Difference between revisions of "1990 AHSME Problems/Problem 10"
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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> | ||
| − | The best angle for cube viewing is centered on the corner, meaning three of the six faces are visible. So, therefore, the answer is just counting the number of cubes on the three faces, which is 331 or <math>\fbox{D}.</math> | + | The best angle for cube viewing is centered on the corner, meaning three of the six faces are visible. So, therefore, the answer is just counting the number of cubes on the three faces, which is <math>3 \times 10^2</math> for the inside parts of the faces, plus <math>3 \times 10</math> for the edges, plus <math>1</math> for the single shared cube in the corner, giving a total of <math>331</math> or <math>\fbox{D}.</math> |
== See also == | == See also == | ||
Latest revision as of 21:02, 2 July 2021
Problem
An 
wooden cube is formed by gluing together 
unit cubes. What is the greatest number of unit cubes that can be seen from a single point?
Solution
![[asy]import three; unitsize(1cm);size(100); draw((0,0,0)--(0,1,0),red); draw((0,0,1)--(0,0,0)--(1,0,0)--(1,0,1),red); draw((0,0,1)--(1,0,1),red); draw((0,0,1)--(0,1,1)--(0,1,0)--(1,1,0)--(1,0,0),red); draw((1,1,0)--(1,1,1),red); draw((0,1,1)--(1,1,1)--(1,0,1),red); [/asy]](http://latex.artofproblemsolving.com/7/f/1/7f14f925c508d3156122e3ba9c7c4581669707a1.png)
The best angle for cube viewing is centered on the corner, meaning three of the six faces are visible. So, therefore, the answer is just counting the number of cubes on the three faces, which is 
for the inside parts of the faces, plus 
for the edges, plus 
for the single shared cube in the corner, giving a total of 
or 
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
