Consider two polygons $ P$ and $ Q$. We want to cut $ P$ into some smaller polygons and put them together in such a way to obtain $ Q$. We can translate the pieces but we can not rotate them or reflect them. We call $ P,Q$ equivalent if and only if we can obtain $ Q$ from $ P$(which is obviously an equivalence relation). a) Let $ P,Q$ be two rectangles with the same area(their sides are not necessarily parallel). Prove that $ P$ and $ Q$ are equivalent. b) Prove that if two triangles are not translation of each other, they are not equivalent. c) Find a necessary and sufficient condition for polygons $ P,Q$ to be equivalent.
Problem
Source: Iranian National Olympiad (3rd Round) 2007
Tags: rotation, geometry, geometric transformation, reflection, rectangle, geometry proposed
elmagodelosnumeros
11.09.2007 03:29
This is a classical problem in a field of plane geometry known as transformation theory. But without rotation perhaps is a little harder. Here is my solution to (a). (I think the lemma 2 divides in two cases and is incomplete). \bf{Lemma 1:} \normalfont A rectangle is equivalent to a square with same area. Proof: Let $ a=AB, b=BC$ the sides of the rectangle $ ABCD$, and consider three different cases. Case I: ($ a=b$). In this case the rectangle is a square and the result is immediately. Case II: ($ b<a\leq 4b$). Let $ s=\sqrt{ab}$ the side of the square $ AB'C'D'$ with equal area, where $ B'$ lies on $ AB$ and $ D'$ lies on $ AD$. The line $ BD'$ cut the sides $ CD$ and $ B'C'$ in $ K$ and $ L$, respectively. Clearly the triangles $ KBB'$ is similar to $ D'LD$ and $ BCL$ is similar to $ KC'D'$. From the similar $ KB'B\sim D'AB$ we have \centerline{$ {\frac{B'K}{AD'}=\frac{B'B}{AB}\Rightarrow HK=\frac{AD'\cdot AB}{AB}=\frac{s(a-s)}{a}= s-b}$} The hypotesis, $ a\leq 4b$ implies $ s^{2}=ab\leq 4b^{2}$ so $ s-b\leq b$. Therefore $ K$ lies inside the segment $ B'C'$ and we have \centerline{$ KBB'\simeq D'LD$ and $ BCL\simeq KC'D'$}. Case III: ($ 4b<a$). Divide consecutively to half the base $ AB$ and duplicate the altitude $ BC$ of the rectangle until obtain the cases I or II. \bf{Lemma 2:} \normalfont Two squares with same area are equivalent. Proof: Let $ ABCD$ and $ AB'C'D'$ two squares with equal area, where the second is a rotation by an arbitrary angle $ 0<\phi<90$ from the first. Line $ B'C'$ cuts $ CD$ on $ E$, $ F$ is the foot of the perpendicular to $ CD$ from $ C'$, $ ED$ cuts $ C'D'$ on $ G$ and the parallel on $ G$ to $ DA$ cuts $ D'A$ on $ H$. $ I$ is the point inside $ AD$ so $ AI=C'F$ and the parallel on $ I$ to $ AB$ cuts $ D'A$ on $ J$. $ K$ is the point inside $ AB'$ so $ AK=C'G$ and the parallel on $ K$ to $ B'C'$ cuts $ AB$ on $ L$. Parallel on $ B'$ to $ BC$ cuts $ KL$ and $ CD$ on $ M$ and $ N$, respectively. It is easy to prove that $ EC'G\simeq LKA$, $ D'HG\simeq KMB'$ and $ AIJ\simeq B'NE$, so we can translate these triangles without rotation. One can takes points $ X$, $ Y$ and $ Z$ from segments $ IJ$, $ DG$ and $ BC$ to construct a pentagon $ GHJXY$ congruent to $ B'MLBZ$, with $ B'Z$ parallel to $ AB$. Finally the residues $ XIDY$ and $ B'ZCN$ are two rectangles with the same area an parallel sides, so by the lemma 1 these are equivalent. \bf{Proof of the problem:} \normalfont Since cutting for figures is an equivalence relation, from the lemma 1 and lemma 2 we obtain the equivalence of any two rectangles with the same area.