Let $P$ and $P^{\prime }$ be two parallelograms with equal area, and let their sidelengths be $a,$ $b$ and $a^{\prime },$ $b^{\prime }.$ Assume that $a^{\prime }\leq a\leq b\leq b^{\prime },$ and moreover, it is possible to place the segment $b^{\prime }$ such that it completely lies in the interior of the parallelogram $P.$ Show that the parallelogram $P$ can be partitioned into four polygons such that these four polygons can be composed again to form the parallelogram $% P^{\prime }.$
Problem
Source: IMO LongList 1959-1966 Problem 22
Tags: geometry, parallelogram, vector, IMO Shortlist, IMO Longlist
07.09.2004 01:54
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08.09.2004 16:24
If ABCD whith AB=b and BD=a is one of the two parallelograms, let E on BD a point so that AE=b'. This is possible since we know that b'>b but b' < AD (the biggest of the two diagonals). Then if C' is the midpoint of ED, let's take a point D' on AD s.t. AD'=C'D' in this way we have bisected the parallegram whith a segmente long b'. Now we translate D'DDC' of the vector DA, s.t. D'-->A" and C'-->B". After this we take a point A' on A"B" s.t. D'A'=a' this is possible since, as the two parallelograms are equivalent, a' is greater than the altitude wrt D'C' of the parallelogram A"D'C'B" and a' < a = D'A". If G=A'D'^AB, let D"F the translation of A'G of the vector AD. If one translate A"D'A' of the vector D'C' we obtain the parallegram A'B'C'D' equivalent to ABCD and having sides a', b'. The parallelogram ABCD have been partitioned into four polygons and these four polygons have been composed again to form the parallelogram A'B'C'D', as requested. Perhaps the picture attached, make more clear the concept.
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