55th IMO Team Selection Test

We prove that the desired value of $\alpha$ is $\alpha =\frac{2}{51}$.

Assume that some $\alpha <\frac{2}{51}$ satisfies the condition of the problem. Choose nonnegative integer $k\ge 0$ such that $\frac{1}{51\times 2^k}\le \alpha <\frac{1}{51\times 2^{k-1}}$. Consider $51\times 2^k$ pumpkins each having weight $\frac{1}{51\times 2^k}$ tons. For any distribution of pumpkins in 50 boxes there exists a box with at least two pumpkins and therefore the weight of this box equals at least $\frac{1}{51\times 2^{k-1}}>\alpha$, a contradiction.

We show that $\alpha =\frac{2}{51}$ satisfies the condition of the problem. Let the number of pumpkins be $m$. Take $m$ empty boxes and put a pumpkin in every one of them. If the two lightest boxes contain in common not more than $\frac{2}{51}$ tons of pumpkins then gather all pumpkins of these two boxes into one of them and remove the empty box. When this operation terminates let the number of boxes be $n$ and they contain $x_1\le x_2\le \cdots \le x_n$ tons of pumpkins. Thus $x_1+x_2>\frac{2}{51}$ and hence $x_2>\frac{1}{51}$. Therefore $$1=x_1+x_2+\cdots +x_n>\frac{2}{51}+(n-2)\times \frac{1}{51},$$ implying $n<51$. We have that all the pumpkins are distributed in no more than 50 boxes and there are no more than $\frac{2}{51}$ tons of pumpkins in each box. This completes the proof.