2021 AIME II Problems/Problem 10
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 .
- Let be the plane that is determined by the centers of the spheres, as shown in the black points. Clearly, the side-lengths of the black dashed triangle are and
- Plane is tangent to the spheres at the green points. Therefore, the blue dashed line segments are the radii of the spheres.
- We can conclude all of the following:
- The four black dashed line segments all lie in plane
- The four green solid line segments all lie in plane
- By symmetry, since planes and are reflections of each other about plane the three planes are concurrent to line
- The red point is the foot of the perpendicular from the smallest sphere's center to line
~MRENTHUSIASM (by Geometry Expressions)
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 Pythagorus, , and we know that and . We know that and must be parallel, using ratios we realize that . Apply the Pythagorean theorem to , , so .
Solution 2 (Coordinates Bash)
Let's try to see some symmetry. We can use an -plane to plot where the circles are. The two large spheres are externally tangent, so we'll make them at and . The center of the little sphere would be since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that (since wouldn't make sense). Now, we draw a line through the little sphere and the origin. It also intersects because of the symmetry we created.
lies on the plane too, so these two lines must intersect. The point at where it intersects is . We can use the distance formula again to find that . Therefore, they intersect at . Since the little circle's -coordinate is and the intersection point's -coordinate is , we get . Therefore, our answer to this problem is .
Solution 3 (Similar Triangles and Pythagorean Theorem)
This solution refers to the Diagram section.
As shown below, let be the centers of the spheres (where sphere is the smallest) and be their respective points of tangency to plane Suppose is the foot of the perpendicular from to line so is the perpendicular bisector of We wish to find
As the intersection of planes and is line we know that both and must intersect line Furthermore, 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. Let the intersection of and line be which must also be the intersection 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 is a rectangle, we have Applying the Pythagorean Theorem to right gives from which the answer is
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