Let $a,b,c$ be non-zero real numbers. The lines $ax + by = c$ and $bx + cy = a$ are perpendicular and intersect at a point $P$ such that $P$ also lies on the line $y=2x$. Compute the coordinates of point $P$.

[i]2016 CCA Math Bonanza Individual #6[/i]
1. **Determine the slopes of the lines:**
   The given lines are \( ax + by = c \) and \( bx + cy = a \). The slopes of these lines are:
   \[
   \text{slope of } ax + by = c \text{ is } -\frac{a}{b}
   \]
   \[
   \text{slope of } bx + cy = a \text{ is } -\frac{b}{c}
   \]

2. **Use the perpendicularity condition:**
   Since the lines are perpendicular, the product of their slopes is \(-1\):
   \[
   \left( -\frac{a}{b} \right) \left( -\frac{b}{c} \right) = 1 \implies \frac{a}{c} = 1 \implies a = c
   \]

3. **Substitute \( c = -a \) into the equations:**
   The equations become:
   \[
   ax + by = -a
   \]
   \[
   bx + cy = a
   \]
   Substituting \( c = -a \) into the second equation:
   \[
   bx - ay = a
   \]

4. **Add the equations to eliminate \( y \):**
   \[
   ax + by + bx - ay = -a + a
   \]
   \[
   (a + b)x + (b - a)y = 0
   \]
   \[
   x(a + b) = y(a - b)
   \]
   \[
   y = \left( \frac{a + b}{a - b} \right)x
   \]

5. **Use the condition that \( P \) lies on \( y = 2x \):**
   Since \( P \) lies on \( y = 2x \), we have:
   \[
   \frac{a + b}{a - b} = 2
   \]
   Solving for \( b \):
   \[
   a + b = 2(a - b)
   \]
   \[
   a + b = 2a - 2b
   \]
   \[
   3b = a
   \]
   \[
   b = \frac{1}{3}a
   \]

6. **Substitute \( b = \frac{1}{3}a \) into the equations:**
   The equations become:
   \[
   ax + \frac{1}{3}ay = -a
   \]
   \[
   \frac{1}{3}ax - ay = a
   \]

7. **Simplify the equations:**
   Dividing both equations by \( a \):
   \[
   x + \frac{1}{3}y = -1
   \]
   \[
   \frac{1}{3}x - y = 1
   \]

8. **Solve the system of equations:**
   Multiply the second equation by 3 to eliminate the fraction:
   \[
   x + \frac{1}{3}y = -1
   \]
   \[
   x - 3y = 3
   \]
   Subtract the first equation from the second:
   \[
   (x - 3y) - (x + \frac{1}{3}y) = 3 - (-1)
   \]
   \[
   -3y - \frac{1}{3}y = 4
   \]
   \[
   -\frac{10}{3}y = 4
   \]
   \[
   y = -\frac{6}{5}
   \]
   Substitute \( y = -\frac{6}{5} \) back into the first equation:
   \[
   x + \frac{1}{3} \left( -\frac{6}{5} \right) = -1
   \]
   \[
   x - \frac{2}{5} = -1
   \]
   \[
   x = -1 + \frac{2}{5}
   \]
   \[
   x = -\frac{3}{5}
   \]

The final answer is \(\boxed{\left( - \frac{3}{5}, - \frac{6}{5} \right)}\)