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### triangle in a square has area less than 1/2

Posted: Sun Feb 08, 2015 10:57 pm
Take a square of length \$n\$ and choose \$(n+1)^2\$ points inside the square. Prove that, you can choose some \$3\$ points so that the triangle made by them has an area at most \$\dfrac12\$.

### Re: triangle in a square has area less than 1/2

Posted: Sun Feb 15, 2015 9:51 pm
I think \$(n+1)^2\$ is more than sufficient, we just need \$n^2+1\$.

### Re: triangle in a square has area less than 1/2

Posted: Mon Feb 19, 2018 12:59 pm
Masum wrote:
Sun Feb 08, 2015 10:57 pm
Take a square of length \$n\$ and choose \$(n+1)^2\$ points inside the square. Prove that, you can choose some \$3\$ points so that the triangle made by them has an area at most \$\dfrac12\$.
It is a nice problem.
Sketch of solution:
Take the convex hull of points . Denote it by H.Apparently,the hull has area at most n^2 and perimeter 4n.Now suppose that hull has k points on its boundary. So definitely n^2+2n+1-k points in the interior of the hull.Now triangulate the hull .This use exactly 2 (n^2+2n)-k triangles. So one has area at most n^2/{2 (n^2+2n)-k} .Next there must exist 2 consecutive edges such that if side length r a,b then a+b/2 <= 4n/k
So this triangle will have area at most 8n^2/k^2
SO some triangle will have area at most minimum of the above mentioned 2 triangles. The first one is increasing wrt k while the 2nd one is decreasing.So that area will be maximized when both will be equal.Then k=4n
The rest is just calculation

### Re: triangle in a square has area less than 1/2

Posted: Mon Feb 19, 2018 1:00 pm
Silly me.Couldn't latex for shortage of time :"(

### Re: triangle in a square has area less than 1/2

Posted: Mon Feb 19, 2018 2:27 pm
SN.Pushpita wrote:
Mon Feb 19, 2018 12:59 pm
Masum wrote:
Sun Feb 08, 2015 10:57 pm
Take a square of length \$n\$ and choose \$(n+1)^2\$ points inside the square. Prove that, you can choose some \$3\$ points so that the triangle made by them has an area at most \$\dfrac12\$.
It is a nice problem.
Sketch of solution:
Take the convex hull of points H. Apparently,the hull has area at most n^2 and perimeter 4n.Now suppose that hull has k points on its boundary .Now triangulate the hull .This use exactly 2 (n^2+2n)-k triangles. So one has area at most n^2/{2 (n^2+2n)-k} .Next there must exist 2 consecutive edges such that if side length r a,b then a+b/2 <= 4n/k
So this triangle will have area at most 8n^2/k^2
SO some triangle will have area at most minimum of the above mentioned 2 triangles. The first one is increasing wrt k while the 2nd one is decreasing.So that area will be maximized when both will be equal.Then k=4n
The rest is just calculation
I didn't uderstand the first line of the skectch of solution.

### Re: triangle in a square has area less than 1/2

Posted: Tue Feb 20, 2018 11:32 am
samiul_samin wrote:
Mon Feb 19, 2018 2:27 pm
SN.Pushpita wrote:
Mon Feb 19, 2018 12:59 pm
Masum wrote:
Sun Feb 08, 2015 10:57 pm
Take a square of length \$n\$ and choose \$(n+1)^2\$ points inside the square. Prove that, you can choose some \$3\$ points so that the triangle made by them has an area at most \$\dfrac12\$.
It is a nice problem.
Sketch of solution:
Take the convex hull of points H. Apparently,the hull has area at most n^2 and perimeter 4n.Now suppose that hull has k points on its boundary .Now triangulate the hull .This use exactly 2 (n^2+2n)-k triangles. So one has area at most n^2/{2 (n^2+2n)-k} .Next there must exist 2 consecutive edges such that if side length r a,b then a+b/2 <= 4n/k
So this triangle will have area at most 8n^2/k^2
SO some triangle will have area at most minimum of the above mentioned 2 triangles. The first one is increasing wrt k while the 2nd one is decreasing.So that area will be maximized when both will be equal.Then k=4n
The rest is just calculation
I didn't uderstand the first line of the skectch of solution.
Well ,read about convex hull then. I have also edited some lines to increase understability .