# Beautiful Problem about pairwisely non-similar n-gons.

Let n be an integer (n>2). Show that there exists an infinite number of pairwisely non-similar inscribed n-gons, lengths of all sides and diagonals and areas of each of which are integers.

My intuition says it's done by induction. The base n=3 is easy: consider right triangles with both legs and hypotenuses being even integers. As there are infinite pythagorean triples, all triangles match. Now I'm completely stuck on how to proceed with bigger n's. Tried doing induction, butfor biggers ns, theres just so many sides and diagonals!

It seems that the solution follows from the following known

Lemma. There exists a dense subset $A$ of the unit circle $\mathbb T=\{z\in\mathbb C:|z|=1\}$ such that a distance between each two points of the set $A$ is rational.

Proof. Define a map $f: \mathbb R\to\mathbb T$ as $f(x)=\left(\frac{x+i}{x-i}\right)^2$. Since $f$ is continuous, $\mathbb Q$ is dense in $\mathbb R$, and $\mathbb T\setminus f(\mathbb R)$ is a singleton, $f(\mathbb R)$ is dense in $\mathbb T$ and

$$|f(x)-f(y)|=\left|\frac{4(xy+1)(x-y)}{(x^2+1)(y^2+1)}\right|,$$ for each points $x,y\in\mathbb R$, we can put $A=f(\mathbb Q)$.

Update: Yes, ‘singleton’ is just a wise name for one-point set. :-) If I remember it right, in our case it consists of a point $z_0=(1,0)$ and the inverse of a map $\frac{x+i}{x-i}$ is something like a projection of $\mathbb T\setminus\{z_0\}$ from the point $z_0$ onto a line $\{0\}\times\mathbb R\subset\mathbb C$.

The line with $|f(x)-f(y)|$ is just a simplification of the expression $$|f(x)-f(y)|=\sqrt{\left|\left(\frac{x+i}{x-i}\right)^2-\left(\frac{y+i}{y-i}\right)^2\right|}.$$ Since the expression inside “abs” is a quotient of two polynomials on $x,y$ with integer coefficients, it should be a rational number provided $x$ and $y$ are rational numbers. This implies the rationality of lengths of sides and diagonals of a polygon $P$ with vertices $f(x_1),\dots f(x_n)$ for each rational numbers $x_1,\dots, x_n$. The area of the polygon $P$ is rational too, because it can be dissected into triangles by diagonals from the vertex $f(x_1)$, and each of these triangle has a rational area $S$ (because it has sides $a,b,c$ of rational length and $S=\frac{abc}{4R}$, where $R$ is radius of the circumcircle, that is $R=1$). So a suitable homothetic image of $P$ has integer sides, diagonals and area.

At last, the infinity of mutually not-equal such the polygons (and so mutually non-similar, because all of them have the equal radius $R=1$ of circumcircle) follows from the fact that since the map $f$ is continuous, then one side $[f(x),f(y)]$ of such the polygon can be made arbitrary small by taking points $x,y$ such that $|x-y|$ is small.

• This looks amazing! We haven't passed complex numbers, but I ve read its basics. One thing though: why is T\f(R) a singleton and what is a singleton? Set of a single element? Thanks! – Jackie Poehler Dec 14 '14 at 10:03
• And pleasee,could you please explain the line with abs(f(x)-f(y)) and how we conclude what we need( infinity of n gons? Thanks a lot, this is very cool. – Jackie Poehler Dec 14 '14 at 10:11
• @JackiePoehler See the update of the answer. – Alex Ravsky Dec 14 '14 at 10:58