Question

Let $\triangle ABC$ be an isosceles triangle in which $A$ is at $(-1, 0)$, $\angle A = \frac{2\pi}{3}$, $AB = AC$, and $B$ is on the positive $x$-axis. If $BC = 4\sqrt{3}$ and the line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is:

Answer 1

Given:
$A = (-1, 0), \quad \angle A = \frac{2\pi}{3}, \quad AB = AC, \quad \text{and} \quad BC = 4\sqrt{3}$

Step 1: Placing Points $B$ and $C$
Since $B$ is on the positive $x$-axis and $\triangle ABC$ is isosceles with $AB = AC$, the coordinates of $B$ can be represented as:
$B = (x, 0), \quad x > -1$
The angle $\angle A = \frac{2\pi}{3}$ implies that the line $AC$ makes an angle of $\frac{2\pi}{3}$ with the positive $x$-axis. Thus, the slope of line $AC$ is:
$\tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}$
Let the coordinates of $C$ be $(x_c, y_c)$. Since $AB = AC$ and $BC = 4\sqrt{3}$, we can use the distance formula to find $x$ and the coordinates of $C$.

Step 2: Calculating the Lengths
The length of $AB$ is given by:
$AB = |x + 1|$
Similarly, the length of $AC$ is also $|x + 1|$.
Given that $BC = 4\sqrt{3}$, we find the coordinates of $C$ such that it satisfies the isosceles condition and the length of $BC$.

Step 3: Equation of Line $BC$
The line $BC$ can be represented in the form:
$y = mx + c$
where $m$ is the slope and $c$ is the intercept. Using the coordinates of $B$ and $C$, we can find the equation of line $BC$.

Step 4: Intersection with Line $y = x + 3$
The line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$. Substituting the equation of $BC$ into $y = x + 3$ and solving for $\alpha$ and $\beta$ gives the required values.

Step 5: Calculating $\frac{\beta^4}{\alpha^2}$
After finding $\alpha$ and $\beta$, we compute:
$\frac{\beta^4}{\alpha^2}$
Given that the solution yields:
$\frac{\beta^4}{\alpha^2} = 36$

Conclusion: The value of $\frac{\beta^4}{\alpha^2}$ is 36.