We call a pair of polygons, $p$ and $q$, nesting if we can draw one inside the other, possibly after rotation and/or reflection; otherwise we call them non-nesting. Let $p$ and $q$ be polygons. Prove that if we can find a polygon $r$, which is similar to $q$, such that $r$ and $p$ are non-nesting if and only if $p$ and $q$ are not similar.
Problem
Source: Canada RepĂȘchage 2018/2
Tags: geometry, geometric transformation, rotation, reflection
10.04.2018 18:53
Anyone? By the way, are the polygons $p$ and $q$ given?
10.04.2018 19:03
arbitrary polygons
11.04.2018 08:29
Does this polygon $r$ has to hold for every pair of $p$ and $q$?
11.04.2018 16:01
I think you are misunderstanding the question because of their grammatical error. Rephrasing the problem. Prove: If and only if $p$ and $q$ are not similar you can find a polygon $r$ similar to $q$ such that $p$ and $r$ are non-nesting.
11.04.2018 18:54
We need to prove that: $p$ and $q$ are similar if and only if $p$ and $r$ are nesting for every polygon $r$ that is similar to $q$. The only if part is obvious (by some homothety.) For the if part, suppose $p,q$ are polygons that $p$ and $r$ are nesting for every polygon $r$ that is similar to $q$. For each $c\in \mathbb{R}^+$, let $q_c$ denote the polygon that is similar to $q$ that has area equal to $c$ times the area of $q$. We get that $q_c$ and $p$ are nesting for all $c\in \mathbb{R}^+$. Note that there exists a positive real $c_0$ that the area of $q_{c_0}$ is equal to the area of $p$. The only way one can completely fit inside one another is so that they're exactly the same, i.e. $q_{c_0}\equiv p$. This gives $p\equiv q_{c_0}\sim q$, done.