Difference between revisions of "2019 AIME I Problems/Problem 15"
(Note: I put my solution as Solution 1 as I feel like it is better organized and clearer than the other one.) |
(→Solution 1) |
||
Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
− | + | <asy> | |
size(8cm); | size(8cm); | ||
pair O, A, B, P, O1, O2, Q, X, Y; | pair O, A, B, P, O1, O2, Q, X, Y; | ||
Line 22: | Line 22: | ||
draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); | draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); | ||
− | dot(" | + | dot("$O$", O, S); |
− | dot(" | + | dot("$A$", A, A); |
− | dot(" | + | dot("$B$", B, B); |
− | dot(" | + | dot("$P$", P, dir(70)); |
− | dot(" | + | dot("$Q$", Q, dir(200)); |
− | dot(" | + | dot("$O_1$", O1, SW); |
− | dot(" | + | dot("$O_2$", O2, SE); |
− | dot(" | + | dot("$X$", X, X); |
− | dot(" | + | 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>. | 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>. | ||
Revision as of 19:57, 15 March 2019
Contents
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 the requested sum is .
(Solution by TheUltimate123)
Solution 2
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 hav e . So, we have . As a result, our answer is .
Solution By BladeRunnerAUG (Fanyuchen20020715).
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.