Prove that there exists some positive real number $\lambda$ such that for any $D_{>1}\in\mathbb{R}$, one can always find an acute triangle $\triangle ABC$ in the Cartesian plane such that $A, B, C$ lie on lattice points; $AB, BC, CA>D$; $S_{\triangle ABC}<\frac{\sqrt 3}{4}D^2+\lambda\cdot D^{4/5}$.
Problem
Source: China TST 2023 - Test 2 Problem 6
Tags: combinatorics, combinatorial geometry, China TST, analytic geometry, lattices
dragon-YBW
19.03.2023 04:38
The fourth line should have side lengths greater than D right？
JG666
19.03.2023 04:40
Thank you @above and @below for pointing out my mistake!
a22886
19.03.2023 05:08
Quote: $A, B, C$ lie on lattice points; ${\color{red}AB,BC,CA}>D$; $S_{\triangle ABC}<\frac{\sqrt 3}{4}D^2+\lambda\cdot D^{4/5}$.
R8kt
19.03.2023 17:40
The inequalities should probably be reversed, since now the problem is trivial.
CANBANKAN
22.03.2023 09:06
The problem is equivalent to finding integers $x,y$ such that
$$D^2/4 \le x^2+y^2 \le D^2/4 +10D^{4/5}$$and
$$\{\sqrt{3}y\}, 1-\{\sqrt{3}x\}< 100D^{-1/5} $$For I can let (0,0) be the first vertex, (2x,2y) be the second vertex and $(x-\sqrt{3}y+\{\sqrt{3}y\}, y+\sqrt{3}x+(1-\{\sqrt{3}x\})$ be the third vertex then all side lengths are at least D and at most $D+100D^{-1/5}$
Claim: For any integer $k$, if I sort $\{ a\sqrt{3}\}_{1\le a\le 100k}$, and add 0,1 into our sorted list, then the difference of any two adjacent terms is at most $1/k$.
Proof: Focus on $y$ where $x+y \sqrt{3}= (2+\sqrt{3})^l$ (*)
Induct on $k$. Take the $100(2-\sqrt{3})k$ existing points whose adjacent differences are at most $(2+\sqrt{3})/k$. Add at most 5 points between each interval, so in the end the $a$ will be at most $(100k+5y)(2-\sqrt{3}) < 120k (2-\sqrt{3})< 100k $. This proves our claim.
Now we pick $x \in [D/2-100D^{1/5},D/2]$ such that $ 1-\{\sqrt{3}x\}< 100D^{-1/5}$ and $y_0 = \lceil \sqrt{(D/2)^2-x^2} \rceil$. Note $y_0=O(D^{3/5})$ so $0\le x^2+y_0^2-D^2\le 100D^{3/5}$. By the lemma, picking some $y \in [y_0, y_0+100D^{1/5}]$ gives $\{ y\sqrt{3}\} < 100D^{-1/5}$ and $x^2+y^2-D^2/4\in [0,100D^{4/5}]$.
PureRun89
12.04.2023 11:50
CANBANKAN wrote:
The problem is equivalent to finding integers $x,y$ such that
$$D^2/4 \le x^2+y^2 \le D^2/4 +10D^{4/5}$$and
$$\{\sqrt{3}y\}, 1-\{\sqrt{3}x\}< 100D^{-1/5} $$For I can let (0,0) be the first vertex, (2x,2y) be the second vertex and $(x-\sqrt{3}y+\{\sqrt{3}y\}, y+\sqrt{3}x+(1-\{\sqrt{3}x\})$ be the third vertex then all side lengths are at least D and at most $D+100D^{-1/5}$
Claim: For any integer $k$, if I sort $\{ a\sqrt{3}\}_{1\le a\le 100k}$, and add 0,1 into our sorted list, then the difference of any two adjacent terms is at most $1/k$.
Proof: Focus on $y$ where $x+y \sqrt{3}= (2+\sqrt{3})^l$ (*)
Induct on $k$. Take the $100(2-\sqrt{3})k$ existing points whose adjacent differences are at most $(2+\sqrt{3})/k$. Add at most 5 points between each interval, so in the end the $a$ will be at most $(100k+5y)(2-\sqrt{3}) < 120k (2-\sqrt{3})< 100k $. This proves our claim.
Now we pick $x \in [D/2-100D^{1/5},D/2]$ such that $ 1-\{\sqrt{3}x\}< 100D^{-1/5}$ and $y_0 = \lceil \sqrt{(D/2)^2-x^2} \rceil$. Note $y_0=O(D^{3/5})$ so $0\le x^2+y_0^2-D^2\le 100D^{3/5}$. By the lemma, picking some $y \in [y_0, y_0+100D^{1/5}]$ gives $\{ y\sqrt{3}\} < 100D^{-1/5}$ and $x^2+y^2-D^2/4\in [0,100D^{4/5}]$.
Hi bro I found that the statement $AB,BC,CA>D$ requires $x^2+y^2-2\{\sqrt{3} y\} (x+\sqrt{3} y)+2(1-\{\sqrt{3} x\})(\sqrt{3} x-y)>0.25 D^2$ (This is the distance of $(2x,2y)$ and the other point) How did you ensure this in your solution?
CANBANKAN
08.05.2023 05:24
Hi, sorry for the late reply. In the future please PM me. One easy way to guarantee this is to have $D^2/4 + 1000D^{4/5} \le x^2+y^2 \le D^2/4 + 1010 D^{4/5}$ as the first line instead. As long as I have bounded the lengths to an error of $O(D^{1/5})$ I am fine.
Oksutok
16.11.2024 16:42
https://arxiv.org/pdf/1912.07566