Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Original problem had a diagram of the cube.)
(→‎Solution)
Line 29: Line 29:
  
 
== 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>.
+
Without loss of generality, suppose the cube has side length <math>1</math>. By the Pythagorean Theorem, the side length of the tetrahedron is <math>\sqrt2</math>.  Therefore, each face of the tetrahedron has area <math>\frac{\left(\sqrt{2}\right)^2 \cdot \sqrt{3}}{4} = \frac{\sqrt{3}}{2}</math> and its total surface area is <math>4\cdot\frac{\sqrt3}{2}=2\sqrt3</math>. The surface area of the cube is <math>6\cdot1^2=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{(\textbf{B})\ \sqrt3}</math>.
  
−
-aopspandy
+
-aopspandy, edited by j314andrews
  
 
== See also ==
 
== See also ==

Revision as of 12:30, 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.

[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]

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

Solution

Without loss of generality, suppose the cube has side length $1$. By the Pythagorean Theorem, the side length of the tetrahedron is $\sqrt2$. Therefore, each face of the tetrahedron has area $\frac{\left(\sqrt{2}\right)^2 \cdot \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ and its total surface area is $4\cdot\frac{\sqrt3}{2}=2\sqrt3$. The surface area of the cube is $6\cdot1^2=6$, so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{(\textbf{B})\ \sqrt3}$.

-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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_16&oldid=257649"