This is a math problem from the anime Assassination Classroom Season 2 Episode 12,
which is supposed to be the last question in the math exam.
Given a cube with size length of \(a\). Denote the domain \(D\) as the set of all points in the cube that is closer to the center of the cube than to any of its vertices. Find the volume of \(D\).
The following is an illustration of \(D\):
The anime gives us two directions for finding the solution, brute force by Asano and observation-based by Karma.
Asano's brute force solution:
By observation, \(D\) should be composed by 8 of these smaller solids.
Using Pythagoras theorem, we have \(s = \sqrt{2 \times (a / 4)^2} = a / \sqrt{8} \)
We can further separate the solid into \(3\) tetrahedrons (dark grey) and a right hexagonal pyramid (light grey).
Volume of 3 tetrahedrons \(= (a / 4)^2 / 2 \times (a / 2) / 3 \times 3 = a^3 / 64\)
Base area of pyramid = \(3\sqrt{3}s^2/2 = 3\sqrt{3}a^2/16\)
height of pyramid = half of the distance between center and vertex = \(\sqrt{3 \times (a / 2)^2} / 2 = \sqrt{3}a / 4\)
Volume of pyramid = \((3\sqrt{3}a^2/16 \times \sqrt{3}a/4) / 3 = 3a^3/64\)
Volume of solid = \(a^3 / 64 + 3a^3/64 = a^3/16\)
Finally, we have volume of \(D\) = \(8 \times a^3/16 = a^3/2\)
\(\blacksquare\)
Brute force solution looks very rigorous though... Let's take a look of Karma's obserbation-based (power of friendship) solution:
Notice that switching the position of the center and the vertex does not change the answer.
Using this symmetric property, we can deduce that the volume of the solid is half the volume of the cube.
Thus, we can further duduce that the volume of \(D\) is half the volume of the cube, which is \(a^3/2\).
\(\blacksquare\)
Overall, there is nothing wrong using a valid brute force geometric construction for this problem, but a solution using symmetry could be a lot cleaner and simpler.
