# How many points you should draw in the square at least？

There is a square, which side length is $2$, To ensure there exists a triangle in the square, with an area less than $0.5$, how many points should you draw in the square at least.

the goal is for all quantification of possible point arrangements, there must exist a triple of points, which triangle area less than $\frac{1}{2}$.

use pigeonhole principle，it's easy to prove the ceiling is 10; I think 9 points is the answer, but i can't prove it. Please give your answer and proof to help me, thank you.

-
Do I understand your question correctly? You are asking for the number of points such that no matter how this many points are arranged within the square, there must exist a triple of points whose convex hull has size less than $\frac12$. The important part here seems the “for all” quantification of possible point arrangements, which is not very clear in your post. – MvG Jan 24 '14 at 8:10
Yes， your explaination is correct, for all quantification of possible point arrangements, there must exist a triple of points, which triangle area less than $\frac{1}{2}$. thank you. – yang_bigarm Jan 26 '14 at 7:18
@yang_bigarm the configuration above is not valid, and I suspect the answer is 7 – Zackkenyon Jan 28 '14 at 11:07

## 1 Answer

For a square whose vertices are $(\pm 1,\pm 1)$, the area of every triangle formed by any three of the six points $(\pm 1,\pm 1), (0,\pm\frac 12)$ is not smaller than $\frac 12$.

Zhenbing Zeng and Liangyu Chen prove that for any seven points in a unit square there exist three points whose area is not greater than a constant $0.083859\cdots$.

It follows from these that the answer is $\color{red}{7}$.

You might be interested in the Heilbronn triangle problem (a problem to place $n\ge 3$ points in a disk (square, equilateral triangle, etc.) of unit area so as to maximize the area of the smallest of the $\binom n3$ triangles determined by the $n$ points.)

-