Difference between revisions of "1996 AHSME Problems/Problem 10"

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==Problem==
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How many line segments have both their endpoints located at the vertices of a given cube?
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<math> \text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56 </math>
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==Solution 1==
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There are <math>8</math> choices for the first endpoint of the line segment, and <math>7</math> choices for the second endpoint, giving a total of <math>8\cdot 7 = 56</math> segments.  However, both <math>\overline{AB}</math> and <math>\overline{BA}</math> were counted, while they really are the same line segment.  Every segment got double counted in a similar manner, so there are really <math>\frac{56}{2} = 28</math> line segments, and the answer is <math>\boxed{D}</math>.
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In shorthand notation, we're choosing <math>2</math> endpoints from a set of <math>8</math> endpoints, and the answer is <math>\binom{8}{2} = \frac{8!}{6!2!} = 28</math>.
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==Solution 2==
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Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
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A cube has <math>12</math> edges:  Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.
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A cube has <math>6</math> square faces, each of which has <math>2</math> facial diagonals, for a total of <math>6\cdot 2 = 12</math>.
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A cube has <math>4</math> spacial diagonals:  each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
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Thus, there are <math>12 + 12 + 4 = 28</math> segments, and the answer is <math>\boxed{D}</math>.
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=9|num-a=11}}
 
{{AHSME box|year=1996|num-b=9|num-a=11}}

Revision as of 20:48, 18 August 2011

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 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AHSME Problems and Solutions
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