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# Tetrahedron Problem | AMC-10A, 2011 | Problem 24

Try this beautiful problem from Geometry:Tetrahedron box from AMC-10A, 2011. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Tetrahedron.

## Tetrahedron Problem – AMC-10A, 2011- Problem 24

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

• $4\sqrt 5+3\sqrt 2$
• $4\sqrt 5+3\sqrt 2$
• $\frac{1}{6}$
• $7\sqrt 3+4\sqrt 2$
• $\frac{1}{8}$

### Key Concepts

Geometry

Tetrahedron

Pythagoras

Answer: $\frac{1}{6}$

AMC-10A (2011) Problem 24

Pre College Mathematics

## Try with Hints

We know that if we split a regular tetrahedron then it will split into eight tetrahedra that have lengths of $\frac{1}{2}$. The volume of a regular tetrahedron can be found using base area and height. Let us assume that the side length of the above tetrahedron is 1, its base area is $\frac{\sqrt{3}}{4}$, and its height can be obtaine using Pythagoras’ Theorem.Can you find out the height …..

can you finish the problem……..

Therefore the height will be $\sqrt{1^2-(\frac{\sqrt 3}{3})^2}$=$\frac{\sqrt2}{\sqrt 3}$ And the volume will be

$\frac{1}{3} .\frac{\sqrt 3}{4}.\frac{\sqrt 2}{\sqrt 3$=$\frac{\sqrt 2}{12}$

Now the actual side length of the Tetrahedron=$\sqrt 2$.Therefore the actual volume is $\frac{\sqrt 2}{12}.{\sqrt 2}^3=\frac{1}{3}$

can you finish the problem……..

There are eight small tetrahedron, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, $\frac{4}{8}=\frac{1}{2}$ of the large tetrahedra will not be inside the other large tetrahedra.

Therefore  the volume of the region formed by the intersection of the tetrahedron is $\frac{1}{3} \times \frac{1}{3}$=$\frac{1}{6}$

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