Difference between revisions of "1996 AHSME Problems/Problem 10"
Talkinaway (talk | contribs) (Created page with "==See also== {{AHSME box|year=1996|num-b=9|num-a=11}}") |
Ishankhare (talk | contribs) (→Solution) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
+ | How many line segments have both their endpoints located at the vertices of a given [[cube]]? | ||
+ | |||
+ | <math> \text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56 </math> | ||
+ | |||
+ | __TOC__ | ||
+ | ===Solution 1=== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | A cube has <math>4</math> spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex. | ||
+ | |||
+ | Thus, there are <math>12 + 12 + 4 = 28</math> segments, and the answer is <math>\boxed{D}</math>. | ||
+ | |||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=9|num-a=11}} | {{AHSME box|year=1996|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 07:42, 25 February 2016
Problem
How many line segments have both their endpoints located at the vertices of a given cube?
Contents
Solution 1
There are choices for the first endpoint of the line segment, and choices for the second endpoint, giving a total of segments. However, both and were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really line segments, and the answer is .
In shorthand notation, we're choosing endpoints from a set of endpoints, and the answer is .
Solution 2
Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
A cube has 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 square faces, each of which has facial diagonals, for a total of .
A cube has spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
Thus, there are segments, and the answer is .
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.