Question

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Hint:

When solving such problems, always check the triangle inequality condition to ensure the distances are valid.

Answer 1

Step 1: Understand the geometry of the problem.
The problem involves a circle with a diameter of 13 metres. Let the centre of the circle be \( O \). The two gates \( A \) and \( B \) are diametrically opposite on the boundary of the circle. Therefore, \( OA = OB = \frac{13}{2} = 6.5 \) metres, as the radius of the circle is half the diameter.
Step 2: Set up the equation.
Let the pole be erected at a point \( P \) on the boundary of the circle. Let the distances from the pole to gates \( A \) and \( B \) be \( PA = x \) and \( PB = y \). We are given that: \[ |x - y| = 7 \, \text{metres}. \]
Step 3: Apply the distance formula.
Since \( P \), \( A \), and \( B \) lie on the boundary of the circle, the points form a triangle. By the triangle inequality, the difference in distances from \( P \) to \( A \) and \( P \) to \( B \) must be less than the length of the diameter of the circle. Therefore, we have the inequality: \[ |x - y|<13. \]
Step 4: Solve the problem.
The given condition \( |x - y| = 7 \) is valid because it is less than the diameter. So, it is possible to erect the pole, and the distances from the two gates can be determined by solving the system of equations: \[ x - y = 7 \quad \text{or} \quad y - x = 7. \] Given that \( x + y = 13 \), we can solve for \( x \) and \( y \). Solving the system of equations, we get: \[ x = 10 \, \text{metres} \quad \text{and} \quad y = 3 \, \text{metres} \quad \text{(or vice versa)}. \]