Difference between revisions of "2021 AIME II Problems/Problem 10"
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<math>\ell</math> lies on the plane too, so these 2 lines must intersect. The point at where it intersects is -24a, 0, and 23a. We can use the distance formula again to find that a = <math>\dfrac{36}{23}</math>. Therefore, they intersect at <math>\left(-\dfrac{864}{23},0,-36\right)</math>. Since the little circle's x coordinate is -24 and the intersection point is <math>\dfrac{864}{23}</math>, we get <math>\dfrac{864}{23}</math> - 24 = <math>\dfrac{312}{23}</math>. Therefore, our answer to this problem is 312 + 23 = <math>\boxed{335}</math>. | <math>\ell</math> lies on the plane too, so these 2 lines must intersect. The point at where it intersects is -24a, 0, and 23a. We can use the distance formula again to find that a = <math>\dfrac{36}{23}</math>. Therefore, they intersect at <math>\left(-\dfrac{864}{23},0,-36\right)</math>. Since the little circle's x coordinate is -24 and the intersection point is <math>\dfrac{864}{23}</math>, we get <math>\dfrac{864}{23}</math> - 24 = <math>\dfrac{312}{23}</math>. Therefore, our answer to this problem is 312 + 23 = <math>\boxed{335}</math>. | ||
| + | |||
| + | ~Arcticturn | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=9|num-a=11}} | {{AIME box|year=2021|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:40, 28 March 2021
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 
.
Solution 1
The centers of the three spheres form a 49-49-72 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 72 side of this triangle. Take its midpoint 
, which is 36 away from the midpoint of the 72 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 B 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 AC= 
, and we know 
. We know that 
must be parallel, using ratios we realize that 
. Apply Pythagorean theorem on triangle BCD; 
, so 312 + 23 = 
-Ross Gao
Solution 2 (Coord Bash)
Let's try to see some symmetry. We can use a coordinate plane to plot where the circles are. The 2 large spheres are externally tangent, so we'll make them at 0, -36, 0 and 0, 36, 0. The center of the little sphere would be x, 0, and -23 since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that x is -24 (since 24 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 2 lines must intersect. The point at where it intersects is -24a, 0, and 23a. We can use the distance formula again to find that a = 
. Therefore, they intersect at 
. Since the little circle's x coordinate is -24 and the intersection point is 
, we get - 24 = 
. Therefore, our answer to this problem is 312 + 23 = .
~Arcticturn
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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
