Difference between revisions of "1990 AHSME Problems/Problem 9"
(Created page with "== 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 <math>\text...") |
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Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | 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 <math>\fbox{B}</math> | ||
== See also == | == See also == |
Revision as of 04:22, 4 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
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: This gives
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.