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 "1996 AHSME Problems/Problem 10"

m (wikify)
Line 29: Line 29:
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Revision as of 13:07, 5 July 2013

Problem

How many line segments have both their endpoints located at the vertices of a given cube?

$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$

Solution

Solution 1

There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments. However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{D}$.

In shorthand notation, we're choosing $2$ endpoints from a set of $8$ endpoints, and the answer is $\binom{8}{2} = \frac{8!}{6!2!} = 28$.

Solution 2

Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.

A cube has $12$ edges: Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.

A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$.

A cube has $4$ spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.

Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{D}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_10&oldid=56477"