Golden Ratio in Isosceles Triangle

Source

Kadir Altintas has kindly communicated to me the problem below:

Golden Ratio in Isosceles Triangle,source

Problem

In an isosceles $\Delta ABC,$ $AB=AC$ and $OI=2r,$ where $O$ is the circumcenter, $I$ the incenter, and $r$ the inradius of the triangle.

Golden Ratio in Isosceles Triangle,illustration

Prove that $\displaystyle \sqrt{\frac{AC}{BC}}=\varphi,$ the Golden Ratio.

Solution

Let's denote $a=BC$ and $b=AC,$ and let $R$ be the circumradius of $\Delta ABC.$ We shall consider two pairs of similar triangles:

Golden Ratio in Isosceles Triangle,proof

$\Delta ACE\sim\Delta AFI:$ $\displaystyle \frac{a/2}{b}=\frac{r}{R+2r},$ or, $\displaystyle \frac{b}{a}=\frac{R+2r}{2r}.$

$\Delta ADO\sim\Delta AFI:$ $\displaystyle \frac{b/2}{R}=\frac{b-a/2}{R+2r},$ or, $\displaystyle \frac{b}{a}=\frac{R}{R-2r}.$

Thus we have $\displaystyle \frac{b}{a}=\frac{R+2r}{2r}=\frac{R}{R-2r},$ so that $R^2-2Rr-4r^2=0.$ Taking the positive root, $\displaystyle \frac{R}{r}=\sqrt{5}+1.$ It follows that

$\displaystyle\begin{align} \frac{b}{a}&=\frac{R}{R-2r}=\frac{\sqrt{5}+1}{\sqrt{5}-1}\\ &=\frac{(\sqrt{5}+1)^2}{4}=\left(\frac{\sqrt{5}+1}{2}\right)^2=\varphi^2. \end{align}$

 

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