Difference between revisions of "2019 AIME I Problems/Problem 15"
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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>. | 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 | + | 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>. |
Revision as of 21:58, 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 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 .
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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