2021 Fall AMC 10A Problems/Problem 22
Contents
Problem
Inside a right circular cone with base radius and height are three congruent spheres with radius . Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is ?
Diagram
~MRENTHUSIASM
Solution 1 (Cross Sections and Angle Bisectors)
We can take half of a cross section of the sphere, as such: Notice that we chose a cross section where one of the spheres was tangent to the lateral surface of the cone at .
To evaluate , we will find and in terms of ; we also know that , so with this, we can solve . Firstly, to find , we can take a bird's eye view of the cone: Note that is the centroid of equilateral triangle . Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from to ; this is because medians cut each other into a to ratio. This equilateral triangle has a side length of , therefore it has an altitude of length ; two thirds of this is , so To evaluate in terms of , we will extend past point to at point . is similar to . Also, is the angle bisector of . Therefore, by the angle bisector theorem, . Also, , so , so . This means that We have that and that , so . We also were given that . Therefore, we have This is a simple linear equation in terms of . We can solve for to get
~ihatemath123
Solution 2 (Cross Sections and Areas)
Denote by , , the centers of three spheres.
Because three congruent spheres are tangent to the base of the cone, the plane formed by , , (denoted as ) is parallel to the base, with the distance .
Let be the point that the sphere with center meets the base of the cone at. Hence, .
Because three congruent spheres are mutually externally tangent to each other, is equilateral, with side length .
Let be the center of the base, be the vertex of the base. Let line and plane intersect at point . By symmetry, is the center . Hence, .
Let be the point that the sphere with center meets the side of the cone at. Hence, .
Let line and the base intersect at point .
Hence, we only need to analyze the following 2-d geometry problem: In with , , , there is an interior point whose distances to , , , are , , and , respectively. What is ?
Now, we solve this problem.
We compute the area of in two ways.
First, we have
Second, we have
These two approaches to compute should give me the same number. Hence,
~Steven Chen (www.professorchenedu.com)
Solution 3 (Coordinate Geometry)
We will use coordinates. WLOG, let the coordinates of the center of the base of the cone be the origin. Then, let the center of one of the spheres be . Note that the distance between this point and the plane given by is . Thus, by the point-to-plane distance formula, we have Solving for yields .
~ Leo.Euler
Video Solution by TheBeautyofMath
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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