Difference between revisions of "2021 Fall AMC 10A Problems/Problem 22"
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Revision as of 17:44, 23 November 2021
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
?
Solution 1 (Coordinates)
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
Solution 2 (Cross section & angle bisector)
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:
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.