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
19.03.2023 04:38
The fourth line should have side lengths greater than D right?
19.03.2023 04:40
Thank you @above and @below for pointing out my mistake!
19.03.2023 05:08
Quote: $A, B, C$ lie on lattice points; ${\color{red}AB,BC,CA}>D$; $S_{\triangle ABC}<\frac{\sqrt 3}{4}D^2+\lambda\cdot D^{4/5}$.
19.03.2023 17:40
The inequalities should probably be reversed, since now the problem is trivial.
22.03.2023 09:06
12.04.2023 11:50
CANBANKAN wrote:
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?
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.
16.11.2024 16:42
https://arxiv.org/pdf/1912.07566