2011 AMC 12A Problems/Problem 15: Difference between revisions

Revision as of 20:03, 17 July 2020

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 is 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 $2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}$

Video Solution

https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S

2011 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

@@ 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>
+==Video Solution==
+https://www.youtube.com/watch?v=u23iWcqbJlE
+~Shreyas S
 == See also ==

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2011 AMC 12A Problems/Problem 15: Difference between revisions

Revision as of 20:03, 17 July 2020

Contents

Problem

Solution

Video Solution

See also