Volume and surface area of a drilled out cube (BM01 2010/11 Contest Question 2)

Question

Let $s$ be an integer greater than $6$. A solid cube of side $s$ has a square hole of side $x < 6$ drilled directly through from one face to the opposite face (so the drill removes a cuboid). The volume of the remaining solid is numerically equal to the total surface area of the remaining solid. Determine all possible integer values of $x$.

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I've figured out the first bit, setting the equations. Would it be easier if I said "Find the solutions of $s$ and $x$ such that $s^3-x^2s=6s^2-2x^2+4xs$"? (bearing in mind that $s$ must be $>6$ and an integer and $x$ must be $<6$).

Answer 1

So we just need to write an equation and solve for $x$. The remaining volume is easy: It's just $s^3-sx^2$. The surface area is a little more complicated, but not too hard. We have $6s^2-2x^2$ for the outside surface area and $4sx$ for the inside surface area.

Hence we have the equation

$$s^3-sx^2=6s^2-2x^2+4sx$$

which is the same as the one you got.

Now solving the quadratic for $x$, we get

$$\begin{align} x&=\frac{-4s\pm\sqrt{16s^2-4(s-2)(6s^2-s^3)}}{2(s-2)} \\&=\frac{-4s\pm\sqrt{4s^2(4-(s-2)(6-s))}}{2(s-2)} \\&=\frac{-2s\pm s\sqrt{4-(s-2)(6-s)}}{s-2} \\&=\frac{-2s\pm s\sqrt{s^2-8s+16}}{s-2} \\&=\frac{-2s\pm s(s-4)}{s-2} \end{align}$$

Note that we can discard the negative branch because it gives a negative result for $x$. Hence $$\begin{align} x&=\frac{-2s+ s(s-4)}{s-2} \\&=\frac{s^2-6s}{s-2} \\&=s-\frac{4s}{s-2} \end{align}$$

This leaves us the problem to solve for what integers $s>6$ does $(s-2)\mid 4s$. To do this, note that $4(s-2)<4s$ and $5(s-2)=5s-10$ is greater than $4s$ whenever $s>10$, so it suffices to check the integers $s=7,8,9,10$.

Doing this we get the single solution $s=10$ which gives us $x=5$.

Answer 2

Initial area of the cube: $s^3$

Area removed: $s\cdot x^2$

Area remaining:$ s^3-sx^2$

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Initial suface area of cube: $6s^2$

We remove $2x^2$ from the surface area but then add $4sx$ surface area. so we have surface area: $6s^2-2x^2+4sx$.

Hence we want $s^3-sx^2=6s^2-2x^2+4sx$, that is $(s-2)x^2+(4s)x+(6s^2-s^3)$. Which is a quadratic equation in $x$.

It has solutions: $x= \frac{-4 s\pm \sqrt{16 s^2-4 (s-2) (6 s^2-s^3)}}{2 (s-2)} $

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Inspecting this function on $s$:The negative branch is negative when $s>6$ The positive branch of the function is $0$ when $s=6$ and it is an increasing function in the interval $(6,\infty)$. If you subsititute for values between $1$ and $10$ you will find that the only solution is with $s=10$ which gives $x=5$. And since it is increasing you are done.