Difference between revisions of "2021 AIME I Problems/Problem 13"
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Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively. Set <math>X</math> as the projection of <math>O</math> onto <math>O_1O_2</math>, and denote by <math>Y</math> the intersection of <math>AB</math> with <math>O_1O_2</math>. Note that <math>\ell = XY</math>. Now recall that<cmath>d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.</cmath>Furthermore, note that<cmath>\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}</cmath>Substituting the first equality into the second one and subtracting yields<cmath>2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,</cmath>which rearranges to the desired. | Denote by <math>O_1</math> and <math>O_2</math> the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively. Set <math>X</math> as the projection of <math>O</math> onto <math>O_1O_2</math>, and denote by <math>Y</math> the intersection of <math>AB</math> with <math>O_1O_2</math>. Note that <math>\ell = XY</math>. Now recall that<cmath>d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.</cmath>Furthermore, note that<cmath>\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}</cmath>Substituting the first equality into the second one and subtracting yields<cmath>2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,</cmath>which rearranges to the desired. | ||
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+ | ==Solution 3== | ||
+ | WLOG assume <math>\omega</math> is a line. Note the angle condition is equivalent to the angle between <math>AB</math> and <math>\omega</math> being <math>60^\circ</math>. We claim the angle between <math>AB</math> and <math>\omega</math> is fixed as <math>\omega</math> varies. | ||
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+ | Proof: Perform an inversion at <math>A</math>, sending <math>\omega_1</math> and <math>\omega_2</math> to two lines <math>\ell_1</math> and <math>\ell_2</math> intersecting at <math>B'</math>. Then <math>\omega</math> is sent to a circle tangent to lines <math>\ell_1</math> and <math>\ell_2</math>, which clearly intersects <math>AB'</math> at a fixed angle. Therefore the angle between <math>AB</math> and <math>\omega</math> is fixed as <math>\omega</math> varies. | ||
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+ | Now simply take <math>\omega</math> to be a line. If <math>\omega</math> intersects <math>\omega_1</math> and <math>\omega_2</math> and <math>X,Y</math>, respectively, and the circles' centers are <math>O_1</math> and <math>O_2</math>, then the projection of <math>O_2</math> to <math>O_1X</math> at <math>F</math> gives that <math>O_2FO_1</math> is a <math>30\text{-}60\text{-}90</math> triangle. Therefore,<cmath>O_1O_2=2O_1F=2(O_1X-O_2Y)=2(961-625)=\boxed{672}.</cmath> | ||
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+ | ~spartacle | ||
+ | |||
+ | ==Solution 4== | ||
+ | Suppose we label the points as shown below: https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn. By radical axis, the tangents to <math>\omega</math> at <math>D</math> and <math>E</math> intersect on <math>AB</math>. Thus <math>PDQE</math> is harmonic, so the tangents to <math>\omega</math> at <math>P</math> and <math>Q</math> intersect at <math>X \in DE</math>. Moreover, <math>OX \parallel O_1O_2</math> because both <math>OX</math> and <math>O_1O_2</math> are perpendicular to <math>AB</math>, and <math>OX = 2OP</math> because <math>\angle POQ = 120^{\circ}</math>. Thus<cmath>O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}</cmath>by similar triangles. | ||
+ | |||
+ | ~mathman3880 | ||
==Video Solution== | ==Video Solution== |
Revision as of 20:33, 22 April 2021
Contents
Problem
Circles and with radii and , respectively, intersect at distinct points and . A third circle is externally tangent to both and . Suppose line intersects at two points and such that the measure of minor arc is . Find the distance between the centers of and .
Solution
Let and be the center and radius of , and let and be the center and radius of .
Since extends to an arc with arc , the distance from to is . Let . Consider . The line is perpendicular to and passes through . Let be the foot from to ; so . We have by tangency and . Let . Since is on the radical axis of and , it has equal power with respect to both circles, so since . Now we can solve for and , and in particular, We want to solve for . By the Pythagorean Theorem (twice): Therefore, .
Solution 2 (Official MAA, Unedited)
Denote by , , and the centers of , , and , respectively. Let and denote the radii of and respectively, be the radius of , and the distance from to the line . We claim thatwhere . This solves the problem, for then the condition implies , and then we can solve to get .
Denote by and the centers of and respectively. Set as the projection of onto , and denote by the intersection of with . Note that . Now recall thatFurthermore, note thatSubstituting the first equality into the second one and subtracting yieldswhich rearranges to the desired.
Solution 3
WLOG assume is a line. Note the angle condition is equivalent to the angle between and being . We claim the angle between and is fixed as varies.
Proof: Perform an inversion at , sending and to two lines and intersecting at . Then is sent to a circle tangent to lines and , which clearly intersects at a fixed angle. Therefore the angle between and is fixed as varies.
Now simply take to be a line. If intersects and and , respectively, and the circles' centers are and , then the projection of to at gives that is a triangle. Therefore,
~spartacle
Solution 4
Suppose we label the points as shown below: https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzRiM2JjYThjYmZlY2ViZGI0ODhjYzE4YzMyMmM0M2QyOTZlMmU5LmpwZw==&rn=MTU4ODUxMDg3XzczMDI0ODE4MTAwNjA5N184NDQzMjQxMjM3MDQ2NzQ5NjM4X24uanBn. By radical axis, the tangents to at and intersect on . Thus is harmonic, so the tangents to at and intersect at . Moreover, because both and are perpendicular to , and because . Thusby similar triangles.
~mathman3880
Video Solution
Who wanted to see animated video solutions can see this. I found this really helpful.
P.S: This video is not made by me. And solution is same like below solutions.
≈@rounak138
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.