Let $ riangle 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:

Question

Let $	riangle 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:

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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.

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:

Correct Answer: 36

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