OrthocentreH: It is the point of intersection of altitude of a triangle.The points O,G,H are collinear. If `L_(1)=a_(1)x+b_(1)y+c_(1)+0` etc. then any line through the intersection of `L_(1)` and `L_(2)` i.e. `L_(1)+lamdaL_(2)=0` and perpendiculart to `L_(3)` is
`(L_(1))/(a_(1)a_(3)+b_(1)b_(3))=(L_(2))/(a_(2)a_(3)+b_(2)b_(3))`........(A)
The vetices of a triangle are `A(p,p tan alpha),B(q,q tan beta),C(r,r tan gamma)`. If circumcentre O of triangle ABC is at the origin and `H(barx,bary)` be its orthocentre, then show that
`(barx)/(bary)=(cos alpha+cos beta+cos gamma)/(sin alpha+sin beta+singamma)`

Answer

To solve the problem, we need to show that the ratio of the coordinates of the orthocenter \( H \) of triangle \( ABC \) is given by:

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

where the vertices of the triangle are given as:
- \( A(p, p \tan \alpha) \)
- \( B(q, q \tan \beta) \)
- \( C(r, r \tan \gamma) \)

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

### Step 2: Calculate the circumradius \( R \)
Since the circumcenter \( O \) is at the origin, we can find the circumradius \( R \) using the distance formula from the origin to each vertex.

For vertex \( A \):
\[ OA = \sqrt{p^2 + (p \tan \alpha)^2} = \sqrt{p^2(1 + \tan^2 \alpha)} = p \sec \alpha \]

For vertex \( B \):
\[ OB = \sqrt{q^2 + (q \tan \beta)^2} = \sqrt{q^2(1 + \tan^2 \beta)} = q \sec \beta \]

For vertex \( C \):
\[ OC = \sqrt{r^2 + (r \tan \gamma)^2} = \sqrt{r^2(1 + \tan^2 \gamma)} = r \sec \gamma \]

### Step 3: Set the circumradius equal for all vertices
Since all vertices lie on the circumcircle, we have:
\[ p \sec \alpha = q \sec \beta = r \sec \gamma = R \]

### Step 4: Express \( p, q, r \) in terms of \( R \)
From the above equations, we can express \( p, q, r \) as:
\[ p = R \cos \alpha, \quad q = R \cos \beta, \quad r = R \cos \gamma \]

### Step 5: Find the coordinates of the orthocenter \( H \)
The orthocenter \( H \) can be calculated using the formula for the coordinates of the orthocenter in terms of the vertices:
\[ H_x = A_x + B_x + C_x = p + q + r \]
\[ H_y = A_y + B_y + C_y = p \tan \alpha + q \tan \beta + r \tan \gamma \]

Substituting \( p, q, r \):
\[ H_x = R \cos \alpha + R \cos \beta + R \cos \gamma = R(\cos \alpha + \cos \beta + \cos \gamma) \]
\[ H_y = R \tan \alpha \cos \alpha + R \tan \beta \cos \beta + R \tan \gamma \cos \gamma \]
Using \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ H_y = R(\sin \alpha + \sin \beta + \sin \gamma) \]

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

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