Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Find the answer

Given that OO is a regular octahedron, that CC is the cube whose vertices are the centers of the faces of OO, and that the ratio of the volume of OO to that of CC is mn\frac{m}{n}, where mm and nn are relatively prime integers, find m+nm+n.

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

Solution

1. Set the side length of the octahedron to 1 for simplicity.
- The volume of a regular octahedron with side length a a is given by:
VO=23a3 V_O = \frac{\sqrt{2}}{3} a^3
- For a=1 a = 1 :
VO=23 V_O = \frac{\sqrt{2}}{3}

2. Determine the height of the octahedron.
- An octahedron can be divided into two pyramids with a square base.
- The height of each pyramid can be found using a 45-45-90 triangle:
Height=22 \text{Height} = \frac{\sqrt{2}}{2}

3. Find the side length of the cube.
- The vertices of the cube are the centers of the faces of the octahedron.
- The center of each face of the octahedron is the circumcenter of an equilateral triangle.
- The height of an equilateral triangle with side length 1 is:
Height=32 \text{Height} = \frac{\sqrt{3}}{2}
- The circumcenter is 13\frac{1}{3} of the way up the height of the triangle:
Distance from base to circumcenter=36 \text{Distance from base to circumcenter} = \frac{\sqrt{3}}{6}
- The height of the pyramid above the circumcenter is:
Height above circumcenter=33 \text{Height above circumcenter} = \frac{\sqrt{3}}{3}

4. Calculate the side length of the cube.
- The side length of the cube is the distance between two adjacent face centers of the octahedron.
- This distance is the hypotenuse of a right triangle with legs 23\frac{\sqrt{2}}{3} and 33\frac{\sqrt{3}}{3}:
Side length of cube=(23)2+(33)2=29+39=59=53 \text{Side length of cube} = \sqrt{\left(\frac{\sqrt{2}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2} = \sqrt{\frac{2}{9} + \frac{3}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}

5. Calculate the volume of the cube.
- The volume of a cube with side length s s is s3 s^3 :
VC=(53)3=5527 V_C = \left(\frac{\sqrt{5}}{3}\right)^3 = \frac{5\sqrt{5}}{27}

6. Find the ratio of the volumes.
- The ratio of the volume of the octahedron to the volume of the cube is:
Ratio=VOVC=235527=232755=272155=27215555=271075=91025 \text{Ratio} = \frac{V_O}{V_C} = \frac{\frac{\sqrt{2}}{3}}{\frac{5\sqrt{5}}{27}} = \frac{\sqrt{2}}{3} \cdot \frac{27}{5\sqrt{5}} = \frac{27\sqrt{2}}{15\sqrt{5}} = \frac{27\sqrt{2}}{15\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{27\sqrt{10}}{75} = \frac{9\sqrt{10}}{25}

7. **Simplify the ratio and find m+n m+n .**
- The simplified ratio is:
92 \frac{9}{2}
- Therefore, m=9 m = 9 and n=2 n = 2 :
m+n=9+2=11 m+n = 9+2 = 11

The final answer is 11 \boxed{11} .

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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.