Find max value - (Triangles, geometric progressions, greatest integer(floor)function)

fardeen_gen · #1 $Old$ May 21st, 2009, 12:04 PM

If for $r > 1$ , three successive terms of a Geometric Progression with common ratio $r$ represent sides of a triangle, then find the maximum value of $(\lfloor 2r \rfloor + \lfloor - r \rfloor)$

Media_Man · #2 $Old$ May 22nd, 2009, 12:30 PM

Take some triangle with side lengths $a,ar,ar^2$ for some $a>0$ and $r>1$ . Without loss of generality, let $a=1$ and find the allowable values of $r$ for a triangle to exist:

(i) $1+r>r^2$ true for all $r>1$ but $r<\phi$ for $\phi$ the golden ratio $\approx 1.618$
(ii) $1+r^2>r$ true for all $r>1$
(iii) $r+r^2>1$ true for all $r>1$

Therefore, $r\in(1,\phi)$ to satisfy this condition. It can be shown that:

$\lfloor 2r \rfloor + \lfloor -r \rfloor = \left\{ \begin{array}{rcl} 0 & \mbox{if} & 1<r<1.5 \\1 & \mbox{if} & 1.5\leq r<2 \\2 & \mbox{if} & 2\leq r<2\phi \end{array}\right.$

Therefore the max value of this function on the allowable interval is 2.

QED