Berkeley Math Circle Monthly Contest 6

Solution:

Let us position our coordinate system so that $A$ is the origin. Let $B = (a, b)$ and $C = (c, d)$. Then the formula for the area of a triangle with given vertex coordinates gives $\frac{1}{2} = \frac{1}{2}|ad - bc|$, i.e. $ad - bc = \pm 1$.

Let $H = (x, y)$ be the orthocenter of $\triangle ABC$. The condition $CH \perp AB$ can be expressed in vector form as
$$0 = \overrightarrow{AB} \cdot \overrightarrow{CH} = \langle a, b \rangle \cdot \langle x - c, y - d \rangle$$
or
$$a x + b y = a c + b d.$$
Similarly, the condition $BH \perp AC$ can be expressed as
$$c x + d y = a c + b d.$$
Adding $d$ times equation (2) to $-b$ times equation (3)$:$
$$(a d - b c) x = d(a c + b d) - b(a c + b d).$$
Since $ad - bc = \pm 1$, it follows that $x$ is an integer. Similarly, we can prove that $y$ is an integer as well.