<s>[INST] Harvard MIT math question Let P be a point selected uniformly at random in the cube $[0, 1]^3$. The plane parallel to x+y+z = 0
passing through P intersects the cube in a two-dimensional region R. Let t be the expected value of
the perimeter of R. If $t^2$ can be written as $\frac{a}{b}$, where a and b are relatively prime positive integers,compute 100a + b
Let me know if the solution below is right?
Let $x_1,y_1,z_1$ be any point P in the cube.
The plane that passes through the point parallel to x+y+z 0
is ($x+y+z=x_1+y_1+z_1 = t$)
This plane cuts the x axis at say (x,0,0) then x = t and the point is (t,0,0)
This plane cuts the y axis at say (0,y,0) then y = t and the point is (0,t,0)
This plane cuts the z axis at say (0,0,z) then z = t and the point is (0,0,t)
The region of intersection is an equilateral triangle whose perimeter is given to be $P = 3\sqrt{2t^2} = 3\sqrt{2}t$
Was it a fluke that i hit the correct answer or is the rationale correct?
$E(P) = 3\sqrt{2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{\sqrt{3}-x_1-y_1} (x_1+y_1+z_1)dz_1dy_1dx_1 = \frac{11}{2\sqrt{2}}$.  After extraction the answer is $12108$
 [/INST] If you take a unit cube and try to cut it along a plane $x + y + z = r$, then you will notice that the two-dimensional region can be two different shapes.
More precisely, if $r \leq 1$ or $r \geq 2$ (which are symmetric via $x, y, z \mapsto 1 - x, 1 - y, 1 - z$), then the region $R$ is a triangle with side length $\sqrt 2 r$ or $\sqrt 2 (3 - r)$, respectively. Thus $R$ has perimeter $3\sqrt 2 r$ and area $\frac {\sqrt 3} 2 r^2$ when $r \leq 1$ (and symmetrically when $r \geq 2$).
If $1 < r < 2$, then the region $R$ will be a hexagon, whose angles are all $120^\circ$ and whose side lengths are $\sqrt 2(r - 1)$ and $\sqrt 2(2 - r)$, alternatively. In this case, $R$ has perimeter $3\sqrt 2$ and area $\frac{\sqrt 3}2 (-2r^2 + 6r - 3)$.
Since the point $P$ is chosen uniformly in the cube, the probability density of the plane being $x + y + z = r$ is proportional to the area of $R$. Therefore the expected value of the perimeter is calculated as
$$t = \left(2\int_0^1 3\sqrt 2 r \cdot \frac {\sqrt 3} 2 r^2 dr + \int_1^2 3\sqrt 2 \cdot \frac{\sqrt 3}2 (-2r^2 + 6r - 3)dr\right)/\left(2\int_0^1 \frac {\sqrt 3} 2 r^2 dr + \int_1^2 \frac{\sqrt 3}2 (-2r^2 + 6r - 3)dr\right) = \frac {11} 4 \sqrt 2.$$ Therefore $t^2 = \frac{121}{8}$ and the final answer is $12108$.
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