1997 AJHSME Problems/Problem 17
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(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 segments within the cube. 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 , 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 | |
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