Difference between revisions of "1980 AHSME Problems/Problem 16"
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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. | 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. | ||
| + | |||
| + | <asy> | ||
| + | import three; | ||
| + | unitsize(1cm); | ||
| + | size(200); | ||
| + | currentprojection=orthographic(1/2,1/3,2/5); | ||
| + | draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black); | ||
| + | draw((0,0,0)--(0,0,1),black); | ||
| + | draw((0,1,0)--(0,1,1),black); | ||
| + | draw((1,1,0)--(1,1,1),black); | ||
| + | draw((1,0,0)--(1,0,1),black); | ||
| + | draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,black); | ||
| + | draw((0,0,0)--(0,1,1),black+dashed); | ||
| + | draw((0,0,0)--(1,0,1),black+dashed); | ||
| + | draw((0,0,0)--(1,1,0),black+dashed); | ||
| + | draw((0,1,1)--(1,0,1),black+dashed); | ||
| + | draw((0,1,1)--(1,1,0),black+dashed); | ||
| + | draw((1,0,1)--(1,1,0),black+dashed); | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | </asy> | ||
<math>\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2</math> | <math>\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2</math> | ||
| + | == 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>. | ||
| − | + | -aopspandy | |
| − | |||
== See also == | == See also == | ||
Latest revision as of 14:59, 28 June 2025
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.
![[asy] import three; unitsize(1cm); size(200); currentprojection=orthographic(1/2,1/3,2/5); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black); draw((0,0,0)--(0,0,1),black); draw((0,1,0)--(0,1,1),black); draw((1,1,0)--(1,1,1),black); draw((1,0,0)--(1,0,1),black); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,black); draw((0,0,0)--(0,1,1),black+dashed); draw((0,0,0)--(1,0,1),black+dashed); draw((0,0,0)--(1,1,0),black+dashed); draw((0,1,1)--(1,0,1),black+dashed); draw((0,1,1)--(1,1,0),black+dashed); draw((1,0,1)--(1,1,0),black+dashed); [/asy]](http://latex.artofproblemsolving.com/3/1/6/3169e51fda81f66bcf382bc14056b446014d2cd6.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
