Find The range of $r/R$.

Question

Given a triangle $ABC$ with angle $A=90^{\circ}$. Let $M$ be the midpoint of $BC$. If the inradii of the triangles $ABM$ and $ACM$ are $r$ and$\ R$ respectively, then find the range of $\dfrac rR$ .

Answer 1

$D$ and $d$ the middle of $AB$ and $AC$, $BM=MC$, the inradius of $ABM$ is:

$$r= \frac{2*AD*DM}{AB+AM+BM}$$

$$R= \frac{2*Ad*dM}{AC+AM+BM}$$

$$r/R=\frac{AD*DM}{AB+AM+BM} X \frac{AC+AM+BM}{Ad*dM} $$

Answer 2

For a right triangle, the median bisecting the hypotenuse, is half the hypotenuse itself. Considering the legs of your triangle as b and c, and the hypotenuse, a [as per convention]. We have:

$$r = \frac{2(Area \: of \: ABM)}{Perimeter \: of\: ABM} = \frac{2(Area\: of\: ABM)}{c+a}$$ $$R = \frac{2(Area \: of \: ACM)}{Perimeter \: of\: ACM} = \frac{2(Area\: of\: ACM)}{b+a}$$

Considering that a median bisects the triangle into two parts of equal area, we divide the terms:

$$\frac{r}{R} = \frac{2(Area \: of \: ABM)(b+a)}{2(Area \: of\: ACM)(c+a)} = \frac{b+a}{c+a}$$

That I think is the best we can get, apart from some other algebraic manipulation you may involve. And your question is framed such that it has, of course, symmetric solutions, so take care of that.