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Difference between revisions of "1957 AHSME Problems/Problem 17"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(A) }24 \text{ in.}}</math>
 
  
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To travel to an adjacent vertex, the fly must move <math>3</math> inches. Furthermore, because a cube only has <math>8</math> vertices, the fly can travel at most <math>8</math> times before coming to a vertex for a second time (as a consequence of the [[Pigeonhole Principle]]). Thus, the fly can travel at most for 24 inches. It is possible to reach this maximum by zig-zagging along the edges of the cube such that the fly walks over three of the four edges of each of the four side faces of the cube. Thus, our answer is <math>\boxed{\textbf{(A) }24 \text{ in.}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 08:37, 25 July 2024

Problem

A cube is made by soldering twelve $3$-inch lengths of wire properly at the vertices of the cube. If a fly alights at one of the vertices and then walks along the edges, the greatest distance it could travel before coming to any vertex a second time, without retracing any distance, is:

$\textbf{(A)}\ 24\text{ in.}\qquad \textbf{(B)}\ 12\text{ in.}\qquad \textbf{(C)}\ 30\text{ in.}\qquad \textbf{(D)}\ 18\text{ in.}\qquad\textbf{(E)}\ 36\text{ in.}$

Solution

To travel to an adjacent vertex, the fly must move $3$ inches. Furthermore, because a cube only has $8$ vertices, the fly can travel at most $8$ times before coming to a vertex for a second time (as a consequence of the Pigeonhole Principle). Thus, the fly can travel at most for 24 inches. It is possible to reach this maximum by zig-zagging along the edges of the cube such that the fly walks over three of the four edges of each of the four side faces of the cube. Thus, our answer is $\boxed{\textbf{(A) }24 \text{ in.}}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AHSME Problems and Solutions

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