I want to construct triangle $ABC$ and it is given:

- Segment $BC$
- $\angle BAC = \theta$
- There is a circle through $B,C$, the orthocenter $H$ of $\triangle ABC$ and the centroid $G$ of $\triangle ABC$.

I noticed that with $\angle ABC$, we have that $\angle BHC = 180^{\circ} – \theta$ and so we can actually construct the circle that goes through $BCHG$. I also thought about the homothety centered at $G$ that takes $H$ and leads to $O$ but I'm not quite sure how to use it.

## Best Answer

Here is an approach. I want you to be aware of the variations. The approach remains the same but be aware that positions of orthocenter and circumcenter would change depending on whether $\theta$ is acute or obtuse or right and whether the triangle is acute, obtuse or right. For example, if $\theta$ is acute, it is not necessary that $\angle BHC = 180^\circ - \theta~$. If it is an obtuse triangle, $\angle BHC = \theta$. So $H$ can either be on the minor arc of the other circle or on the major arc. If $\theta$ is obtuse, $A$ itself will be on the minor arc of the circumcircle.

The diagram that I have shown assumes $\theta$ is acute and for the measurement of $BC$ and $\theta$ I took, $\triangle ABC$ turned out to be an obtuse triangle.

First, to both sides of segment $BC$, draw lines at point $B$ and $C$ measuring angle $\left(90^\circ - \frac{\theta}{2}\right)$ to $BC$. Say the lines intersect at $A_1$ and $A'$. Now draw circumcircles to $\triangle A_1BC$ and $\triangle A'BC$, we call the circles $S$ and $S'$ respectively and their centers $O$ and $O'$.

Point $A$ that we seek is on one of these circles. If we choose circle $S$ for point $A$ then $H$ and $G$ are on $S'$.

If $M$ is the midpoint of $BC$, points $A_1, O, M, O'$ and $A'$ must be collinear. Also, we know that Euler line divides the circumcenter, centroid and orthocenter $O, G$ and $H$ in the ratio $~GH = 2 OG$. So using power of point $O$ with respect to circle $S'$,

$OG \cdot OH = OX \cdot OA' \implies 3 OG^2 = OX \cdot OA'$

Next, from $O$, we draw an arc with length $OG$ and it will intersect circle $S'$ at two points. We pick one of these points as point $G$. We then draw a line through $M$ and $G$ which is one of the medians. Its intersection with circle $S$ gives point $A$ and we have $\triangle ABC$.

After initial construction, you can look to develop other approaches using a few other relationships, for example we know $AH = 2 OM$. I will leave them to you.