2023 AMC 10A Problem 13

Call the positions $A$ (Abdul), $C$ (Chiang), $B$ (Bharat), with $AC = 48$ and $\angle ABC = 60^\circ$.

All points $B$ on one side of line $AC$ from which $AC$ subtends $60^\circ$ lie on a single circle through $A$ and $C$ (inscribed angles on the same arc are equal). Since $AB$ is a chord of this circle, it is longest when it is a diameter, which happens when $\angle ACB = 90^\circ$.

In that position triangle $ABC$ is a $30$-$60$-$90$ triangle with the right angle at $C$, the $60^\circ$ angle at $B$, and $AC = 48$ opposite the $60^\circ$ angle. So
$$AB = \frac{AC}{\sin 60^\circ} = \frac{48}{\sqrt3/2} = \frac{96}{\sqrt3} = 32\sqrt3,$$
and $AB^2 = 1024 \cdot 3 = 3072$.

The answer is $\boxed{\textbf{(C)}\ 3072}$.

A fixed segment viewed at a fixed angle is the locus definition of a circular arc. Turning "as far as possible" into "longest chord of that circle" converts an optimization into the single fact that the diameter is the longest chord. Equivalently, the law of sines gives $AB = 2R \sin C$, maximized when $\sin C = 1$.

By the extended law of sines, $\frac{AB}{\sin C} = \frac{AC}{\sin B} = \frac{48}{\sin 60^\circ} = 32\sqrt3$. So $AB = 32\sqrt3 \sin C \le 32\sqrt3$, with equality at $C = 90^\circ$.

• Assuming the triangle must be equilateral or isosceles and answering 48^2 = 2304-type values, or maximizing the wrong side.
• Computing $AB = 48 \cdot \frac{\sqrt3}{2}$ (treating $AC$ as the hypotenuse), which gives $1728$, choice (A).