Difference between revisions of "2011 AMC 12A Problems/Problem 15"
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− | Let <math> ABCDE </math> be the pyramid with <math> ABCD </math> as the square base. Let <math> O </math> and <math> M </math> be the center of square <math> ABCD </math> and the midpoint of | + | Let <math> ABCDE </math> be the pyramid with <math> ABCD </math> as the square base. Let <math> O </math> and <math> M </math> be the center of square <math> ABCD </math> and the midpoint of side <math> AB </math> respectively. Lastly, let the hemisphere be tangent to the triangular face <math> ABE </math> at <math> P </math>. |
Notice that <math> \triangle EOM </math> has a right angle at <math> O </math>. Since the hemisphere be tangent to the triangular face <math> ABE </math> at <math> P </math>, <math>\angle EPO </math> is also <math> 90^{\circ} </math>. Hence <math> \triangle EOM </math> is similar to <math>\triangle EPO </math>. | Notice that <math> \triangle EOM </math> has a right angle at <math> O </math>. Since the hemisphere be tangent to the triangular face <math> ABE </math> at <math> P </math>, <math>\angle EPO </math> is also <math> 90^{\circ} </math>. Hence <math> \triangle EOM </math> is similar to <math>\triangle EPO </math>. |
Revision as of 19:45, 9 August 2011
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 be tangent to the triangular face at , is also . Hence is similar to .
The length of the square base is thus
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 |