Difference between revisions of "2021 AIME I Problems/Problem 13"
Sugar rush (talk | contribs) m (boxed the final answer) |
|||
Line 104: | Line 104: | ||
The requested distance <cmath>O_{1}O_{2}=O_{1}T+O_{2}T=u+v</cmath> is therefore equal to <math>2\cdot(961-625)=\boxed{672}</math>. | The requested distance <cmath>O_{1}O_{2}=O_{1}T+O_{2}T=u+v</cmath> is therefore equal to <math>2\cdot(961-625)=\boxed{672}</math>. | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/gN7Ocu3D62M ~Math Problem Solving Skills | ||
+ | |||
==Video Solution== | ==Video Solution== |
Revision as of 14:35, 13 June 2022
Contents
[hide]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 (Properties of Radical Axis)
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 (Linearity)
Let and be the centers of and , respectively, and let be the center of . Then, the distance from to the radical axis of is equal to . Let and the orthogonal projection of onto line . Define the function by Then By Linearity of Power of a Point, Notice that and , thus since is nonzero.
Solution 3
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 4 (Quick)
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
Solution 5 (Official MAA)
Like in other solutions, let be the center of with its radius; also, let and be the centers of and with and their radii, respectively. Let line intersect line at , and let , , . Because the lines and are perpendicular, lines and meet at a angle. Applying the Law of Cosines four times:
Adding the first and fourth equations, then subtracting the second and third equations gives us
Since lies on the radical axis of and , the power of point with respect to either circle is
Hence which simplifies to
The requested distance is therefore equal to .
Video Solution
https://youtu.be/gN7Ocu3D62M ~Math Problem Solving Skills
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.