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Difference between revisions of "1997 AJHSME Problems/Problem 17"

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==Solution 2==
 
==Solution 2==
  
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There are <math>8</math> 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 <math>\frac{8\cdot7}{2} = 28</math> segments connecting the eight points.  The division by <math>2</math> is necessary because you counted both the segment from <math>A</math> to <math>B</math> and the segment from <math>B</math> to <math>A</math>.
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There are <math>8</math> vertices on a cube(A, B, C, D, E, F, G, H).  If you pick one vertice, it will have 7 line segments associated with it, so you will have <math>\frac{8\cdot7}{2} = 28</math> segments within the cube.  The division by <math>2</math> is necessary because you counted both the segment from <math>A</math> to <math>B</math> and the segment from <math>B</math> to <math>A</math>.
  
 
But not all <math>28</math> of these segments are diagonals.  Some are edges.  There are <math>4</math> edges on the top, <math>4</math> edges on the bottom, and <math>4</math> edges that connect the top to the bottom.  So there are <math>12</math> edges total, meaning that there are <math>28 - 12 = 16</math> segments that are not edges.  All of these segments are diagonals, and thus the answer is <math>\boxed{E}</math>.
 
But not all <math>28</math> of these segments are diagonals.  Some are edges.  There are <math>4</math> edges on the top, <math>4</math> edges on the bottom, and <math>4</math> edges that connect the top to the bottom.  So there are <math>12</math> edges total, meaning that there are <math>28 - 12 = 16</math> segments that are not edges.  All of these segments are diagonals, and thus the answer is <math>\boxed{E}</math>.
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==Solution 3==
 
==Solution 3==
  
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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.
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Consider picking one point on the corner of the cube.  That point has <math>3</math> "x-like" diagonals , 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{E}</math>.
 
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>.
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 19:50, 26 May 2021

Problem

A cube has eight vertices (corners) and twelve edges. A segment, such as $x$, which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have?

[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution 1

On each face, there are $2$ diagonals like $x$. There are $6$ faces on a cube. Thus, there are $2\times 6 = 12$ 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 $4$ bottom vertices of the cube, there is a different "y-like" diagonal. So there are $4$ "y-like" diagonals.

This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\boxed{E}$.

Solution 2

There are $8$ vertices on a cube(A, B, C, D, E, F, G, H). If you pick one vertice, it will have 7 line segments associated with it, so you will have $\frac{8\cdot7}{2} = 28$ segments within the cube. The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$.

But not all $28$ of these segments are diagonals. Some are edges. There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom. So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges. All of these segments are diagonals, and thus the answer is $\boxed{E}$.

Solution 3

Consider picking one point on the corner of the cube. That point has $3$ "x-like" diagonals , and $1$ "y-like" diagonal that ends on the opposite vertex. Thus, each vertex has $3 + 1 = 4$ diagonals associated with it. There are $8$ vertices on the cube, giving a total of $8 \cdot 4 = 32$ diagonals.

However, each diagonal was counted as both a "starting point" and an "ending point". So there are really $\frac{32}{2} = 16$ diagonals, giving an answer of $\boxed{E}$.

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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