Find the length of segment $GM$ in the triangle below

Question

In a triangle $ABC$, right at $B$, $AB=a$. Through its barycenter $G$ draw $MN$ perpendicular to $AC$, such that $M$ and $N$ are on $BC$ and $AC$. If $GN=b$, calculate $GM$.
(Answer: $ \frac{a^2}{9b}$)

I tried to use the similarity of triangles but I didn't finish.

$\triangle CNM \sim \triangle CAB$

$\dfrac{CM}{AC}=\dfrac{b+c}{a}=\dfrac{CN}{BC}$

$GB = 2GO$

$BO=CO=AO$

Answer 1

EDIT – much simpler answer, original answer below.

Draw point $D$ on $AB$ such that $BD={1\over3}BA$. Line $GD$ is then parallel to $BC$: let $L$ be the point where it intersects the perpendicular from $B$ to $AC$, we have $BL=GM=c$.

If $K$ is the projection of $D$ onto $BL$, then $BK={1\over3}BH=GN=b$. In triangle $BDL$ we have then: $$ BD:BK=BL:BD,\quad \text{whence:}\quad c={a^2\over 9b}. $$

ORIGINAL ANSWER.

Drop perpendicular $BH$ onto $AC$. Note that $BH=3b$. Compute $AH=\sqrt{a^2-9b^2}$, $CH=9b^2/\sqrt{a^2-9b^2}$ and $$ OH={CH-AH\over2}={18b^2-a^2\over2\sqrt{a^2-9b^2}}. $$ We have then $HN={2\over3}OH$. Produce $$ to meet at $S$ the parallel to $$ passing through $B$. Triangle $$ is then similar to $$: from $=−=2−$, and $=$ we obtain $$ (2b-c):HN=BH:CH, $$ which gives: $$ c={a^2\over9b}. $$

Answer 2

$OC=OB=OA$

$a=2\cos{\theta}\cdot OC$

$3OG=OC$

$b=OG\cos{(2\theta)}$

$ON=OG\sin(2\theta)=(1/3)OC\sin(2\theta)$

$b+c=\cot(\theta)(ON+OC)$

$b+c=\cot(\theta)OC[(1/3)\sin(2\theta)+1]$

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$a=6b\cos(\theta) \sec{2\theta}$

$2a \cos^2{\theta} -6b\cos(\theta)-a=0$

$\cos(\theta)=\frac{3b\pm \sqrt{9b^2+2a^2}}{2a}$

$\cos^2(\theta)=\frac{18b^2+2a^2\pm6b\sqrt{9b^2+2a^2}}{4a^2}$

$\sin^2(\theta)=\frac{2a^2-18b^2\mp 6b\sqrt{9b^2+2a^2}}{4a^2}$

$OC=\frac{a^2}{3b\pm6b\sqrt{9b^2+2a^2}}$

Now that sine, cosine and OC are known in terms of a and b, values can be substituted in the expression from $b+c$ ultimately allowing solving for $c$.