2023 AMC 10A Problems/Problem 13

Abdul and Chiang are standing $48$ feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chaing measures $60^\circ$ . What is the square of the distance (in feet) between Abdul and Bharat?

$\textbf{(A) }\frac1728\qquad\textbf{(B) }2601\qquad\textbf{(C) }3072\qquad\textbf{(D) }4608\qquad\textbf{(E) }6912$

Solution 1

Let $\theta=\angle ACB$ and $x=\overline{AB}$ .

By the Law of Sines, we know that $\dfrac{\sin\theta}x=\dfrac{\sin60^\circ}{48}=\dfrac{\sqrt3}{96}$ . Rearranging, we get that $x=\dfrac{\sin\theta}{\frac{\sqrt3}{96}}=32\sqrt3\sin\theta$ where $x$ is a function of $\theta$ . We want to maximize $x$ .

We know that the maximum value of $\sin\theta=1$ , so this yields $x=32\sqrt3\implies x^2=\boxed{\text{(C) }3072.}$

A quick checks verifies that $\theta=90^\circ$ indeed works.

~Technodoggo

Solution 2 (no law of sines)

Help with the diagram please?

Let us begin by circumscribing the two points A and C so that the arc it determines has measure $120$ . Then the point B will lie on the circle, which we can quickly find the radius of by using the 30-60-90 triangle formed by the radius and the midpoint of segment $\overline{AC}$ . We will find that $r=16*\sqrt3$ . Due to the triangle inequality, $\overline{AB}$ is maximized when B is on the diameter passing through A, giving a length of $32*\sqrt3$ and when squared gives $\boxed{\text{(C) }3072}$ .

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2023 AMC 10A Problems/Problem 13

Solution 1

Solution 2 (no law of sines)