Difference between revisions of "1980 AHSME Problems/Problem 16"
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== Solution == | == Solution == | ||
− | We assume the side length of the cube is <math>1</math>. The side length of the tetrahedron is <math>\sqrt2</math>, so the surface area is <math>4\times\frac{2\sqrt3}{4}=2\sqrt3</math>. The surface area of the cube is <math>6\times1\times1=6</math>, so the ratio of the surface area of the cube to the surface area of the tetrahedron is <math>\frac{6}{2\sqrt3}=\boxed{\sqrt3}</math> | + | We assume the side length of the cube is <math>1</math>. The side length of the tetrahedron is <math>\sqrt2</math>, so the surface area is <math>4\times\frac{2\sqrt3}{4}=2\sqrt3</math>. The surface area of the cube is <math>6\times1\times1=6</math>, so the ratio of the surface area of the cube to the surface area of the tetrahedron is <math>\frac{6}{2\sqrt3}=\boxed{\sqrt3}</math>. |
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+ | -aopspandy | ||
== See also == | == See also == |
Latest revision as of 19:03, 18 June 2021
Problem
Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
Solution
We assume the side length of the cube is . The side length of the tetrahedron is , so the surface area is . The surface area of the cube is , so the ratio of the surface area of the cube to the surface area of the tetrahedron is .
-aopspandy
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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