2023 AMC 10A Problem 13

Call the three people $A$ (Abdul), $B$ (Bharat) and $C$ (Chiang). We know $AC = 48$ and the angle at $B$ is $60^\circ$. We want $AB$ as long as possible.

1. Try an easy position first. If the triangle is equilateral, $AB = 48$ and $AB^2 = 2304$. That is not a choice, so $B$ can do better.
2. Where can $B$ stand? A fact from circles: every point on the same arc of a circle sees a chord at the same angle. So all the spots (on one side of $AC$) where $A$ and $C$ appear $60^\circ$ apart form one arc of a fixed circle through $A$ and $C$.
3. Longest distance. $A$ and $B$ are both on that circle, so $AB$ is a chord. The longest chord of a circle is a diameter. So $B$ should stand at the point directly across the circle from $A$.
4. What does that triangle look like? An angle that stands on a diameter is a right angle, so now $\angle ACB = 90^\circ$. With $60^\circ$ at $B$ and $90^\circ$ at $C$, the angle at $A$ is $30^\circ$. This is a $30$-$60$-$90$ triangle.
5. Use the $30$-$60$-$90$ side ratios $1 : \sqrt3 : 2$. The side opposite $60^\circ$ is $AC = 48$, which is the "$\sqrt3$" side. So one "unit" is $\frac{48}{\sqrt3} = 16\sqrt3$, and the hypotenuse is
$$AB = 2 \times 16\sqrt3 = 32\sqrt3.$$
6. Square it: $AB^2 = 32^2 \times 3 = 1024 \times 3 = 3072$.

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

"A fixed segment seen at a fixed angle" always describes an arc of a circle. That turns "as far as possible" into "the longest chord of that circle," which is the diameter. Look for this whenever a problem fixes two points and one angle and asks for the largest or smallest distance.

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 $\boxed{\textbf{(C)}\ 3072}$.