2021 AIME II Problems/Problem 10
Contents
Problem
Two spheres with radii and one sphere with radius are each externally tangent to the other two spheres and to two different planes and . The intersection of planes and is the line . The distance from line to the point where the sphere with radius is tangent to plane is , where and are relatively prime positive integers. Find .
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
This solution refers to the Diagram section.
As shown below, let be the centers of the spheres (where sphere has radius ) and be their respective points of tangency to plane Let be the plane that is determined by and Suppose is the foot of the perpendicular from to line so is the perpendicular bisector of We wish to find Note that:
- In we get and
- Both and lie in plane Both and lie in plane
- By symmetry, since planes and are reflections of each other about plane the three planes are concurrent to line
- Since and it follows that from which and are coplanar.
Now, we focus on cross-sections and
- In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.
Clearly, cross-section intersects line at exactly one point. Furthermore, as the intersection of planes and is line we conclude that and must intersect line at the same point. Let be the point of concurrency of and line
- In cross-section let be the foot of the perpendicular from to line and be the foot of the perpendicular from to
We have the following diagram: In cross-section since as discussed, we obtain by AA, with the ratio of similitude Therefore, we get or
In cross-section note that and Applying the Pythagorean Theorem to right we have Moreover, since and we obtain so that by AA, with the ratio of similitude Therefore, we get or
Finally, note that and Since quadrilateral is a rectangle, we have Applying the Pythagorean Theorem to right gives from which the answer is
~MRENTHUSIASM
Solution 2 (Pythagorean Theorem)
The centers of the three spheres form a -- triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the side of this triangle. Take its midpoint , which is away from the midpoint of the side, and connect these two midpoints.
Now consider the point at which the plane is tangent to the small sphere, and connect with the small sphere's tangent point . Extend through until it hits the ray from through the center of the small sphere (convince yourself that these two intersect). Call this intersection , the center of the small sphere , we want to find .
By Pythagoras, , and we know that and . We know that and must be parallel, using ratios we realize that . Apply the Pythagorean theorem to , , so .
~Ross Gao
Solution 3 (Proportion)
This solution refers to the Diagram section. The isosceles triangle of centers ( is the center of sphere of radii ) has sides and
Let be the midpoint .
The isosceles triangle of points of tangency has sides and
Let be the midpoint
The height is
The tangents of the half-angle between the planes is so vladimir.shelomovskii@gmail.com, vvsss
Video Solution by Interstigation
~Interstigation
See Also
2021 AIME II (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 | ||
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