Difference between revisions of "1985 AIME Problems/Problem 15"
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== Solution == | == Solution == | ||
Note that gluing two of the given polyhedra together along a hexagonal face (rotated <math>60^\circ</math> from each other) yields a [[cube (geometry) | cube]], so the volume is <math>\frac12 \cdot 12^3 = 864</math>. | Note that gluing two of the given polyhedra together along a hexagonal face (rotated <math>60^\circ</math> from each other) yields a [[cube (geometry) | cube]], so the volume is <math>\frac12 \cdot 12^3 = 864</math>. | ||
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+ | Image: [[Media:AoPS_AIME_1985.png]] | ||
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== See also == | == See also == | ||
{{AIME box|year=1985|num-b=14|after=Last Question}} | {{AIME box|year=1985|num-b=14|after=Last Question}} |
Revision as of 23:50, 31 December 2013
Problem
Three 12 cm 12 cm squares are each cut into two pieces and , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon, as shown in the second figure, so as to fold into a polyhedron. What is the volume (in ) of this polyhedron?
Solution
Note that gluing two of the given polyhedra together along a hexagonal face (rotated from each other) yields a cube, so the volume is .
Image: Media:AoPS_AIME_1985.png
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |