A problem on a triangle's inradius and circumradius .

Question

I'm trying to solve the following problem :

In $△ABC$, $AB = AC, BC = 48$ and inradius $r = 12$. Find the circumradius $R$.

Here is a figure that I drew : ( note : it was not given in the question so there may be some mistakes )

I don't know how to solve it , am I missing any relation between inradius , circumradius and sides of a isosceles triangle?

EDIT: Is there a simple solution without using trigonometry ?

Answer 1

Denote the center of incircle by $O$ and of circumcircle by $O'$. It's easy to caclulate that $$\angle ABC=\angle ACB= 2\angle OBC=2\arctan\frac{r}{BC/2}=2\arctan\frac{1}{2}$$ Thus we can calculate the height $h$ with base $BC$ is $$h=\frac{BC}{2}\tan\angle ABC=24\tan(2\arctan\frac{1}{2})=24\times\frac{2\times\frac{1}{2}}{1-(\frac{1}{2})^2}=32$$ By symmetry, $O'$ shall lie on the height $h$. Consider the property of circumcircle that $O'A=O'B=O'C=R$ $$O'B^2=O'C^2=(\frac{BC}{2})^2+(h-R)^2=R^2$$ which gives the solution $$R=25$$

Answer 2

Let $M$ be the midpoint of $BC$, let $P$ be the point where the perpendicular from $O$ meets the side $AB$, and let $|PA|=:x$. Since the two tangent segments from $B$ to the incircle have equal length it follows that $|PB|=24$; therefore $|AB|=24+x$, and $|AO|^2= 12^2+x^2$. It follows that $$(24+x)^2=24^2+\bigl(12+\sqrt{12^2+x^2}\bigr)^2\ .$$ Solving for $x$ gives $x=16$, whence $|AB|=40$, $|AO|=20$, $|AM|=32$.

Now let $|MK|=:y$. Then $\sqrt{24^2+y^2}=32-y$, which enforces $y=7$. It follows that $R=32-7=25$.

Answer 3

$AI=k$, $\triangle AIF\sim\triangle ABD$ $\Longrightarrow$ $AB=2k$ , $AD=k+12$

$AB^{2}=BD^{2}+AD^{2}$ $\Longrightarrow$ $k=20$ , $AF=s-a=16$ , $AT=20$

$\triangle AOT\sim\triangle AIF$ $\Longrightarrow$ $\dfrac{AO}{AT}=\dfrac{AI}{AF}=\dfrac{5}{4}$ $\Longrightarrow$ $AO=25$