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Difference between revisions of "2011 AMC 12A Problems/Problem 15"

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== Solution ==
 
== Solution ==
  
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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 length <math> AB </math> respectively. Lastly, let the hemisphere be tangent to the triangular face <math> ABE </math> at <math> P </math>.
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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 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 $2$ rests on the base of a square pyramid of height $6$. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

$\textbf{(A)}\ 3\sqrt{2} \qquad \textbf{(B)}\ \frac{13}{3} \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ \frac{13}{2}$

Solution

Let $ABCDE$ be the pyramid with $ABCD$ as the square base. Let $O$ and $M$ be the center of square $ABCD$ and the midpoint of side $AB$ respectively. Lastly, let the hemisphere be tangent to the triangular face $ABE$ at $P$.

Notice that $\triangle EOM$ has a right angle at $O$. Since the hemisphere be tangent to the triangular face $ABE$ at $P$, $\angle EPO$ is also $90^{\circ}$. Hence $\triangle EOM$ is similar to $\triangle EPO$.

$\frac{OM}{2} = \frac{6}{EP}$

$OM = \frac{6}{EP} \times 2$

$OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2}$

The length of the square base is thus $2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}$

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
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