\section*{Problem 3 - 071033}

Inge says to her sister Monika:

"We performed calculations on a square pyramid in math class yesterday and obtained the same numerical values for the volume and the surface area.

I still remember that all the numerical values were natural numbers, but I can't recall what they were."

"Which numerical values did you mean when you said 'all numerical values'?"

"I meant the numerical values of the side length of the base, the height, the volume, and the surface area of the pyramid."

"Were these quantities labeled with matching units, for example, were the lengths in cm, the surface area in \(\mathrm{cm}^{2}\), and the volume in \(\mathrm{cm}^{3}\)?"

"Yes, that's how it was."

From these statements, Monika can reconstruct the problem. How can this be done?
}

We assume a straight square pyramid, otherwise the problem is not uniquely solvable.

Let \(a \neq 0\) be the length of the side of the base and \(h \neq 0\) be the height of the pyramid. Then the volume \(V\) is given by \(V=\frac{1}{3} \cdot a^{2} \cdot h\) and the surface area is

\[
A=a^{2}+4 \cdot \frac{1}{2} \cdot a \cdot \sqrt{h^{2}+\left(\frac{a}{2}\right)^{2}}=a^{2}+a \cdot \sqrt{4 h^{2}+a^{2}}
\]

From \(V=A\) it follows that \(\frac{1}{3} \cdot a^{2} \cdot h=a^{2}+a \cdot \sqrt{4 h^{2}+a^{2}}\), or after dividing by \(\frac{a}{3}\) and rearranging, \(a \cdot(h-3)=3 \cdot \sqrt{4 h^{2}+a^{2}}\). Squaring yields

\[
a^{2} \cdot(h-3)^{2}=9 \cdot\left(4 h^{2}+a^{2}\right) \quad \text { or } \quad a^{2} \cdot\left(h^{2}-6 h+9\right)=36 h^{2}+9 a^{2}
\]
which, after subtracting \(9 a^{2}\) and dividing by \(h\), gives the equation \(a^{2} \cdot h-6 a^{2}=36 h\), thus \(h \cdot\left(a^{2}-36\right)=6 a^{2}\) and hence \(h=\frac{6 a^{2}}{a^{2}-36}\).

Since \(h\) is a natural number, the denominator \(a^{2}-36\) must be a divisor of the numerator \(6 a^{2}\). Therefore, \(a^{2}-36\) must also be a divisor of \(6 a^{2}-6 \cdot\left(a^{2}-36\right)=216\).

For every divisor \(t\) of 216 that is divisible by 2 but not by 4, \(t+36\) would be even but not divisible by 4, and thus not a perfect square. Similarly, we can also exclude the divisors \(t\) of 216 that are divisible by 3 but not by 9, as in that case \(t+36\) would not be a perfect square \(a^{2}\).

The remaining divisors are 1, 9, 27, 4, 36, 108, 8, 72, and 216. Of these, only \(t=108\) satisfies that \(t+36\) is a perfect square, namely \(t+36=144=12^{2}=a^{2}\). Therefore,

\[
a=12 \quad \text { and } \quad h=\frac{6 a^{2}}{a^{2}-36}=\frac{6 \cdot 12^{2}}{12^{2}-36}=\frac{6 \cdot 12}{12-3}=8
\]

Indeed, for these values of the side length of the base \(a=12\) and the height of the pyramid \(h=8\), the volume of the pyramid is \(V=\frac{1}{3} \cdot a^{2} \cdot h=384\) and the surface area is

\[
A=a^{2}+a \cdot \sqrt{4 h^{2}+a^{2}}=144+12 \cdot \sqrt{256+144}=144+12 \cdot 20=384=V
\]