Archimedes has a two-pan balance with arms of different lengths. Aiming to weigh two kilograms of sugar, he proceeded as follows: he placed a one-kilogram weight on the left pan and sugar on the other side until the balance was level. Then, he emptied both pans, placed the one-kilogram weight on the right pan and sugar on the left pan until the pans were balanced. By adding the two quantities of sugar from these weighings, did he obtain less than two kilograms, more than two kilograms, or exactly two kilograms? Note: For balance to occur, the weights on the pans must be inversely proportional to the lengths of the corresponding arms.

![](https://cdn.mathpix.com/cropped/2024_05_01_40ea5c1759d4b041c087g-01.jpg?height=268&width=343&top_left_y=1599&top_left_x=959)
Solution

Denoting the length of the left arm by $p$, the right arm by $q$, and the amounts of sugar weighed in the first and second weighings by $x$ and $y$, respectively. Due to the balance, we can write

$$
\begin{aligned}
& 1 \cdot p=x \cdot q \\
& y \cdot p=1 \cdot q
\end{aligned}
$$

Therefore, the total amount of sugar weighed is

$$
\begin{aligned}
\frac{p}{q}+\frac{q}{p} & =\left(\frac{\sqrt{p}}{\sqrt{q}}-\frac{\sqrt{q}}{\sqrt{p}}\right)^{2}+2 \\
& >2
\end{aligned}
$$

since the arms have different lengths, we have $\left(\frac{\sqrt{p}}{\sqrt{q}}-\frac{\sqrt{q}}{\sqrt{p}}\right)^{2}>0$.