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

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