Given a right triangle with `/_A =90^@`. Let M be the mid-point of BC. If the radii of the triangle `ABM and ACM` are `r_1 and r_2` then find the range of `r_1/r_2`.

Explanation

In the figure, D is mid-point of BC. From the geometry, `AD = BD = CD = a//2`
In `DeltaABD, r_(1) = (Delta_(1))/(s_(1))`
`rArr r_(1) = ((1)/(2) ((a)/(2))h)/(((a)/(2) + (a)/(2) c)/(2))`
`rArr r_(1) = (ah)/(2(a + c))`

In `Delta ACD, r_(2) = (Delta_(2))/(s_(2))`
`rArr r_(2) = ((1)/(2) ((a)/(2))h)/(((a)/(2) + (a)/(2) + b)/(2))`
`rArr r_(2) = (ah)/(2(a + b))`
`rArr (r_(1))/(r_(2)) = (a + b)/(a + c)`
`= (2R sin A + 2 R sin B)/(2R sin A + 2R sin C)`
`= (1 + sin B)/(1 + sin C) " " ("as " A = 90^(@))`
`= (1+ sin B)/(1+ cos B) " " ("as " C = 90^(@) - B)`
When B aproaches to `90^(@), (r_(1))/(r_(2))` approaches to 2
When B approaches to `0^(@), (r_(1))/(r_(2))` approaches to `(1)/(2)`
`:. (r_(1))/(r_(2)) in ((1)/(2), 2)`