Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 3"
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== Problem == | == Problem == | ||
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| + | A cube that is one inch wide has | ||
| + | had its eight corners shaved off. The cube’s vertices | ||
| + | have been replaced by eight congruent equilateral triangles, | ||
| + | and the square faces have been replaced by six | ||
| + | congruent octagons. If the combined area of the eight | ||
| + | triangles equals the area of one of the octagons, what is | ||
| + | that area? (Each octagonal face has two different edge | ||
| + | lengths that occur in alternating order.) | ||
== Solution == | == Solution == | ||
| + | <math>\frac{2\sqrt{3}}{1+2\sqrt{3}}=\frac{12-2\sqrt{3}}{11}</math> | ||
== See also == | == See also == | ||
{{UNCO Math Contest box|year=2016|n=II|num-b=2|num-a=4}} | {{UNCO Math Contest box|year=2016|n=II|num-b=2|num-a=4}} | ||
| − | [[ | + | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 03:02, 13 January 2019
Problem
A cube that is one inch wide has had its eight corners shaved off. The cube’s vertices have been replaced by eight congruent equilateral triangles, and the square faces have been replaced by six congruent octagons. If the combined area of the eight triangles equals the area of one of the octagons, what is that area? (Each octagonal face has two different edge lengths that occur in alternating order.)
Solution
See also
| 2016 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
