Difference between revisions of "2019 AIME I Problems/Problem 15"
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==Solution 4== | ==Solution 4== | ||
− | Note that the tangents to | + | Note that the tangents to <math>\omega</math> 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 ZAQ</math> and <math>\angle ZBA = \angle ZQB</math>, we have |
<cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath> | <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+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>. | 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+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>. |
Revision as of 11:39, 19 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 , and the requested sum is .
(Solution by TheUltimate123)
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 hav e . So, we have . As a result, our answer is .
Solution By BladeRunnerAUG (Fanyuchen20020715).
Solution 4
Note that the tangents to 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 .
Solution 5 (Lazy)
First we solve for with PoAP, . 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 .
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 |
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