A problem on a triangle's inradius and circumradius . 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 ?
 A: 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
$$R=25$$
A: 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 $R=32-7=25$.
A: 
$AI=k$, $\triangle AIF\sim\triangle ABD$ $\Longrightarrow$ $AB=2k$ , $AD=k+12$    
$AB^{2}=BD^{2}+AD^{2}$ $\Longrightarrow$ $k=20$ , $AF=s-a=16$ , $AT=20$   
$\triangle AOT\sim\triangle AIF$ $\Longrightarrow$ $\dfrac{AO}{AT}=\dfrac{AI}{AF}=\dfrac{5}{4}$ $\Longrightarrow$ $AO=25$
A: The standard formula for the inradius is r = area/s,where s is the semi-perimeter.Letting x = the triangle sides that are equal,we have s=(1/2)(x+x+48)=
x+24.The area is calculated  using Heron"s formula = sqrt(s(s-a)(s_b)(s_c)).In
your problem,this becomes sqrt((x+24)(24)(24)(x-24)= 24sqrt(x^2 - 576).
Substituting,we get:  12 = 24sqrt(x^2 - 576)/(x + 24).Dividing by 12,and cross-
multiplying, x+24 = 2sqrt(x^2 - 576).Squaring,(x+24)^2 = 4(x^2 - 576),or
x^2 + 48x + 576 = 4(x^2 - 576) = 4x^2 - 2304.Expanding and transposing,
3x^2 - 48x - 2880=0. Dividing by 3,x^2 - 16x - 960 =0.Solving for x,
2x = 16 + sqrt(256 + 3840),or 2x=16+sqrt(4096),and x=8 + 32 = 40.Then the
area is 24sqrt(1600 - 576) = 24sqrt(1024)=24*32 = 768.A formula for the
circumradius is R = abc/(4*area).In your problem,this becomes:
R = (40)(40)(48)/((4)(8)(8)(12)) = (5)(%)(4)/(4) = 25.      Ed Gray 
