t (a, b, c)= min{b/a,c/b} where a <= b <= c , sidelengths

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t (a, b, c)= min{b/a,c/b} where a <= b <= c , sidelengths

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Source: Germany Federal - Bundeswettbewerb Mathematik 2010, round 2, p1

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#1 h Apr 14, 2020, 7:45 PM

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Let $a, b, c$ be the side lengths of an non-degenerate triangle with $a \le b \le c$ . With $t (a, b, c)$ denote the minimum of the quotients $\frac{b}{a}$ and $\frac{c}{b}$ . Find all values that $t (a, b, c)$ can take.

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#2 h Apr 15, 2020, 6:03 PM

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Denote $x=\dfrac{b}{a};y=\dfrac{c}{b}$ .
$x\ge1;y\ge1\Longrightarrow t(a,b,c)=\min\{x,y\}\ge1$ .
Results: $b=ax;c=axy$ .
The condition for a non-degenerate triangle is:
$a+b>c \Longleftrightarrow 1+x>xy\Longleftrightarrow x(y-1)<1$ .

For $x=1$ or $y=1$ results $t(a,b,c)=1$ .
For $x>1$ and $y>1$ we consider $2$ cases:

$\textbf{Case 1: }x\le y$ .
$x(y-1)<1\Longrightarrow x(x-1)\le x(y-1)<1\Longrightarrow$
$\Longrightarrow x^2-x-1<0\Longrightarrow t(a,b,c)=x<\dfrac{1+\sqrt5}{2}$ .

$\textbf{Case 2: }x\ge y$ .
$x(y-1)<1\Longrightarrow y(y-1)\le x(y-1)<1\Longrightarrow$
$\Longrightarrow y^2-y-1<0\Longrightarrow t(a,b,c)=y<\dfrac{1+\sqrt5}{2}$ .

Results: $1\le t(a,b,c)<\dfrac{1+\sqrt5}{2}$ .

$\forall k\in\left[1,\dfrac{1+\sqrt5}{2}\right)$ ,
exists a triangle for which $t(a,b,c)=k$ .
Just take $\dfrac{b}{a}=\dfrac{c}{b}=k$ .

Hence, the possible set of the values $t(a,b,c)$ is the interval
$I=\left[1,\dfrac{1+\sqrt5}{2}\right)$ .

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