geometrytriangles
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 ?
From the figure
one imediately derives the equations $${\rho\over h}={a\over 2R}\ ,\qquad{\rm i.e.,}\qquad a h=600\ ,$$ $$h+\rho+d=2R\ ,\qquad{\rm i.e.,}\qquad h+d=38\ ,$$ $$2R d=a^2 \ ,\qquad{\rm i.e.,}\qquad 50d = a^2\ .$$ Eliminating $h$ and $d$ one obtains a cubic equation for $a$, two of whose solutions are natural numbers. Given $a$ and $d$ one computes $s=\sqrt{4R^2-a^2}$, and the base $b$ of the triangle is $b=2\sqrt{a^2-d^2}$.
The formula for length of $AG = \frac{2}{3}\cdot\frac{1}{2}\sqrt{AC^2-BC^2+AB^2}$. Similarly, the length of $BG=\frac{2}{3}\cdot\frac{1}{2}\sqrt{AB^2+BC^2-AC^2}$. Now, we can find the circumradius of the triangle is $$\frac{AB\cdot BG\cdot AG}{4\sqrt{s(s-AB)(s-BG)(s-AG)}}$$, where $s=\frac{AG+BG+AB}{2}$.
If you are given $AB,BC,AC$, surely you cannot get the wrong answer unless there is a calculation error.
Best Answer
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 $$R=25$$