Let $c$ be the length of the base of an isosceles triangle, $a$ the length of the legs, $R$ the radius of the circumscribed circle and $r$ the radius of the incircle of the triangle.
Find $\frac{a}{c}$ if $\frac{r}{R}=k$
I've done: \begin{align} S&=\frac{abc}{4R}\\ S&=pr\\ \frac{a^2c}{4R}&=\frac{2a+c}{2}r \end{align} And don't know how to continue.
On
We know that $r=4R\sin{(\dfrac{A}{2})} \sin{(\dfrac{B}{2})} \sin{(\dfrac{C}{2})}.$ where A, B and C are the angles of the triangles. And $\dfrac{a}{c}=\tan{A}$, where A is the equal angles or base angles which can be proofed using sine rule.
Now $\frac{r}{R}=k=8\sin{(\dfrac{A}{2})}\cos{(\dfrac{A}{2})}$, $\therefore k=4\sin{A}$. Now you should express $\sin{A}$in terms of $\tan{A}$, and then find the values of $\tan{A}$ in terms of k and it is done.
I think you mean $a=b$. Thus, $2a>c$ or $\frac{a}{c}>\frac{1}{2}$ and $$k=\frac{r}{R}=\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}=\frac{16S^2}{2abc(a+b+c)}=$$ $$=\frac{(a+b-c)(a+c-b)(b+c-a)}{2abc}=\frac{(2a-c)c^2}{2a^2c}=\frac{\frac{2a}{c}-1}{\frac{2a^2}{c^2}}.$$ Hence, $$\frac{2ka^2}{c^2}-\frac{2a}{c}+1=0,$$ which gives $$\frac{a}{c}=\frac{1\pm\sqrt{1-2k}}{2k}.$$