Maths Olympiad Prep

Track / Stage 8 / 104 of 180 #1804 of 1964

Problem 1804

IMO Shortlist mid-range; USAMO P2/P5
Combinatorics Difficulty 8.5 Prove it BMO 2010 Shortlist · Balkan Mathematical Olympiad · 2010

A grasshopper jumps on the plane from an integer point (point with all integer coordinates) to another integer point according to the following rules: His first jump is of length 98\sqrt{98}, his second jump is of length 149\sqrt{149}, his next jump is of length 98\sqrt{98}, and so on, alternatively. What is the least possible odd number of moves in which the grasshopper could return to his starting point?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since the only representations of 9898 and 149149 as sums of two squares are 72+727^2 + 7^2 and 72+1027^2 + 10^2, we conclude that every odd move of the grasshopper is of the form (x,y)(x±7,y±7)(x, y) \rightarrow (x \pm 7, y \pm 7) and every even one - of the form (x,y)(x±7,y±10)(x, y) \rightarrow (x \pm 7, y \pm 10) or (x±10,y±7)(x \pm 10, y \pm 7).
Let the starting point be (0,0)(0,0). We need the grasshopper to get in an even number of moves to some of the points (±7,±7)(\pm 7, \pm 7), e.g. to (7,7)(7,7).

After any pair of two consecutive moves the grasshopper gets from the point (a,b)(a,b) to the point (c,d)(c,d), where c{a,a±14}c \in \{a, a \pm 14\}, d{b±17,b±3}d \in \{b \pm 17, b \pm 3\} or c{a±17,a±3}c \in \{a \pm 17, a \pm 3\}, d{b,b±14}d \in \{b, b \pm 14\}. It means that after any pair of consecutive moves one of the coordinates remains the same modulo 1414, and another changes by 33 modulo 1414. This implies that each coordinate may obtain a value equivalent to 77 modulo 1414, in particular precisely the value 77, only after at least 77 pairs of moves, e.g. 0+3+3+3+3+3+3+37(mod14)0+3+3+3+3+3+3+3 \equiv 7 \pmod{14}.

To obtain a similar result for another coordinate one needs at least another 77 pairs of moves. Therefore, one needs at least 2×14=282 \times 14 = 28 moves to get the point (7,7)(7,7), and totally at least 28+1=2928 + 1 = 29 moves to get the initial point (0,0)(0,0).

An example of such 2929 moves is the following: all 1515 odd moves are as (x7,y7)(x-7, y-7), and 1414 even moves consist of six moves as (x+10,y+7)(x+10, y+7), six moves as (x+7,y+10)(x+7, y+10), one move (x+10,y7)(x+10, y-7) and one move (x7,y+10)(x-7, y+10).

29\boxed{29}

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.