2022 AIME I Problems/Problem 10
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
Solution 1
We let be the plane that passes through the spheres and and be the centers of the spheres with radii and . We take a cross-section that contains and , which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, . Since is a trapezoid, we can drop an altitude from to to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is and let the distance from to be . Then we have .
We have because of the rectangle, so . Squaring, we have . Subtracting, we get . We also notice that since we had means that and since we know that , .
We now look at our second diagram.
. Since , we have . Using Pythagorean theorem, . Therefore,
~KingRavi
Solution 2
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 3
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.