Question Number 75104 by chess1 last updated on 07/Dec/19
Answered by mr W last updated on 07/Dec/19
Commented by mr W last updated on 08/Dec/19
![let b=side length of pentagon α=54° 2r+2r cot α=b ⇒r=(b/(2(1+cot α))) AD=((b tan 72°)/2)−(r/(sin α))−a (((b tan 72°)/2)−(r/(sin α))−a−2r cos α)^2 +(2r sin α−(a/2))^2 =r^2 [r(1+cot α)tan 72°−(r/(sin α))−a−2r cos α]^2 +(2r sin α−(a/2))^2 =r^2 let λ=(a/r) [(1+cot 54)tan 72°−(1/(sin 54))−λ−2 cos 54]^2 +(2 sin 54−(λ/2))^2 =1 ⇒λ=(a/r)≈2.0845](https://www.tinkutara.com/question/Q75154.png)
$${let}\:{b}={side}\:{length}\:{of}\:{pentagon} \\ $$$$\alpha=\mathrm{54}° \\ $$$$\mathrm{2}{r}+\mathrm{2}{r}\:\mathrm{cot}\:\alpha={b} \\ $$$$\Rightarrow{r}=\frac{{b}}{\mathrm{2}\left(\mathrm{1}+\mathrm{cot}\:\alpha\right)} \\ $$$${AD}=\frac{{b}\:\mathrm{tan}\:\mathrm{72}°}{\mathrm{2}}−\frac{{r}}{\mathrm{sin}\:\alpha}−{a} \\ $$$$\left(\frac{{b}\:\mathrm{tan}\:\mathrm{72}°}{\mathrm{2}}−\frac{{r}}{\mathrm{sin}\:\alpha}−{a}−\mathrm{2}{r}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left(\mathrm{2}{r}\:\mathrm{sin}\:\alpha−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left[{r}\left(\mathrm{1}+\mathrm{cot}\:\alpha\right)\mathrm{tan}\:\mathrm{72}°−\frac{{r}}{\mathrm{sin}\:\alpha}−{a}−\mathrm{2}{r}\:\mathrm{cos}\:\alpha\right]^{\mathrm{2}} +\left(\mathrm{2}{r}\:\mathrm{sin}\:\alpha−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${let}\:\lambda=\frac{{a}}{{r}} \\ $$$$\left[\left(\mathrm{1}+\mathrm{cot}\:\mathrm{54}\right)\mathrm{tan}\:\mathrm{72}°−\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{54}}−\lambda−\mathrm{2}\:\mathrm{cos}\:\mathrm{54}\right]^{\mathrm{2}} +\left(\mathrm{2}\:\mathrm{sin}\:\mathrm{54}−\frac{\lambda}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\lambda=\frac{{a}}{{r}}\approx\mathrm{2}.\mathrm{0845} \\ $$