Baltic Way 2019 · Problem 5

Algebra

The $2m$ numbers

1\cdot2,\ 2\cdot3,\ 3\cdot4,\ \ldots,\ 2m(2m+1)

are written on a blackboard, where $m\ge2$ is an integer. A move consists of choosing three numbers $a,b,c$ , erasing them from the board and writing the single number

\frac{abc}{ab+bc+ca}.

After $m-1$ such moves, only two numbers will remain on the blackboard. Supposing one of these is $\frac43$ , show that the other is larger than 4.

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Solution

Denoting the new number

g = \frac{abc}{ab + bc + ca},

its reciprocal is

\frac{1}{g} = \frac{ab + bc + ca}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.

The sum of all reciprocals is therefore invariantly

\sum_{k=1}^{2m} \frac{1}{k(k+1)} = \sum_{k=1}^{2m} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{2m+1} = \frac{2m}{2m+1}.

The other remaining number $x$ can then be calculated from the equation

\frac{1}{x} + \frac{3}{4} = \frac{2m}{2m+1},

leading to

x = \frac{4(2m+1)}{2m-3} = 4 + \frac{16}{2m-3} > 4.

Contest context

Results from Baltic Way 2019

11 teams

Mean score: 2.7 / 5
Scores of 4 or 5: 6 / 11
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
St. Petersburg	5 / 5
Poland	5 / 5
Estonia	5 / 5
Lithuania	5 / 5
Germany	5 / 5
Norway	5 / 5
Finland	0 / 5
Denmark	0 / 5
Sweden	0 / 5
Latvia	0 / 5
Iceland	0 / 5

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