Let $\Omega$ be a circle in a plane. $2022$ pink points and $2565$ blue points are placed inside $\Omega$ such that no point has two colors and no two points are collinear with the center of $\Omega$. Prove that there exists a sector of $\Omega$ such that the angle at the center is acute and the number of blue points inside the sector is greater than the number of pink points by exactly $100$. (Note: such sector may contain no pink points.)
Problem
Source: 2022 Thailand MO Day 1 P3
Tags: combinatorics
CANBANKAN
15.06.2022 09:35
Toss the situation onto a complex plane, where the circle is the unit circle. Then only the arguments of these points matter. This problem is a textbook exercise of discrete continuity.
We divide the circle into five sectors of length $\frac{2\pi}{5}$ such that no point is on the border. Then the total number of blue points minus red points in the five sectors is 543. This means there exists one sector with at least 100 points.
Take this sector and sort the points $p_i$ inside the sector by $arg(p_i)$ where the center of the circle is 0. Call the points $p_1, \cdots, p_k$. Let $x_j=1$ if $p_j$ is blue and $x_j=-1$ otherwise. Let $S_j=x_1+\cdots+x_j$. We are given $S_0=0, S_k>100$ and $|S_{i+1}-S_i|=1$, so there exists $t$ such that $S_t=100$. The sector containing $p_1, \cdots p_t$ but not $p_{t+1}$ works.
iammocnhan
15.11.2022 18:56
Let $O$ be the center of $\omega$. Since $\text{blue point}-\text{pink point}=543$, by Pigenonhole principle there's a quadrant $I$ that $\text{blue point}-\text{pink point}\geq 101$ Now, considering the quadrant $I$. Assume $Ox$ is starting ray of $I$, and $Oy$ is the ending ray. We denote $A_1,A_2,...,A_n$ respectively by the point in $I$ clockwisely. Let $f(i)$ be the difference between blue points and pink points in the sector made by $Ox,OA_i$, with assuming $OA_{n+1} \equiv Oy$ Firstly, $f(1)=0,f(n+1) \geq 101$ and we see that $|f(i+1)-f(i)|=1 \forall 0\leq i \leq n$ Therefore, $\exists 0 \leq j \leq n: f(j)=100$ So, the sector that made by $Ox,OA_j$ is satisfied.
Saucepan_man02
05.11.2024 16:53
Nice exercise of discrete continuity: Divide the plane into 5 equal sectors. By PHP, some sector contains $\# (\text{Blue points}) - \# (\text{Red points}) \ge 100$. Equality holds, we are done. Assume it doesnt. Then, we consider the points in a counter-clock-wise direction from the point having angle with positive x-axis. Consider the sum of first $k$ points to be $S_k$ in the scan. We add $+1$ if we see a blue, $-1$ if we see a red. Notice that: $S_{net} > 100$ and $S_0=0$. By Discrete Continuity, for some $j$, $S_j = 100$ and we are done.