Difference between revisions of "2011 AMC 12A Problems/Problem 15"
(→See also) |
|||
Line 22: | Line 22: | ||
The length of the square base is thus <math>2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}</math> | The length of the square base is thus <math>2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=u23iWcqbJlE | ||
+ | ~Shreyas S | ||
== See also == | == See also == |
Revision as of 21:03, 17 July 2020
Contents
Problem
The circular base of a hemisphere of radius rests on the base of a square pyramid of height
. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?
Solution
Let be the pyramid with
as the square base. Let
and
be the center of square
and the midpoint of side
respectively. Lastly, let the hemisphere be tangent to the triangular face
at
.
Notice that has a right angle at
. Since the hemisphere is tangent to the triangular face
at
,
is also
. Hence
is similar to
.
The length of the square base is thus
Video Solution
https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.