2021 AIME I Problems/Problem 13
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 .
Quick, Olympiad-Style Solution
Let and . The solution relies on the following key claim:
Claim. is a harmonic quadrilateral.
Proof. Using the radical axis theorem, the tangents to circle at and are concurrent with line at the radical center, implying that our claim is true by harmonic quadrilateral properties.
Thus, we deduce the tangents to at and are concurrent at with line . Denote by the centers of respectively.
Now suppose . Note that giving us the pairs of similar triangles We thereby obtain since , where denotes the radius of , and as well. It follows that
Solution 1 (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 with , , and their radii, respectively. 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, Notice that and , thus Dividing both sides by (which is obviously nonzero as is nonzero) gives us so . Since and , the answer is .
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 below. 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 , , , where the length splits as . Because the lines and are perpendicular, lines and meet at a angle.
Applying the Law of Cosines to , , , and gives
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 .
Solution 6 (Geometry)
Let circle tangent circles and respectively at distinct points and . Let be the centers (the radii) of and respectively. WLOG Let be the point of such, that Let be the midpoint be the radical axes of and
Then is radical center of and
In is bisector of is median
and are collinear.
is cyclic (in is diameter as they intercept the same arc in
Since and , the answer is .
vladimir.shelomovskii@gmail.com, vvsss
Solution 7
We are not given the radius of circle , but based on the problem statement, that radius isn't important. We can set to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be so that the line containing the center of , call it , and , call it , is perpendicular to the line containing the center of , call it and . Let and . Also, let the projections of and onto line be and , respectively.
By Pythagorean Theorem on , we get Note that since , . So, . We now get two more equations from Pythag: From subtracting and , . Rearranging yields . Plugging in our result from , .
~sml1809
Solution 8 (Cheese)
Let the circle be infinitely big (a line). Then for it to be split into an arc of , must intersect at a with line .
Notice the 30-60-90 triangle in the image. .
Thus, the distance between the centers of and is
picture by afly
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution
~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.