Difference between revisions of "1985 AIME Problems/Problem 15"

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== Problem ==
 
== Problem ==
Three 12 cm 12 cm [[square (geometry) | squares]] are each cut into two pieces <math>A</math> and <math>B</math>, as shown in the first figure below, by joining the [[midpoint]]s of two adjacent sides. These six pieces are then attached to a [[regular polygon | regular]] [[hexagon]], as shown in the second figure, so as to fold into a [[polyhedron]]. What is the [[volume]] (in <math>\mathrm{cm}^3</math>) of this polyhedron?
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Three 12 cm <math>\times</math>12 cm [[square (geometry) | squares]] are each cut into two pieces <math>A</math> and <math>B</math>, as shown in the first figure below, by joining the [[midpoint]]s of two adjacent sides. These six pieces are then attached to a [[regular polygon | regular]] [[hexagon]], as shown in the second figure, so as to fold into a [[polyhedron]]. What is the [[volume]] (in <math>\mathrm{cm}^3</math>) of this polyhedron?
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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>.

Revision as of 01:28, 21 January 2007

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?

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 $\frac12 \cdot 12^3 = 864$.

See also