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 18:03, 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 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
