Geometric Geometry

Since $\Delta ABC$ is right angled we know that $AC$ is the longest side, so we can let the progression $|AB|, |BC|, |AC|$ be $a, ar, ar^{2}$ for some reals $a \gt 0, r \gt 1$ . Then by Pythagoras we have that

$|AC|^{2} = |AB|^{2} + |BC|^{2} \Longrightarrow (ar^{2})^{2} = a^{2} + (ar)^{2} \Longrightarrow a^{2}(r^{4} - r^{2} - 1) = 0$ .

As $a \ne 0$ we must then have that $r^{4} - r^{2} - 1 = (r^{2})^{2} - r^{2} - 1 = 0$ , and thus that

$r^{2} = \dfrac{1 \pm \sqrt{5}}{2}$ . As $r^{2}$ is necessarily positive we take the positive root as the answer, i.e.,

$r^{2} = \dfrac{1 + \sqrt{5}}{2} = \boxed{1.618}$ to 3 decimal places.

$\frac{AB}{BC}=\frac{BC}{AC}$
$\tan(C)=\cos(C)$
$\sin(C)=\cos^2(C)=1-\sin^2(C)$
$\sin(C)>0$ so $\sin(C)=\frac{-1+√5}{2}$

Answer is $\frac{1}{\sin(C)}=\frac{2}{\sqrt 5-1}=\frac{\sqrt 5+1}{2} \approx 1.618$

let $\large x,y,z$ denote the sides of the triangle with $\large x<y<z$

By Pythagoras theorem we have, $\large x^2+y^2=z^2$

Also, since x,y,z are in G.P we have $\large y^2=xz$

Thus we get, $\large x^2+xz=z^2$

or, $\large (\dfrac{z}{x})^2-\dfrac{z}{x}-1=0$

on solving we get, $\large \dfrac{z}{x}=\dfrac{1 \pm \sqrt{5}}{2}$

however since $\large x,y,z>0$ we have $\large \dfrac{z}{x}=\dfrac{1+ \sqrt{5}}{2}=\boxed{\large 1.618}$

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