2023 AMC 10A Problem 13

Points that see segment $AC$ at a fixed $60^\circ$ lie on a circular arc through $A$ and $C$

The farthest such point from Abdul is a diameter endpoint; find the circle’s diameter with the law of sines

Let $A$ be Abdul, $C$ be Chiang with $AC = 48,$ and $B$ be Bharat with $\angle B = 60^\circ.$ Every point seeing $AC$ at $60^\circ$ lies on one circular arc, so all valid $B$ sit on a circle where chord $AC$ subtends $60^\circ.$ The law of sines gives its diameter, $\frac{AC}{\sin 60^\circ} = \frac{48}{\frac{\sqrt3}{2}} = 32\sqrt3.$ Now $AB$ is a chord, and a chord is longest when it’s a diameter. So $AB = 32\sqrt3$ and $AB^2 = 1024 \cdot 3 = 3072.$ Thus, C is the correct answer.