Orthocentre and circumcentre at $(9,5)$ and $(0,0)$

Question

Let $ABC$ be a triangle having respectively orthocentre and circumcentre at $(9,5)$ and $(0,0)$. If the equation of side $BC$ is $2x-y=10$, then find the possible co-ordinates of vertex $A$.
MY ATTEMPT: I found the equation of the perpendicular on $BC$ passing through $H$. So, $A$ is a point on this line. But, then what must I do next. Please help.

Answer 1

The projections of $H$ and $O$ on the $BC$-side are the feet of the $A$-altitude and the midpoint $M_A$ of $BC$. By Euler's theorem, given $O$ and $H$ we are able to locate the centroid $G$. Moreover, we know that the segment joining $A$ and $M_A$ is split by $G$ in two pieces fulfilling $\frac{AG}{GM_A}=2$. So $H,O$ and the $BC$-line fix the position of $A$:

1. Let $M_A$ be the projection of $O$ on $BC$;
2. Let $G=\frac{2O+H}{3}$;
3. $A=\color{red}{3G-2M_A}.$

In our case, we may prove $A=\color{red}{(1;9)}$ with very few computations.

Answer 2

Let $G$ be the centroid of $ABC$. Then, it is well known that $O$, $G$, and $H$ are collinear (on a line called the Euler line of $ABC$) and that $OG:GH=1:2$, so $$G=\frac{2}{3}O+\frac{1}{3}H\,.$$ Let $M_a$ be the midpoint of $BC$ (which is given by the intersection of the line $BC$ and its perpendicular from $O$). Compute $M_a$. Then, we know that $A$, $G$, and $M_a$ are collinear and that $AG:GM_a=2:1$, whence $G=\frac{2}{3}M_a+\frac{1}{3}A$, or $$A=3G-2M_a=2O+H-2M_a\,.$$ Compute $A$. In fact, you can now find $B$ and $C$ (up to permutation) via determining the intersection points between the circle centered at $O$ with radius $OA$ and the line $BC$. If computations involved are done right, with $O=(0,0)$, $H=(9,5)$, and $BC=\big\{(x,y)\in\mathbb{R}^2\,|\,2x-y-10=0\big\}$, we have $M_a=(4,-2)$, $A=(1,9)$, and $$\{B,C\}=\Biggl\{\left(\frac{20-\sqrt{310}}{5},\frac{-10-2\sqrt{310}}{5}\right),\left(\frac{20+\sqrt{310}}{5},\frac{-10+2\sqrt{310}}{5}\right)\Biggr\}\,.$$