## Saturday, December 3, 2005

### That 3rd Dimension. Topic: Space Geometry. Level: AIME.

Problem: (1983 AIME - #11) $ABCD$ is a square side $6\sqrt{2}$. $EF$ is parallel to the square and has length $12\sqrt{2}$. The faces $BCF$ and $ADE$ are equilateral. What is the volume of the solid $ABCDEF$? [Source: http://www.kalva.demon.co.uk/aime/aime83.html]

Solution #1: Here's the generic cut-it-up, add-it-up solution. Divide the solid into three pieces, a triangular prism in the middle plus two tetrahedrons on the ends. Use the Pythagorean Theorem to find the length of the edges of the triangles. We find that the area of the triangular face of the prism is $18\sqrt{2}$. So the volume of the prism is

$(18\sqrt{2})(6\sqrt{2}) = 216$.

Then we have the "height" of the tetrahedrons as the remaining portions of $EF$, or $3\sqrt{2}$ each. So the volume of each is

$\frac{(18\sqrt{2})(3\sqrt{2})}{3} = 36$.

But we have two, making a total volume of $216+36+36 = 288$. QED.

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Comment: This solution is actually rather easy, especially when tackling an AIME #11. Simple applications of the Pythagorean Theorem and space geometry volume formulas kill it quickly. Consider the following solution - a more elegant approach to the problem.

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Solution #2: Consider two of these solids, connected by the square base, but with the corresponding $EF$'s perpendicular (rotate $90^{\circ}$). What shape is this? It's actually a regular tetrahedron with edge length $EF = 12\sqrt{2}$.

Well, we know the formula for the volume of a tetrahedron is $\frac{s^3\sqrt{2}}{6}$ (derive this if you have not in the past), so the volume of the two combined shapes is

$\frac{(12\sqrt{2})^3\sqrt{2}}{6} = 576$.

But we want half of that, so we get $288$, confirming our previous answer. QED.

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Practice Problem #1: Derive the formulas for the volumes of a regular tetrahedron and octahedron.

Practice Problem #2: Find the volume of the inscribed and circumscribed spheres of a regular tetrahedron of side length $1$.