1. Given data

• $A(1,2)$
• $B(\alpha,\beta)$ and $C(\gamma,\delta)$ lie on the line $y=x+4$
• $\angle ABC=\dfrac{\pi}{6}$
• $\angle BAC=\dfrac{2\pi}{3}$

We need to find $\alpha^2+\gamma^2$.

2. Find the third angle

In triangle $ABC$,

So,

Hence, the sides opposite these equal angles are equal:

Therefore, triangle $ABC$ is isosceles with vertex at $A$.

3. Use the fact that $B$ and $C$ lie on the same line

Since both $B$ and $C$ lie on the line

side $BC$ lies on this line.

Because $AB=AC$, point $A$ must lie on the perpendicular bisector of segment $BC$. In an isosceles triangle, the bisector of angle $A$ is also the perpendicular bisector of the base $BC$.

Now,

• slope of line $BC$ is $1$ (since $y=x+4$),
• so slope of the perpendicular bisector is $-1$.

The angle at $A$ is $120^\circ$. Since triangle is symmetric about the angle bisector at $A$, each equal side makes an angle of $60^\circ$ with the bisector.

The bisector through $A(1,2)$ has slope $-1$, so its equation is

4. Find the midpoint of $BC$

Let midpoint of $BC$ be $M$. Since $M$ lies on both:

• the line $BC: y=x+4$,
• the perpendicular bisector: $y=-x+3$,

we solve

Then

So,

5. Find the length $AM$

6. Use triangle geometry to find $BM$

Since $AB=AC$, the line $AM$ is also the altitude and angle bisector. Thus,

Also, $AM \perp BC$, so triangle $ABM$ is right-angled at $M$.

Thus, in right triangle $ABM$,

so

7. Move from midpoint along the line $y=x+4$ to get $B$ and $C$

A direction vector along the line $y=x+4$ is $(1,1)$, whose unit vector is

Hence, moving a distance $BM=\dfrac{3\sqrt3}{\sqrt2}$ from $M$ along the line gives displacement

Therefore,

or vice versa.

So the $x$-coordinates are

(in either order).

8. Compute $\alpha^2+\gamma^2$

Let

Then

Hence,

Now,

So,

9. Final answer

This matches the stored correct answer.