Difference between revisions of "1980 AHSME Problems/Problem 16"

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(Solution)
 
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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>
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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.

$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$


Solution

We assume the side length of the cube is $1$. The side length of the tetrahedron is $\sqrt2$, so the surface area is $4\times\frac{2\sqrt3}{4}=2\sqrt3$. The surface area of the cube is $6\times1\times1=6$, so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{\sqrt3}$.

-aopspandy

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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