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

Latest revision as of 22:14, 5 January 2024

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 1

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

Solution 2 (Answer Choices)

Consider a cross section of the pyramid such that it is a hemisphere inscribed inside of a triangle. Let $A$ be the vertex furthest away from the hemisphere. Let $P$ be a point where the hemisphere is tangent. Let $O$ be the centre of the hemisphere. Triangle $AOP$ is a right triangle, since $P$ is ninety degrees. Apply the pythagorean formula to find side $AP$ as $4\sqrt{2}.$ Since we know the answer must have $\sqrt{2}$ somewhere, we can remove choices B, D and E from selection. Choice C is simply the length of $AP$ , which cannot be the desired length. Thus, the answer must be $\boxed{\textbf{A}}$ .

~ jaspersun

Video Solution

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

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ \frac{13}{2} </math>
-== Solution ==
+== Solution 1 ==
+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 is 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>.
+<math> \frac{OM}{2} = \frac{6}{EP} </math>
+<math> OM = \frac{6}{EP} \times 2 </math>
+<math> OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2} </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>
+== Solution 2 (Answer Choices) ==
+Consider a cross section of the pyramid such that it is a hemisphere inscribed inside of a triangle. Let <math>A</math> be the vertex furthest away from the hemisphere. Let <math>P</math> be a point where the hemisphere is tangent. Let <math>O</math> be the centre of the hemisphere. Triangle <math>AOP</math> is a right triangle, since <math>P</math> is ninety degrees. Apply the pythagorean formula to find side <math>AP</math> as <math>4\sqrt{2}.</math> Since we know the answer must have <math>\sqrt{2}</math> somewhere, we can remove choices B, D and E from selection. Choice C is simply the length of <math>AP</math>, which cannot be the desired length. Thus, the answer must be <math>\boxed{\textbf{A}}</math>.
+~ jaspersun
+==Video Solution==
+https://www.youtube.com/watch?v=u23iWcqbJlE
+~Shreyas S
 == See also ==
 {{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}}
+[[Category:Introductory Geometry Problems]]
+[[Category:3D Geometry Problems]]
+{{MAA Notice}}

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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