Let ABC be a triangle. I be the incentre. If BI = 5 , CI = 4 , BC =a then what are the possible values a?

Question

Let $ABC$ be a triangle. $I$ be the incentre. The opposite side of $A$ is $a$ (i.e $BC=a$). If $BI = 5$ , $CI = 4$ then what are the possible values $a$?

We can always narrow down the value of a by considering the triangle $IBC$. Since the sum of two sides is greater than the third and the difference is smaller than the third we get $$1\lt a\lt 9$$
This is the only thing I've able to figure out. There just must be a better way to be more precise on the value of a.

All help is appreciated

Answer 1

Notice $$\angle IBC + \angle BCI = \frac12(\angle ABC + \angle BCA) = \frac12(180^\circ - \angle CAB) < 90^\circ\\ \implies \angle CIB = 180^\circ - (\angle IBC + \angle BCI ) > 90^\circ$$

$\triangle IBC$ is an obtuse triangle and $a > \sqrt{5^2+4^2} = \sqrt{41}$.

Conversely, when $\sqrt{41} < a < 9$, One can

• construct an obtuse triangle with $BI = 5$, $CI = 4$ and $BC = a$. We have $\angle CIB > 90^\circ$.
• reflect the line $BC$ with respect to $IB$ to obtain a new line $\ell_1$. $\ell_1$ passes through $B$ and is making an angle $\alpha_1 = 2\angle IBC$ with line $BC$.
• reflect the line $BC$ with respect to $IC$ to obtain another line $\ell_2$. $\ell_2$ passes through $C$ and is making an angle $\alpha_2 = 2\angle BCI$ with line $CB$.

Since $\angle CIB > 90^\circ$, $\alpha_1 + \alpha_2 = 2(\angle IBC + \angle BCI) < 180^\circ$. The two lines $\ell_1$ and $\ell_2$ will intersect at some point $A'$ on the same side of $I$ with respect to $BC$. It is easy to see the incenter of $\triangle A'BC$ equals to given $I$.

Combine these, we can conclude the legal range of $a$ is $( \sqrt{41}, 9 )$.