In the triangle $ABC$ it is known that $BC = 5, AC - AB = 3$. Prove that $r <2 <r_a$ .
(here $r$ is the radius of the circle inscribed in the triangle $ABC$, $r_a$ is the radius of an exscribed circle that touches the sides of $BC$).
(Mykola Moroz)
parmenides51 wrote:
In the triangle $ABC$ it is known that $BC = 5, AC - AB = 3$. Prove that $r <2 <r_a$ .
(here $r$ is the radius of the circle inscribed in the triangle $ABC$,
$r_a$ is the radius of an inscribed circle that touches the sides of $BC$).
Maybe $r_a$ is a excircle?
Let $BC=a$, $CA=b$ and $AB=c$, then we need to prove
$r=\frac{1}{2}\sqrt{\frac{(b+c-a)(c+a-b)(a+b-c)}{a+b+c}}=2\sqrt{\frac{c-1}{c+4}}<2$
$r_a=\frac{1}{2}\sqrt{\frac{(a+b+c)(c+a-b)(a+b-c)}{b+c-a}}=2\sqrt{\frac{c+4}{c-1}}>2$, which is obvious!