Difference between revisions of "1997 AJHSME Problems/Problem 17"
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Consider picking one point on the corner of the cube. That point has <math>3</math> "x-like" diagonals that end on each of the three planes of the cube that the vertex is part of, and <math>1</math> "y-like" diagonal that ends on the opposite vertex. Thus, each vertex has <math>3 + 1 = 4</math> diagonals associated with it. There are <math>8</math> vertices on the cube, giving a total of <math>8 \cdot 4 = 32</math> diagonals. | Consider picking one point on the corner of the cube. That point has <math>3</math> "x-like" diagonals that end on each of the three planes of the cube that the vertex is part of, and <math>1</math> "y-like" diagonal that ends on the opposite vertex. Thus, each vertex has <math>3 + 1 = 4</math> diagonals associated with it. There are <math>8</math> vertices on the cube, giving a total of <math>8 \cdot 4 = 32</math> diagonals. | ||
− | However, each diagonal was counted as both a "starting point" and an "ending point". So there are really <math>\frac{32}{2} = 16</math> diagonals, giving an answer of <math>\boxed{ | + | However, each diagonal was counted as both a "starting point" and an "ending point". So there are really <math>\frac{32}{2} = 16</math> diagonals, giving an answer of <math>\boxed{E}</math>. |
== See also == | == See also == |
Revision as of 10:17, 22 July 2018
Contents
[hide]Problem
A cube has eight vertices (corners) and twelve edges. A segment, such as , which joins two vertices not joined by an edge is called a diagonal. Segment is also a diagonal. How many diagonals does a cube have?
Solution 1
On each face, there are diagonals like . There are faces on a cube. Thus, there are diagonals that are "x-like".
Every "y-like" diagonal must connect the bottom of the cube to the top of the cube. Thus, for each of the bottom vertices of the cube, there is a different "y-like" diagonal. So there are "y-like" diagonals.
This gives a total of diagonals on the cube, which is answer .
Solution 2
There are vertices on a cube. If you pick any of these eight points and connect it to any of the other seven points, you will have segments connecting the eight points. The division by is necessary because you counted both the segment from to and the segment from to .
But not all of these segments are diagonals. Some are edges. There are edges on the top, edges on the bottom, and edges that connect the top to the bottom. So there are edges total, meaning that there are segments that are not edges. All of these segments are diagonals, and thus the answer is .
Solution 3
Consider picking one point on the corner of the cube. That point has "x-like" diagonals that end on each of the three planes of the cube that the vertex is part of, and "y-like" diagonal that ends on the opposite vertex. Thus, each vertex has diagonals associated with it. There are vertices on the cube, giving a total of diagonals.
However, each diagonal was counted as both a "starting point" and an "ending point". So there are really diagonals, giving an answer of .
See also
1997 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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