Find the segmento $NO$ in semicircle below

Question

In the graph, calculate: $NO$, if the golden section of $SO$ measures $8u$.($T$ point of tangency).(Answer$4u$)

Could someone explain what this golden section would be and how to use it in this exercise

Answer 1

Dividing $SO$ into two parts by a point $B$ we will have the golden section of $SO = 8u$ $\dfrac{SO}{8}=\dfrac{8}{BO} \implies \dfrac{SO}{8}=\dfrac{8}{SO-8}\implies SO^2-8SO-64 =0\\ \therefore SO=4(1+\sqrt5)=8\underbrace{\dfrac{(1+\sqrt5)}{2}}_{\varphi} \implies \color{red}{SO = 8\varphi}$

$NS = L \\ \triangle CNS: L^2=r^2+h^2(I)\\ \triangle CTN: h^2=TN^2+r^2(II)\\ \triangle BTN : r^2=TN^2+(L-r)^2(III)\\ (I)+(II): TN^2 = L^2-2r^2(IV)\\ (IV)em(III): {r^2}= L^2-2r^2+L^2-2Lr+{r^2} \implies L^2-Lr-r^2=0 \\ \therefore \color{red}{L = r\dfrac{1+\sqrt5}{2}=r \varphi} \\ \triangle CSN: cos \angle NSC=\cos \alpha = \dfrac{r}{L} = \dfrac{r}{r\varphi} = \dfrac{1}{\varphi}\\ \triangle CSD: cos \alpha = \dfrac{4\varphi}{r}\implies \dfrac{1}{\varphi}= \dfrac{4\varphi}{r} \therefore \color{red}{r = 4\varphi^2}\\ Sabendo : \varphi^2=\varphi+1 \implies \varphi = \varphi^2-1\\ NO = NS - SO=r\varphi - 8\varphi = 4\varphi^3-8\varphi = 4\varphi^2.\varphi-8\varphi=4(\varphi+1)\varphi-8\varphi=\\ 4\varphi^2-4\varphi =4(\varphi+1)-4\varphi =\boxed{4u}$

(Solution:by electron)

Answer 2

Let $r$ be the circle radius, $\phi = \frac{1+\sqrt 5}2$, and refer to the Figure above for additional labels (note in particular $CD\perp NS$). Recall also that $\phi^2 -\phi-1 = 0$.

1. Similarity $\triangle ABC \sim \triangle CNT$ yields $$\overline{NC} = r\sqrt \phi$$ and $$\overline{BC} = r\sqrt{\phi+1}=r\phi.$$
2. Similarity $\triangle CNS \sim \triangle CDS$ yields $$\overline{DS} = \frac{r}{\phi}.$$
3. We have $$\overline{OS} = \frac{2r}{\phi} = 8 \phi,$$hence $$r = 4\phi^2=4(\phi+1).$$
4. Finally \begin{eqnarray}\overline{NO} &=& \overline{BC}-\overline{OS}=\\&=& 4\phi(\phi+1)-8\phi =\\ &=& 4\phi^2-4\phi = 4.\end{eqnarray}