Difference between revisions of "1990 AHSME Problems/Problem 9"

(Solution)
 
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draw((0,1,1)--(1,1,1)--(1,0,1),red);
 
draw((0,1,1)--(1,1,1)--(1,0,1),red);
 
</asy>
 
</asy>
This gives <math>\fbox{B}</math>
+
This gives the answer <math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 17:40, 24 February 2016

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 the answer $\fbox{B}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 26 27 28 29 30
All AHSME Problems and Solutions

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