2006 SMT/Geometry Problems/Problem 1

Problem

Given a cube, determine the ratio of the volume of the octahedron formed by connecting the centers of each face of the cube to the volume of the cube.

Solution

$[asy] import three; draw((0,0,0)--(0,0,1),dotted); draw((0,0,1)--(0,1,1)); draw((0,1,1)--(0,1,0)); draw((0,0,0)--(1,0,0),dotted); draw((1,0,0)--(1,1,0)); draw((1,1,0)--(0,1,0)); draw((0,1,0)--(0,0,0),dotted); draw((1,0,0)--(1,0,1)); draw((1,0,1)--(0,0,1)); draw((1,1,0)--(1,1,1)); draw((1,1,1)--(1,0,1)); draw((1,1,1)--(0,1,1)); draw((1,0.5,0.5)--(0.5,1,0.5)); draw((0.5,1,0.5)--(0.5,0.5,1)); draw((1,0.5,0.5)--(0.5,0.5,1)); draw((1,0.5,0.5)--(0.5,0.5,0)); draw((0.5,0.5,0)--(0.5,1,0.5)); draw((0.5,0.5,0)--(0.5,0,0.5),dotted); draw((0.5,0,0.5)--(1,0.5,0.5)); draw((0.5,0.5,0)--(0,0.5,0.5),dotted); draw((0.5,0,0.5)--(0.5,0.5,1)); draw((0.5,0.5,1)--(1,0.5,1)); draw((1,0.5,1)--(1,0.5,0.5)); label("$A$",(0.5,0.5,1),N); label("$B$",(1,0.5,1),NNW); label("$C$",(1,0.5,0.5),SW); [/asy]$

Let the side length of the square be $s$ . Consider $\triangle ABC$ . It's an isosceles right triangle with hypotenuse $AC$ and legs of length $\frac{s}{2}$ . Thus, $AC=\frac{s\sqrt{2}}{2}$ , and the side length of the octahedron is $\frac{s\sqrt{2}}{2}$ .

Now consider the top half of the octahedron. It's a pyramid with a square base of length $\frac{s\sqrt{2}}{2}$ and height $\frac{s}{2}$ . Therefore, its volume is $\frac{1}{3}\left(\frac{s}{2}\right)\left(\frac{s\sqrt{2}}{2}\right)^2=\frac{s^3}{12}$ , and the volume of the entire octahedron is twice this, or $\frac{s^3}{6}$ .

Finally, the ratio of the volume of the octahedron to the volume of the cube is $\frac{\frac{s^3}{6}}{s^3}=\boxed{\frac{1}{6}}$ .

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2006 SMT/Geometry Problems/Problem 1

Problem

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