let(f) be the relation defined by f(n) = The smallest interior angle value of the n sided polygon with perimeter n units with maximum area, for each positive integer n(>2).which of the following are true?
- given two positive integers n,m(<2) with n> m always have that f(n)>f(m)
- Given a positive integers n we always have that the difference between f(n) and f(n+1) is less than 45°.
- there are two distinct positive integers n,m(>2) such that the difference between f(n) and f(m) is less than 1°.
For an $n$-sided polygon with a given perimeter, the regular polygon has the maximum area.
Thus, for a regular $n$-sided polygon, all the interior angles are equal and are equal to $$f(n)=\frac{180^o(n-2)}{n}=180^o(1-\frac2n)$$
Clearly, $f(n)$ increases when $n$ increases.
Now, $$f(n+1)-f(n)=180^o(1-\frac2{n+1}-1+\frac2n)=360^o\frac1{n(n+1)}$$
It can be shown using induction that the above is less than $45^o$ for all $n\ge3$.
Finally, $$f(n)-f(m)=180^o(1-\frac2n-1+\frac2m)=360^o\frac{(n-m)}{nm}$$ Let $$f(n)-f(m)<1^o$$ Let $n=m+1$. Then, $$360^o\frac{(n-m)}{nm}=360^o\frac1{m(m+1)}<1^o$$ $$m^2+m-360>0$$ This inequality clearly has positive integer solutions.