2001 AMC 10 Problems/Problem 21
Contents
Problem
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude
, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
Solution 1 (video solution)
Solution 2
Let the diameter of the cylinder be . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence,
which we solve to find
. Our answer is
.
Solution 3 (Very similar to solution 2 but explained more)
We are asked to find the radius of the cylinder, or so we can look for similarity. We know that
and
, thus we have similarity between
and
by
similarity.
Therefore, we can create an equation to find the length of the desired side. We know that:
Plugging in yields:
Cross multiplying and simplifying gives:
Since the problem asks us to find the radius of the cylinder, we are done and the radius of the cylinder is .
~etvat
Solution 4 (graphical)
Assume that a point on a given diameter of the cone is the point on a two-dimensional representation of the cone as shown in Solution 2. The top point of the cone is thus
and the line that goes through both points is
.
Now we create a second equation. We must choose some point on the line
such that
, which implies that the cylinder’s diameter,
, must be equal to its height,
. Solving yields
, and the radius is thus
.
Solution 5 (Similar to Solution 2)
Like in Solution 2, we draw a diagram.
Let's say you don't know similar triangles but still want to solve this problem.
Trivia
This problem appeared in AoPS's Introduction to Geometry as a challenge problem.
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.