View source for 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 <math> s </math>. Consider <math> \triangle ABC </math>. It's an isosceles right triangle with hypotenuse <math> AC </math> and legs of length <math> \frac{s}{2} </math>. Thus, <math> AC=\frac{s\sqrt{2}}{2} </math>, and the side length of the octahedron is <math> \frac{s\sqrt{2}}{2} </math>. 


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


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

==See Also==
[[2006 SMT/Geometry Problems]]