Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\frac{1}{2}$.

enter image description here

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.

share|improve this question
    
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 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 at 7:18
    
@yang_bigarm the configuration above is not valid, and I suspect the answer is 7 –  Zackkenyon Jan 28 at 11:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.