Difference between revisions of "2011 AMC 12A Problems/Problem 15"
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\textbf{(E)}\ \frac{13}{2} </math> | \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 | + | 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> | + | 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> \frac{OM}{2} = \frac{6}{EP} </math> | ||
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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> | ||
+ | |||
+ | == 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 == | == See also == | ||
{{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}} | {{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:14, 5 January 2024
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 1
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
Solution 2 (Answer Choices)
Consider a cross section of the pyramid such that it is a hemisphere inscribed inside of a triangle. Let be the vertex furthest away from the hemisphere. Let be a point where the hemisphere is tangent. Let be the centre of the hemisphere. Triangle is a right triangle, since is ninety degrees. Apply the pythagorean formula to find side as Since we know the answer must have somewhere, we can remove choices B, D and E from selection. Choice C is simply the length of , which cannot be the desired length. Thus, the answer must be .
~ jaspersun
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 |
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