1990 AHSME Problems/Problem 9

Problem

Each edge of a cube is colored either red or black. Every face of the cube has at least one black edge. The smallest number possible of black edges is

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 5\quad \text{(E) } 6$

Solution

Each black edge can only take care of two adjoining faces, so we know at least three will be needed. Once the first black edge is placed, it is easy to see that three will be sufficient, if they are separated and go in different directions: $[asy]import three; unitsize(1cm);size(100); draw((0,0,0)--(0,1,0),linewidth(2)); draw((0,0,1)--(0,0,0)--(1,0,0)--(1,0,1),red); draw((0,0,1)--(1,0,1),linewidth(2)); draw((0,0,1)--(0,1,1)--(0,1,0)--(1,1,0)--(1,0,0),red); draw((1,1,0)--(1,1,1),linewidth(2)); draw((0,1,1)--(1,1,1)--(1,0,1),red); [/asy]$ This gives $\fbox{B}$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1990 AHSME Problems/Problem 9

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