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 $100r^2$.
[asy] 
// 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]);[/asy]
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 \( \boxed{ 4 } \)