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Construct points $D,E,F$ on $BC,AC,AB$ respectively such that $AD\perp BC,BE\perp AC,CF\perp AB$ . Without loss of generality, let $D(0,0),C(1,0),B(-1,0)$ . Let point $O$ be the orthocentre and let the inradius be $r$ . This means that $G(0,2r)$ . Finally let $A(0,a)$ .

Now, we get the equation of $BA$ to be $y=ax+a$ . This implies that the equation of $CF$ is $y=-\frac{1}{a}x+\frac{1}{a}$ . We know that $G$ lies on $CF$ and has $x$ coordinate 0, so subbing in $x=0$ , we get $G=(0,\frac{1}{a})=(0,2r)\Rightarrow a=\frac{1}{2r}$ .

We also know that the area of triangle is given by $AD\times BC \div 2=a=\frac{1}{2r}$ . Now, we shall find another way to obtain the area so we can get an equation and solve for $r$ . It is also evident that the area of the triangle is given by $BC\times r\div2 +AB\times r=r+r\sqrt{a^2+1}=r+\frac{\sqrt{4r^2+1}}{2}$ . Equating, we have $r+\frac{\sqrt{4r^2+1}}{2}=\frac{1}{2r}\\r\sqrt{4r^2+1}+2r^2=1\\2r^2-1=-r\sqrt{4r^2-1}\\4r^4-4r^2+1=4r^4+r^2\\5r^2=1\\r=\frac{1}{\sqrt{5}}\Rightarrow a=\frac{\sqrt{5}}{2}$ Since $AC=\sqrt{a^2+1}$ , we have $AC=\frac{3}{2}$ and thus, $\frac{AC}{BC}=\frac{\frac{3}{2}}{2}=\frac{3}{4}$ and the answer is 7.

this problem appeared in inmo 2016. so please see the solution either from the official site or from aops