What's the maximum integer value of the segment (IG) joining the incenter to the centroid in the triangle below?

Question

For reference:

Given the obtuse triangle $ABC$, obtuse at $B$, where $IG \parallel BC$, $I$ being the incenter and $G$ the centroid of this triangle and the perimeter of the triangle is $144$. Calculate the maximum integer value of $IG$.

My progress:

My draw and the relationships I found

$IG \parallel BC \implies a = \dfrac{b+c}{2}\\ \dfrac{AI}{IJ} = \dfrac{b+c}{2p}=\dfrac{b+c}{144}\\ c^2+a^2=2BM^2+\dfrac{b^2}{2}$

From angle bissector theorem ($\triangle ABC-AN::$),

$\dfrac{BJ}{CJ}=\dfrac{c}{b}\\ AJ^2 =bc-CJ\cdot BJ \\ \triangle ADK \sim \triangle ACB \implies:\\ \dfrac{AM}{CM}=\dfrac{AK}{BK}=\dfrac{b}{c}$

but I don't see where to fit $IG$..

Answer 1

$ \displaystyle \frac{AI}{IJ} = \frac{b+c}{a} = 2 ~ $ (given $IG \parallel BC$)

$b+c = 2a \implies a = 48 ~ $ (using $a+b+c = 144$)

$ \displaystyle IG = \frac23 \cdot JN = \frac23 \left(BN - BJ \right)$

$ \displaystyle = \frac23 \left(\frac a2 - \frac{ac}{b+c} \right) = \frac{a (b-c)}{3 (b+c)} = \frac{b-c}{6}$

(given $a = 48, b + c = 96$)

Using Triangle inequality, we must have $b - c \lt 48$ so the next biggest value of $(b-c)$ that gives an integer value for $IG$ is $42$ and hence max integer value of $IG$ is $7$

Answer 2

As you find out $2a = b+c$ so $a= 48$ and $b+c= 96$.

By angle bisector theorem we have $$BJ = {ac\over b+c}= {c\over 2}$$

Then $$IG = {2\over 3} JN = {2\over 3} ({a\over 2}-{c\over 2})$$ So $$IG = {b-48\over 3}$$ Since $b< a+c \implies b<72 $ so $$IG < {24\over 3} =8\implies IG\leq 7$$ Value $7$ is achieved at $b = 71$ and $ c = 25$ so $IG_{\max} =7$.