In a Cartesian plane, if both horizontal coordinate and vertical coordinate of a point are rational numbers, we call the point rational point. Otherwise, we call it irrational point. Consider an arbitrary regular pentagon on the Cartesian plane. Please compare the number of rational point and the number of irrational point among the five vertices of the pentagon. Prove your conclusion.
Problem
Source: 2018 China Southeast MO Grade 10 P2, Grade 11 P1
Tags: combinatorics
ThE-dArK-lOrD
30.07.2018 11:10
Can I just say these two numbers have different parity in modulo 2, since they must sum up to 5?
NikoIsLife
30.07.2018 11:15
Are there any other details to this question? It seems vague.
MarkBcc168
30.07.2018 13:45
It asks which number is greater.
SAUDITYA
01.08.2018 14:33
Let's work in $\mathbb{C}$.So the rational points are precisely $\mathbb{Q}[i]$.
Lemma 1: if $x,y,z$ be three distinct non-collinear points in $\mathbb{Q}[i]$ then the circumcenter of $\triangle xyz$ is in $\mathbb{Q}[i]$
Click to reveal hidden textthis follow from the fact that if $t$ is the circumcenter of $\triangle xyz$ then we get two linear equations in $t,\overline t$ with coefficients in $\mathbb{Q}[i]$
Lemma 2: if $\zeta \in \mathbb{Q}[i]$ is a primitive $n^{th}$ root of unity , then $n=1,2$ or $4$
Now, suppose $\{z_1,z_2,\cdots,z_n\}$ is a regular n-gon(n$\ge$ 3) .
Further suppose that $\exists$ three distinct indices $\{i,j,k\} \in \{1,2,\cdots,n\}$ such that $\{z_i,z_j,z_k\} \in \mathbb{Q}[i]$.By lemma 1. the circumcenter,i.e. $z = \frac{\sum_{i=1}^n z_i}{n}$, of the polygon must be in $\mathbb{Q}[i]$.
Now consider the following map $x \mapsto \frac{x-z}{z_i-z}$.It preserves $\mathbb{Q}[i]$ as well as regularity of the polygon
(because it is a composition of translation and rotation).
See that $\{z_1,z_2,\cdots,z_n\}$ maps to a permutation of $\{1,\zeta,\zeta^2,\cdots,\zeta^{n-1}\}$, where $\zeta$ is primitive $n^{th}$ root of unity.
See that $\{z_j,z_k\} = \{\zeta^l,\zeta^m\}, \exists \{l,m\} \in \{1,2,\cdots,n-1\}, \in \mathbb{Q}[i]$.Now by lemma 2. we must have
$\{\frac{n}{gcd(n,l)},\frac{n}{gcd(n,m)}\} =1,2$ or $4$ .if $4\not | n$ then we have a contradiction and so there can't be more than two rational points or else we have $\{1,\zeta^{\frac{n}{4}},\zeta^{\frac{n}{2}},\zeta^{\frac{3n}{4}}\}\in \mathbb{Q}[i]$ and further we can't have more than four rational points.
MarkBcc168
03.08.2018 12:07
The number of irrational points is always larger. To prove that, let the pentagon be $ABCDE$ and assume that three of the vertices are rational points. WLOG, the set of rational points is $\{A,B,C\}$ or $\{A,B,D\}$. Either way, $AB^2$ and $AC^2$ are both rational. But $\tfrac{AB^2}{AC^2} = \left(\tfrac{1+\sqrt{5}}{2}\right)^2 = \tfrac{3+\sqrt{5}}{2}$ which is clearly irrational, contradiction.