Geometry Problem

[fourier jr]

Let f(n) denote the maximum number of right triangles determined by n coplanar points.

Prove f(n)/(n^2) --> infinity as n-->infinity and f(n)/(n^3) --> 0 as n-->infinity.

[bloodhound]

This is for the second part.

 

Instead of finding f(n) explicitly we find g(n) for which [math]f(n) \le g(n)[/math] for all [math]n \ge 3[/math]

 

Here [math]f,g:N \mapsto N[/math] and that [math]f,g \ge 0[/math]

 

Finding a function g(n)

 

The number of right angled triangles formed by n points will be always less than or equal to the total number of triangles formed by those n points.

 

The total number of traingles formed by n points is the number of ways of choosing 3 points out of the n points. i.e [math]g(n)=^{n}C_3[/math]

 

We have

[math]0 \le \frac{{f(n)}}{{n\^3}} \le \frac{{g(n)}}{{n\^3}}[/math]

 

It can be shown that [math]\frac{{g(n)}}{{n\^3}}[/math] converges to 0 as n tends to infinity.

 

Therefore by sandwiching (pinching) [math]\frac{{f(n)}}{{n\^3}}[/math] also must tend to 0

[Dave]

Nicely done, I'd not really thought about it like that.

 

Stumped on the first limit though.

[bloodhound]

actually, when i went through it again, my proof if completely wrong.

 

g(n)/n^3 does not converge to 0 as n tends to infinity. it actually tends to 1/6.

 

so my whole proof collapses.

 

but now we know that if the limit of f(n)/n^3 does exist then its bounded above by 1/6.

 

oh well. back to the drawing board now.

[Freeman]

Let f(n) denote the maximum number of right triangles determined by n coplanar points. Prove f(n)/(n^2) --> infinity as n-->infinity and f(n)/(n^3) --> 0 as n-->infinity.

Lemme get this straight: [math]f(n)/n^2>\infty[/math] as [math]n>\infty[/math] and [math]f(n)/n^3>0[/math] as [math]n>\infty[/math], correct? Or is --> greater than or equal to?

[bloodhound]

---> means tends to . as in a limit.

 

now for the second part.

 

in the first proof i wrote. g(n)/n^3 didnt converge to 0

 

but if u take g(n)=n

 

then [math]f(n) \le g(n)[/math]

[math]\frac{f(n)}{n^3} \le \frac {g(n)}{n^3}[/math]

since g(n)/n^3 = 1/n^2 tends to 0 as n tends to infinity then f(n)/n^3 must also tend to 0.

[fourier jr]

---> means tends to . as in a limit.

 

yeah, sorry I guess I didn't make that clear enough. I haven't learned Latex so I didn't use limit notation.