To solve the given problem, we need to show that:
\[
\frac{\bar{y}}{\bar{x}} = \frac{\sin \alpha + \sin \beta + \sin \gamma}{\cos \alpha + \cos \beta + \cos \gamma}
\]
where the vertices of the triangle are \( A(a, a \tan \alpha) \), \( B(b, b \tan \beta) \), and \( C(c, c \tan \gamma) \), and the circumcenter coincides with the origin.
### Step-by-Step Solution:
1. **Find the Coordinates of the Centroid \( G \)**:
The coordinates of the centroid \( G \) of triangle \( ABC \) can be calculated as:
\[
G\left(\frac{a + b + c}{3}, \frac{a \tan \alpha + b \tan \beta + c \tan \gamma}{3}\right)
\]
2. **Using the Circumcenter**:
Given that the circumcenter coincides with the origin, we have:
\[
O(0, 0)
\]
3. **Finding the Orthocenter \( H(\bar{x}, \bar{y}) \)**:
The centroid \( G \) divides the line segment joining the orthocenter \( H \) and the circumcenter \( O \) in the ratio \( 2:1 \). Therefore, we can express the coordinates of \( H \) as:
\[
\bar{x} = \frac{3}{2} \cdot \frac{a + b + c}{3} = \frac{a + b + c}{2}
\]
\[
\bar{y} = \frac{3}{2} \cdot \frac{a \tan \alpha + b \tan \beta + c \tan \gamma}{3} = \frac{a \tan \alpha + b \tan \beta + c \tan \gamma}{2}
\]
4. **Finding the Distances**:
Since the circumcenter is at the origin, the distances from the circumcenter to each vertex are equal. This gives us:
\[
\sqrt{a^2 + (a \tan \alpha)^2} = \sqrt{b^2 + (b \tan \beta)^2} = \sqrt{c^2 + (c \tan \gamma)^2}
\]
This implies:
\[
a^2 (1 + \tan^2 \alpha) = b^2 (1 + \tan^2 \beta) = c^2 (1 + \tan^2 \gamma)
\]
Using \( 1 + \tan^2 \theta = \sec^2 \theta \), we can rewrite this as:
\[
a^2 \sec^2 \alpha = b^2 \sec^2 \beta = c^2 \sec^2 \gamma
\]
5. **Expressing \( a, b, c \)**:
From the above equality, we can express \( a, b, c \) in terms of a common radius \( r \):
\[
a = r \cos \alpha, \quad b = r \cos \beta, \quad c = r \cos \gamma
\]
6. **Substituting in \( \bar{y} \) and \( \bar{x} \)**:
Now substituting \( a, b, c \) into \( \bar{x} \) and \( \bar{y} \):
\[
\bar{x} = \frac{r \cos \alpha + r \cos \beta + r \cos \gamma}{2} = \frac{r}{2} (\cos \alpha + \cos \beta + \cos \gamma)
\]
\[
\bar{y} = \frac{r \tan \alpha \cdot r \cos \alpha + r \tan \beta \cdot r \cos \beta + r \tan \gamma \cdot r \cos \gamma}{2} = \frac{r^2}{2} \left( \frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta} + \frac{\sin \gamma}{\cos \gamma} \right)
\]
7. **Final Ratio**:
Now, we can find the ratio \( \frac{\bar{y}}{\bar{x}} \):
\[
\frac{\bar{y}}{\bar{x}} = \frac{\frac{r^2}{2} \left( \sin \alpha + \sin \beta + \sin \gamma \right)}{\frac{r}{2} (\cos \alpha + \cos \beta + \cos \gamma)} = \frac{r \left( \sin \alpha + \sin \beta + \sin \gamma \right)}{(\cos \alpha + \cos \beta + \cos \gamma)}
\]
Since \( r \) cancels out, we obtain:
\[
\frac{\bar{y}}{\bar{x}} = \frac{\sin \alpha + \sin \beta + \sin \gamma}{\cos \alpha + \cos \beta + \cos \gamma}
\]
### Conclusion:
Thus, we have shown that:
\[
\frac{\bar{y}}{\bar{x}} = \frac{\sin \alpha + \sin \beta + \sin \gamma}{\cos \alpha + \cos \beta + \cos \gamma}
\]
