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 $\times$12 cm squares are each cut into two pieces $A$ and $B$, 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 $\mathrm{cm}^3$) of this polyhedron?

AIME 1985 Problem 15.png

Solution

Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\circ$ from each other) yields a cube, so the volume is $\frac12 \cdot 12^3 = 864$.

Image: Media:AoPS_AIME_1985.png

See also

1985 AIME (ProblemsAnswer KeyResources)
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