Question

Consider a triangle \( \triangle ABC \) having the vertices \( A(1, 2) \), \( B(\alpha, \beta) \), and \( C(\gamma, \delta) \) and angles \( \angle ABC = \frac{\pi}{6} \) and \( \angle BAC = \frac{2\pi}{3} \). If the points \( B \) and \( C \) lie on the line \( y = x + 4 \), then \( \alpha^2 + \gamma^2 \) is equal to \( \dots \).

Answer 1

Correct Answer: 14

Given that points \( B \) and \( C \) lie on the line \( y = x + 4 \), and the triangle \( ABC \) has specific angles at \( A \), we proceed as follows:

Equation of the Line Passing Through \( A(1, 2) \):
Since \( \angle BAC = \frac{2\pi}{3} \) and \( \angle ABC = \frac{\pi}{6} \), we can find a line through \( A(1,2) \) making an angle \( \frac{\pi}{6} \) with the line \( y = x + 4 \). The slope of the line \( y = x + 4 \) is \( m = 1 \).
The slope of the line through \( A \) that makes an angle of \( \frac{\pi}{6} \) with \( y = x + 4 \) is: \[ m = \frac{1 \pm \tan \frac{\pi}{6}}{1 \mp \tan \frac{\pi}{6}} = \frac{1 \pm \frac{1}{\sqrt{3}}}{1 \mp \frac{1}{\sqrt{3}}}. \]

Simplifying, we get two possible slopes: \[ m = 2 + \sqrt{3} \quad \text{or} \quad m = 2 - \sqrt{3}. \]

Equations for Points \( B \) and \( C \):
Using these slopes, the equations of the lines through \( A(1,2) \) with these slopes are: \[ y - 2 = (2 + \sqrt{3})(x - 1) \quad \text{and} \quad y - 2 = (2 - \sqrt{3})(x - 1). \] We solve each of these with \( y = x + 4 \) to find the coordinates of \( B \) and \( C \).

Solving for \( \alpha \) and \( \gamma \):
- For \( y - 2 = (2 + \sqrt{3})(x - 1) \) and \( y = x + 4 \), we get: \[ x = \frac{4 + \sqrt{3}}{1 + \sqrt{3}}. \] - For \( y - 2 = (2 - \sqrt{3})(x - 1) \) and \( y = x + 4 \), we get: \[ x = \frac{4 - \sqrt{3}}{1 - \sqrt{3}}. \]

Calculating \( \alpha^2 + \gamma^2 \):
\[ \alpha^2 + \gamma^2 = \left( \frac{4 + \sqrt{3}}{1 + \sqrt{3}} \right)^2 + \left( \frac{4 - \sqrt{3}}{1 - \sqrt{3}} \right)^2 = 14. \]

Answer: 14