tfrac{11}2</math>, we establish that <math>PQ=\tfrac{\sqrt{61}}2</math>, from which <math>PQ^2=\tfrac{61}4</math>, and the requested sum is <math>61+4=\boxed{065}</math>. | ||
+ | |||
+ | (Solution by TheUltimate123) | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Firstly we need to notice that <math>Q</math> is the middle point of <math>XY</math>. Assume the center of circle <math>w, w_1, w_2</math> are <math>O, O_1, O_2</math>, respectively. Then <math>A, O_2, O</math> are collinear and <math>O, O_1, B</math> are collinear. Link <math>O_1P, O_2P, O_1Q, O_2Q</math>. Notice that, <math>\angle B=\angle A=\angle APO_2=\angle BPO_1</math>. As a result, <math>PO_1\parallel O_2O</math> and <math>QO_1\parallel O_2P</math>. So we have parallelogram <math>PO_2O_1O</math>. So <math>\angle O_2PO_1=\angle O</math> Notice that, <math>O_1O_2\bot PQ</math> and <math>O_1O_2</math> divide <math>PQ</math> into two equal length pieces, So we have <math>\angle O_2PO_1=\angle O_2QO_1=\angle O</math>. As a result, <math>O_2, Q, O, O_1,</math> lie on one circle. So <math>\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P</math>. Notice that <math>\angle O_1PQ+\angle O_2O_1P=90^{\circ}</math>, we have <math>\angle OQP=90^{\circ}</math>. As a result, <math>OQ\bot PQ</math>. So <math>Q</math> is the middle point of <math>XY</math>. | ||
+ | |||
+ | Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <math>x=\frac{11-\sqrt{61}}{2}=XP</math>. As a result, we have <math>PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math>. So, we have <math>PQ^2=\frac{61}{4}</math>. As a result, our answer is <math>m+n=61+4=\boxed{065}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Solution By BladeRunnerAUG (Fanyuchen20020715). | ||
+ | |||
+ | ==Solution 4== | ||
+ | Note that the tangents to the circles at <math>A</math> and <math>B</math> intersect at a point <math>Z</math> on <math>XY</math> by radical center. Then, since <math>\angle ZAB = \angle ZQA</math> and <math>\angle ZBA = \angle ZQB</math>, we have | ||
+ | <cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath> | ||
+ | so <math>ZAQB</math> is cyclic. But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ} \implies Q</math> is the midpoint of <math>XY</math>. Then, by Power of a Point, <math>PY \cdot PX = PA \cdot PB = 15</math> and it is given that <math>PY+PX = 11</math>. Thus <math>PY, PX = \frac{11 \pm \sqrt{61}}{2}</math> so <math>PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2019|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:01, 2 November 2020
Problem 15
Let be a chord of a circle , and let be a point on the chord . Circle passes through and and is internally tangent to . Circle passes through and and is internally tangent to . Circles and intersect at points and . Line intersects at and . Assume that , , , and , where and are relatively prime positive integers. Find .
Solution 1
Let and be the centers of and , respectively. There is a homothety at sending to that sends to and to , so . Similarly, , so is a parallelogram. Moreover, whence is cyclic. However, so is an isosceles trapezoid. Since , , so is the midpoint of .
By Power of a Point, . Since and , and the requested sum is .
(Solution by TheUltimate123)
Note
One may solve for first using PoAP, . Then, notice that is rational but is not, also . The most likely explanation for this is that is the midpoint of , so that and . Then our answer is . One can rigorously prove this using the methods above
Solution 2
Let the tangents to at and intersect at . Then, since , lies on the radical axis of and , which is . It follows that Let denote the midpoint of . By the Midpoint of Harmonic Bundles Lemma, whence . Like above, . Since , we establish that , from which , and the requested sum is .
(Solution by TheUltimate123)
Solution 3
Firstly we need to notice that is the middle point of . Assume the center of circle are , respectively. Then are collinear and are collinear. Link . Notice that, . As a result, and . So we have parallelogram . So Notice that, and divide into two equal length pieces, So we have . As a result, lie on one circle. So . Notice that , we have . As a result, . So is the middle point of .
Back to our problem. Assume , and . Then we have , that is, . Also, . Solve these above, we have . As a result, we have . So, we have . As a result, our answer is .
Solution By BladeRunnerAUG (Fanyuchen20020715).
Solution 4
Note that the tangents to the circles at and intersect at a point on by radical center. Then, since and , we have so is cyclic. But if is the center of , clearly is cyclic with diameter , so is the midpoint of . Then, by Power of a Point, and it is given that . Thus so and the answer is .
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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