Problem 1265

To solve this problem, we need to verify the given points and determine the coordinates of point $A$ in the triangle $ABC$ with orthocenter $H$.

1. Verify the given points and properties:
- $B = (0,0)$
- $C = (1,2)$
- $H = (5,0)$

2. Use the property of the orthocenter:
The orthocenter $H$ of a triangle is the point where the altitudes intersect. The altitudes are the perpendiculars dropped from each vertex to the opposite side.

3. **Find the equation of the line $BC$:**
The slope of $BC$ is:
$$\text{slope of } BC = \frac{2 - 0}{1 - 0} = 2$$
The equation of the line $BC$ is:
$$y = 2x$$

4. **Find the equation of the altitude from $A$ to $BC$:**
Since the altitude from $A$ to $BC$ is perpendicular to $BC$, its slope is the negative reciprocal of 2, which is $-\frac{1}{2}$. Let $A = (x_1, y_1)$. The equation of the altitude from $A$ to $BC$ is:
$$y - y_1 = -\frac{1}{2}(x - x_1)$$

5. **Find the equation of the altitude from $B$ to $AC$:**
The slope of $AC$ is:
$$\text{slope of } AC = \frac{y_1 - 2}{x_1 - 1}$$
The slope of the altitude from $B$ to $AC$ is the negative reciprocal:
$$\text{slope of altitude from } B = -\frac{x_1 - 1}{y_1 - 2}$$
The equation of the altitude from $B$ to $AC$ is:
$$y = -\frac{x_1 - 1}{y_1 - 2} x$$

6. **Find the coordinates of $A$ using the orthocenter property:**
Since $H = (5,0)$ is the orthocenter, it lies on the altitudes. Substitute $H$ into the altitude equations to find $A$.

For the altitude from $A$ to $BC$:
$$0 - y_1 = -\frac{1}{2}(5 - x_1)$$
Simplifying:
$$-y_1 = -\frac{1}{2}(5 - x_1)$$
$$y_1 = \frac{1}{2}(5 - x_1)$$
$$y_1 = \frac{5 - x_1}{2}$$

For the altitude from $B$ to $AC$:
$$0 = -\frac{x_1 - 1}{y_1 - 2} \cdot 5$$
Simplifying:
$$0 = -\frac{5(x_1 - 1)}{y_1 - 2}$$
Since $5 \neq 0$, we have:
$$x_1 - 1 = 0$$
$$x_1 = 1$$

Substitute $x_1 = 1$ into $y_1 = \frac{5 - x_1}{2}$:
$$y_1 = \frac{5 - 1}{2} = \frac{4}{2} = 2$$

Therefore, $A = (1, 2)$.

7. **Verify the coordinates of $A$:**
Since $A = (1, 2)$ is the same as $C$, this contradicts the problem statement that $A$, $B$, and $C$ are distinct points of the triangle. Therefore, there is no valid point $A$ that satisfies the given conditions.