Difference between revisions of "2022 AIME I Problems/Problem 10"
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== Diagrams == | == Diagrams == | ||
+ | <asy> | ||
+ | size(250); | ||
+ | pair A, B, OA, OB; | ||
+ | B = (0,0); | ||
+ | A = (-\sqrt{560},0); | ||
+ | OB = (0,-8); | ||
+ | OA = (-\sqrt{560,-4); | ||
+ | |||
+ | draw(circle(OB,13)); | ||
+ | draw(circle(OA,11)); | ||
+ | |||
+ | </asy> | ||
==Solution 1== | ==Solution 1== |
Revision as of 12:17, 21 February 2022
Contents
Problem
Three spheres with radii , , and are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at , , and , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that . Find .
Diagrams
size(250); pair A, B, OA, OB; B = (0,0); A = (-\sqrt{560},0); OB = (0,-8); OA = (-\sqrt{560,-4); draw(circle(OB,13)); draw(circle(OA,11)); (Error making remote request. Unknown error_msg)
Solution 1
Let the distance between the center of the sphere to the center of those circular intersections as separately. . According to the problem, we have . After solving we have , plug this back to
The desired value is
~bluesoul
Solution 2
Denote by the radius of three congruent circles formed by the cutting plane. Denote by , , the centers of three spheres that intersect the plane to get circles centered at , , , respectively.
Because three spheres are mutually tangent, , .
We have , , .
Because and are perpendicular to the plane, is a right trapezoid, with .
Hence,
Recall that
Hence, taking , we get
Solving (1) and (3), we get and .
Thus, .
Thus, .
Because and are perpendicular to the plane, is a right trapezoid, with .
Therefore,
In our solution, we do not use the conditio that spheres and are externally tangent. This condition is redundant in solving this problem.
~Steven Chen (www.professorcheneeu.com)
Video Solution
https://www.youtube.com/watch?v=SqLiV2pbCpY&t=15s
~Steven Chen (www.professorcheneeu.com)
Video Solution 2 (Mathematical Dexterity)
https://www.youtube.com/watch?v=HbBU13YiopU
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.