Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.4 National olympiad Find the answer

A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$ . Find $100r^2$ .

Figure (Asymptote source)

// dragon96, replacing
// [img]http://i.imgur.com/08FbQs.png[/img]
size(140); defaultpen(linewidth(.7));
real alpha=10, x=-0.12, y=0.025, r=1/sqrt(3);
path hex=rotate(alpha)*polygon(6);
pair A = shift(x,y)*(r*dir(330+alpha)), B = shift(x,y)*(r*dir(90+alpha)), C = shift(x,y)*(r*dir(210+alpha));
pair X = (-A.x, -A.y), Y = (-B.x, -B.y), Z = (-C.x, -C.y);
int i;
pair[] H;
for(i=0; i<6; i=i+1) {
H[i] = dir(alpha+60*i);}
fill(X--Y--Z--cycle, rgb(204,255,255));
fill(H[5]--Y--Z--H[0]--cycle^^H[2]--H[3]--X--cycle, rgb(203,153,255));
fill(H[1]--Z--X--H[2]--cycle^^H[4]--H[5]--Y--cycle, rgb(255,203,153));
fill(H[3]--X--Y--H[4]--cycle^^H[0]--H[1]--Z--cycle, rgb(153,203,255));
draw(hex^^X--Y--Z--cycle);
draw(H[1]--B--H[2]^^H[3]--C--H[4]^^H[5]--A--H[0]^^A--B--C--cycle, linewidth(0.6)+linetype("5 5"));
draw(H[0]--Z--H[1]^^H[2]--X--H[3]^^H[4]--Y--H[5]);

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. Understanding the structure of the cuboctahedron:
- A cuboctahedron has 6 square faces and 8 equilateral triangle faces.
- Each edge is shared by one square and one triangle.
- Let the side length of the cuboctahedron be $s = 1$ .

2. Volume of the cuboctahedron:
- The cuboctahedron can be decomposed into a central square prism and 4 triangular pyramids.
- The height of the square prism is the distance between the centers of two opposite square faces, which is $\sqrt{2}$ .
- The height of each triangular face (equilateral triangle) is $\frac{\sqrt{3}}{2}$ .
- The height of each triangular pyramid is calculated as:
$\text{Height of pyramid} = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{3}{4} - \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
- Volume of the square prism:
$V_{\text{prism}} = 1^2 \times \sqrt{2} = \sqrt{2}$
- Volume of one triangular pyramid:
$V_{\text{pyramid}} = \frac{1}{3} \times 1 \times \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{6}$
- Total volume of the 4 triangular pyramids:
$V_{\text{4 pyramids}} = 4 \times \frac{\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$
- Total volume of the cuboctahedron:
$V_{\text{cuboctahedron}} = \sqrt{2} + \frac{2\sqrt{2}}{3} = \frac{3\sqrt{2}}{3} + \frac{2\sqrt{2}}{3} = \frac{5\sqrt{2}}{3}$

3. Volume of the octahedron:
- An octahedron can be decomposed into 2 square pyramids.
- Each pyramid has a square base with side length 1 and a height calculated as:
$\text{Height of pyramid} = \sqrt{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$
- Volume of one square pyramid:
$V_{\text{pyramid}} = \frac{1}{3} \times 1^2 \times \frac{1}{\sqrt{2}} = \frac{1}{3\sqrt{2}}$
- Total volume of the octahedron:
$V_{\text{octahedron}} = 2 \times \frac{1}{3\sqrt{2}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$

4. **Finding the ratio $r$ and calculating $100r^2$ :**
- The ratio of the volume of the octahedron to the cuboctahedron:
$r = \frac{V_{\text{octahedron}}}{V_{\text{cuboctahedron}}} = \frac{\frac{\sqrt{2}}{3}}{\frac{5\sqrt{2}}{3}} = \frac{1}{5}$
- Calculating $100r^2$ :
$100r^2 = 100 \left(\frac{1}{5}\right)^2 = 100 \times \frac{1}{25} = 4$

The final answer is $\boxed{ 4 }$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.