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

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 \(\frac{1}{3} \times \frac{1}{3}\)=\(\frac{1}{6}\)