Difference between revisions of "2016 AIME I Problems/Problem 15"
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We can cancel <math>30\cdot 104\cdot y</math> from both sides, finding <math>37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2</math>. Therefore, | We can cancel <math>30\cdot 104\cdot y</math> from both sides, finding <math>37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2</math>. Therefore, | ||
<cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | <cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | ||
+ | |||
+ | ==Solution 3== | ||
+ | <asy> | ||
+ | size(9cm); | ||
+ | import olympiad; | ||
+ | real R1=45,R2=67*R1/37; | ||
+ | real m1=sqrt(R1^2-23.5^2); | ||
+ | real m2=sqrt(R2^2-23.5^2); | ||
+ | pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); | ||
+ | draw(circle(o1,R1)); | ||
+ | draw(circle(o2,R2)); | ||
+ | pair q=(-R1/(R2-R1)*o2.x,0); | ||
+ | pair a=tangent(q,o1,R1,2); | ||
+ | pair b=tangent(q,o2,R2,2); | ||
+ | pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; | ||
+ | pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; | ||
+ | dot(a^^b^^x^^y^^c^^d); | ||
+ | draw(x--y); | ||
+ | draw(a--y^^b--y); | ||
+ | draw(d--x--c); | ||
+ | draw(a--b--c--d--cycle); | ||
+ | draw(x--a^^x--b); | ||
+ | label("$A$",a,NW,fontsize(9)); | ||
+ | label("$B$",b,NE,fontsize(9)); | ||
+ | label("$C$",c,SE,fontsize(9)); | ||
+ | label("$D$",d,SW,fontsize(9)); | ||
+ | label("$X$",x,2*N,fontsize(9)); | ||
+ | label("$Y$",y,3*S,fontsize(9)); | ||
+ | </asy> | ||
+ | First of all, since quadrilaterals <math>ADYX</math> and <math>XYCB</math> are cyclic, we can let <math>\angle DAX = \angle XYC = \theta</math>, and <math>\angle XYD = \angle CBX = 180 - \theta</math>, due to the properties of cyclic quadrilaterals. In addition, let <math>\angle BAX = x</math> and <math>\angle ABX = y</math>. Then, since quadrilateral <math>ABCD</math> is cyclic as well, we have the following sums: | ||
+ | <cmath>\theta + x +\angle XCY + y = 180^{\circ}</cmath> | ||
+ | <cmath>180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}</cmath> | ||
+ | Cancelling out <math>180^{\circ}</math> in the second equation isolating <math>\theta</math> yields <math>\theta = y + \angle XDY + x</math>. Substituting <math>\theta</math> back into the first equation, we obtain | ||
+ | <cmath>2x + 2y + \angle XCY + \angle XDY = 180^{\circ}</cmath> | ||
+ | Since | ||
+ | <cmath>x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}</cmath> | ||
+ | <cmath>x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}</cmath> | ||
+ | we can then imply that <math>\angle DAY = x + y</math>. Similarly, <math>\angle YBC = x + y</math>. So then <math>\angle DXY = \angle YXC = x + y</math>, so since we know that <math>XY</math> bisects <math>\angle DXC</math>, we can solve for <math>DY</math> and <math>YC</math> with Stewart’s Theorem. Let <math>DY = 37n</math> and <math>YC = 67n</math>. Then | ||
+ | <cmath>37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n</cmath> | ||
+ | <cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath> | ||
+ | <cmath>n^2 = \frac{270}{2479}</cmath> | ||
+ | Now, since <math>\angle AYX = x</math> and <math>\angle BYX = y</math>, <math>\angle AYB = x + y</math>. From there, let <math>\angle AYD = \alpha</math> and <math>\angle BYC = \beta</math>. From angle chasing we can derive that <math>\angle YDX = \angle YAX = \beta - x</math> and <math>\angle YCX = \angle YBX = \alpha - y</math>. Now it is clear that <math>\triangle DAY \sim \triangle AYB \sim \triangle YBC</math>. Therefore the length of <math>AY</math> is the geometric mean of the lengths of <math>DA</math> and <math>YB</math> (from <math>\triangle DAY \sim \triangle AYB</math>). However, due to the proportion <math>\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}</math>, the length of <math>AB</math> is the geometric mean of the lengths of <math>DY</math> and <math>YC</math>. | ||
+ | We can now simply use arithmetic to calculate <math>AB^2</math>. | ||
+ | <cmath>AB^2 = DY \cdot YC</cmath> | ||
+ | <cmath>AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}</cmath> | ||
+ | <cmath>AB^2 = \boxed{270}</cmath> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2016|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2016|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:19, 15 October 2016
Problem
Circles and intersect at points and . Line is tangent to and at and , respectively, with line closer to point than to . Circle passes through and intersecting again at and intersecting again at . The three points , , are collinear, , , and . Find .
Solution
Solution 1
By the Radical Axis Theorem concur at point .
Let and intersect at . Note that because and are cyclic, by Miquel's Theorem is cyclic as well. Thus and Thus and , so is a parallelogram. Hence and . But notice that and are similar by Similarity, so . But Hence
Solution 2
First, we note that as and have bases along the same line, . We can also find the ratio of their areas using the circumradius area formula. If is the radius of and if is the radius of , then Since we showed this to be , we see that .
We extend and to meet at point , and we extend and to meet at point as shown below. As is cyclic, we know that . But then as is tangent to at , we see that . Therefore, , and . A similar argument shows . These parallel lines show . Also, we showed that , so the ratio of similarity between and is , or rather We can now use the parallel lines to find more similar triangles. As , we know that Setting , we see that , hence , and the problem simplifies to finding . Setting , we also see that , hence . Also, as , we find that As , we see that , hence .
Applying Power of a Point to point with respect to , we find or . We wish to find .
Applying Stewart's Theorem to , we find We can cancel from both sides, finding . Therefore,
Solution 3
First of all, since quadrilaterals and are cyclic, we can let , and , due to the properties of cyclic quadrilaterals. In addition, let and . Then, since quadrilateral is cyclic as well, we have the following sums: Cancelling out in the second equation isolating yields . Substituting back into the first equation, we obtain Since we can then imply that . Similarly, . So then , so since we know that bisects , we can solve for and with Stewart’s Theorem. Let and . Then Now, since and , . From there, let and . From angle chasing we can derive that and . Now it is clear that . Therefore the length of is the geometric mean of the lengths of and (from ). However, due to the proportion , the length of is the geometric mean of the lengths of and . We can now simply use arithmetic to calculate .
See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.