Animated Solution for Mathematics - Straight Lines: Let $ABC$ 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 :

Welcome, future engineer. Today, we are going to dissect a problem that is a masterclass in coordinate geometry. It is not just about finding an intersection point; it is about orchestrating a symphony of geometric properties to arrive at a beautiful, clean integer.

Imagine you are standing on the Cartesian plane at point $A(-1,0)$. You are looking at an isosceles triangle $ABC$ where the vertex angle at $A$ is a wide, obtuse $120^{\circ }$. The base $BC$ is anchored at a length of $4\sqrt{3}$.

First, we must understand the internal structure of our triangle. Since $△ABC$ is isosceles with $AB=AC$, the base angles must be equal. Given $\angle A=120^{\circ }$, the sum of the remaining angles is $180^{\circ }-120^{\circ }=60^{\circ }$. Thus, $\angle B=\angle C=30^{\circ }$.

This symmetry is our greatest asset. We apply the Sine Rule:

Substituting the known values, we have:

The $\sqrt{3}$ terms cancel out, and we find that $2c=8$, which means $c=4$. Our triangle is perfectly defined.

Now, we place our triangle on the grid. Vertex $B$ lies on the positive x-axis. Since $A$ is at $(-1,0)$ and the distance $AB=4$, $B$ must be at $(-1+4,0)$, which is $(3,0)$.

For vertex $C$, we use the parametric form. The line segment $AC$ has length $4$ and makes an angle of $120^{\circ }$ with the positive x-axis. Thus:

Using $\cos 120^{\circ }=-1/2$ and $\sin 120^{\circ }=\sqrt{3}/2$, we find $C=(-3,2\sqrt{3})$. We have successfully mapped our vertices: $B(3,0)$ and $C(-3,2\sqrt{3})$.

With $B$ and $C$ known, the slope of line $BC$ is:

Using the point-slope form with $B(3,0)$, the equation becomes $y-0=-\frac{1}{\sqrt{3}}(x-3)$, which simplifies to $x+\sqrt{3}y=3$.

We are told this line intersects $y=x+3$ at $(\alpha ,\beta )$. Substituting $y=x+3$ into our line equation, we get $x+\sqrt{3}(x+3)=3$. Solving for $x$ (which is $\alpha$), we find:

Consequently, $\beta =\alpha +3=\frac{6}{\sqrt{3}+1}$.

We need to calculate $\frac{{\beta }^4}{{\alpha }^2}$. This looks terrifying, but let us look closer. We have $\beta =\frac{6}{\sqrt{3}+1}$ and $\alpha =\frac{3(1-\sqrt{3})}{\sqrt{3}+1}$.

When we compute the ratio, the denominators $(\sqrt{3}+1)$ will interact with the powers. Specifically, the expression becomes:

Simplifying this, we get:

The term inside the bracket is a difference of squares: $1-3=-2$. Squaring this gives $4$. So, we have:

And there it is—a perfect, elegant 36. Never fear the algebra; trust the geometry.