Difference between revisions of "2019 AIME I Problems/Problem 15"
(→Solution 2) |
|||
Line 45: | Line 45: | ||
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 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> | + | Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), <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>. | 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>. | ||
Latest revision as of 18:09, 2 March 2021
Problem
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(EGMO 9.17), 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 | |
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.