The vertices of a triangle are `A(p,p tan alphat), B(q,qtan beta),C(r,ttan gamma)`. If the circumcentre O of triangle ABC is at the origin and `H(barx,bary)` be its orthocentre then prove that
`(barx)/(bary)=(cos alpha+cos beta+cos gamma)/(sin alpha+sin beta+sin gamma)`

Answer

To prove that

\[ \frac{\bar{x}}{\bar{y}} = \frac{\cos \alpha + \cos \beta + \cos \gamma}{\sin \alpha + \sin \beta + \sin \gamma} \]

given the vertices of triangle \( A(p, p \tan \alpha) \), \( B(q, q \tan \beta) \), and \( C(r, r \tan \gamma) \) with circumcenter \( O \) at the origin, we will follow these steps:

### Step 1: Determine the coordinates of the vertices
The coordinates of the vertices are given as:
- \( A(p, p \tan \alpha) \)
- \( B(q, q \tan \beta) \)
- \( C(r, r \tan \gamma) \)

### Step 2: Use the circumcenter condition
Since the circumcenter \( O \) is at the origin, the distances from \( O \) to each vertex must be equal to the circumradius \( R \). Therefore, we have:

\[ OA = OB = OC = R \]

Calculating these distances:
- \( OA = \sqrt{p^2 + (p \tan \alpha)^2} = \sqrt{p^2(1 + \tan^2 \alpha)} = p \sec \alpha \)
- \( OB = \sqrt{q^2 + (q \tan \beta)^2} = q \sec \beta \)
- \( OC = \sqrt{r^2 + (r \tan \gamma)^2} = r \sec \gamma \)

Setting these equal to \( R \):
\[ p \sec \alpha = R, \quad q \sec \beta = R, \quad r \sec \gamma = R \]

From these, we can express \( p, q, r \) in terms of \( R \):
\[ p = R \cos \alpha, \quad q = R \cos \beta, \quad r = R \cos \gamma \]

### Step 3: Substitute the values of \( p, q, r \)
Now substituting these values back into the coordinates of the vertices:
- \( A(R \cos \alpha, R \sin \alpha) \)
- \( B(R \cos \beta, R \sin \beta) \)
- \( C(R \cos \gamma, R \sin \gamma) \)

### Step 4: Find the centroid \( G \)
The coordinates of the centroid \( G \) of triangle \( ABC \) are given by:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates:
\[ G\left( \frac{R \cos \alpha + R \cos \beta + R \cos \gamma}{3}, \frac{R \sin \alpha + R \sin \beta + R \sin \gamma}{3} \right) \]

### Step 5: Find the orthocenter \( H \)
The orthocenter \( H \) can be expressed in terms of the centroid \( G \) and the circumcenter \( O \) using the relation:
\[ \bar{x} = 3G_x - 0, \quad \bar{y} = 3G_y - 0 \]
Thus,
\[ \bar{x} = R \left( \cos \alpha + \cos \beta + \cos \gamma \right), \quad \bar{y} = R \left( \sin \alpha + \sin \beta + \sin \gamma \right) \]

### Step 6: Compute \( \frac{\bar{x}}{\bar{y}} \)
Now, we can find the ratio:
\[ \frac{\bar{x}}{\bar{y}} = \frac{R (\cos \alpha + \cos \beta + \cos \gamma)}{R (\sin \alpha + \sin \beta + \sin \gamma)} = \frac{\cos \alpha + \cos \beta + \cos \gamma}{\sin \alpha + \sin \beta + \sin \gamma} \]

### Conclusion
Thus, we have proved that:
\[ \frac{\bar{x}}{\bar{y}} = \frac{\cos \alpha + \cos \beta + \cos \gamma}{\sin \alpha + \sin \beta + \sin \gamma} \]