Difference between revisions of "2021 AIME I Problems/Problem 13"
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Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>. | Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <math>A</math> and <math>B</math>. A third circle <math>\omega</math> is externally tangent to both <math>\omega_1</math> and <math>\omega_2</math>. Suppose line <math>AB</math> intersects <math>\omega</math> at two points <math>P</math> and <math>Q</math> such that the measure of minor arc <math>\widehat{PQ}</math> is <math>120^{\circ}</math>. Find the distance between the centers of <math>\omega_1</math> and <math>\omega_2</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>O_i</math> and <math>r_i</math> be the center and radius of <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>. | Let <math>O_i</math> and <math>r_i</math> be the center and radius of <math>\omega_i</math>, and let <math>O</math> and <math>r</math> be the center and radius of <math>\omega</math>. | ||
Revision as of 05:28, 23 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 1
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 (Inversion)
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 (Radical Axis, Harmonic Quadrilaterals, and Similar Triangles)
Suppose we label the points as shown here. 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.