Difference between revisions of "1980 AHSME Problems/Problem 16"
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draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black); | draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle,black); |
Latest revision as of 17:54, 15 August 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.
Solution
Without loss of generality, suppose the cube has side length . By the Pythagorean Theorem, the side length of the tetrahedron is
. Therefore, each face of the tetrahedron has area
and its total 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, edited by j314andrews
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.