Let $P_1$, $P_2$, $\ldots$, $P_{1993} = P_0$ be distinct points in the $xy$-plane with the following properties: (i) both coordinates of $P_i$ are integers, for $i = 1, 2, \ldots, 1993$; (ii) there is no point other than $P_i$ and $P_{i+1}$ on the line segment joining $P_i$ with $P_{i+1}$ whose coordinates are both integers, for $i = 0, 1, \ldots, 1992$. Prove that for some $i$, $0 \leq i \leq 1992$, there exists a point $Q$ with coordinates $(q_x, q_y)$ on the line segment joining $P_i$ with $P_{i+1}$ such that both $2q_x$ and $2q_y$ are odd integers.
Problem
Source: APMO 1993
Tags: analytic geometry, induction, number theory unsolved, number theory
jin
10.04.2006 15:53
By reduction,easy to see.
anonymouslonely
03.08.2011 18:32
suppose that any point on the considered segments doesn't have the property. take $ x_{i},y_{i} $ the coordinates of $ P_{i} $. if $ x_{i}+y_{i}+x_{i+1}+y_{i+1} $ is even then $ x_{i}+x_{i+1} $ and $ y_{i}+y_{i+1} $ are congruent mod 2 and that means both are even (this case isn't good because the midpoint will have both coordinates integers) or odd but this is a contradiction. then all the sums of this type are odd. $ P_{0} $ has $ x_{0}+y_{0} $ congruent with $ w $ mod 2 and we will say that $ P_{0} $ is red. then $ P_{1} $ has $ x_{1}+y_{1} $ congruent with $ 1-w $ mod 2 and we will say that $ P_{1} $ is blue. using induction we obtain that $ P_{2t} $ are red and $ P_{2t+1} $ are blue. so $ P_{0}=P_{1993} $ and $ P_{1992} $ are of the same colour so the sum of their coordinates is even which is impossible. this is the proof for the existence of $ Q $.
Why-
01.02.2020 10:40
1993 can change into any odd intergers.