# 2023 AMC 10A Problems/Problem 13

## Problem

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 Chiang measures $60^\circ$. What is the square of the distance (in feet) between Abdul and Bharat?

$\textbf{(A) } 1728 \qquad \textbf{(B) } 2601 \qquad \textbf{(C) } 3072 \qquad \textbf{(D) } 4608 \qquad \textbf{(E) } 6912$

## 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 $x=32\sqrt3\implies x^2=\boxed{\textbf{(C) }3072.}$

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

~Technodoggo ~(minor grammar edits by vadava_lx)

## Solution 2 (no law of sines)

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 $r=16\times\sqrt3$. Due to the triangle inequality, $\overline{AB}$ is maximized when B is on the diameter passing through A, giving a length of $32\times\sqrt3$ and when squared gives $\boxed{\textbf{(C) }3072}$.

## Solution 3 (Guessing)

Guess that the optimal configuration is a 30-60-90 triangle, as an equilateral triangle gives an answer of $48^2=2304$, which is not on the answer choices. Its ratio is $\frac{48}{\sqrt{3}}$, so $\overline{AB}=\frac{96}{\sqrt{3}}$.

Its square is then $\frac{96^2}{3}=\boxed{\textbf{(C) }3072}$

~not_slay

~wangzrpi

## Solution 4

We use $A$, $B$, $C$ to refer to Abdul, Bharat and Chiang, respectively. We draw a circle that passes through $A$ and $C$ and has the central angle $\angle AOC = 60^\circ \cdot 2$. Thus, $B$ is on this circle. Thus, the longest distance between $A$ and $B$ is the diameter of this circle. Following from the law of sines, the square of this diameter is $$\left( \frac{48}{\sin 60^\circ} \right)^2 = \boxed{\textbf{(C) 3072}}.$$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

## Solution 5

We can represent Abdul, Bharat and Chiang as $A$, $B$, and $C$, respectively. Since we have $\angle ABC=60^\circ$ and (from other solutions) $\angle BCA=90^\circ$, this is a $30-60-90$ triangle. By the side ratios of a $30-60-90$ triangle, we can infer that $AB=\frac{48\times 2}{\sqrt{3}}$. Squaring AB we get $\boxed{\textbf{(C) 3072}}$.

~ESAOPS

## Solution 6 (Logic)

As in the previous solution, refer to Abdul, Bharat and Chiang as $A$, $B$, and $C$, respectively- we also have $\angle ABC=60^\circ$. Note that we actually can't change the lengths, and thus the positions, of $AB$ and $BC$, because that would change the value of $\angle ABC$ (if we extended either of these lengths, then we could simply draw $AC'$ such that $BC'$ is perpendicular to $AC'$, so $AB$ is unchanged). We can change the position of $AC$ to alter the values of $AC$ and $BC$, but throughout all of these changes, $AB$ remains unvaried. Therefore, we can let $\angle ACB = 90^\circ$.

(What is the justification for all of these assumptions??)

It follows that $\triangle ABC$ is $30$-$60$-$90$, and $BC = \frac{48}{\sqrt{3}}$. $AB$ is then $\frac{96}{\sqrt{3}},$ and the square of $AB$ is $\boxed{\textbf{(C) 3072}}$.

-Benedict T (countmath1)

## Solution 7

$\angle BAC = 90^\circ - 60^\circ = 30^\circ$ (why?)

$$\implies BC = \frac {AB}{2} \implies AB^2 = BC^2+ AC^2 \implies$$

$$AB^2 = \frac {4}{3} \cdot 48^2 = 4 \cdot 48 \cdot 16 \approx 200 \cdot 16 = 3200.$$ We look at the answers and decide: the square of $AB$ is $\boxed{\textbf{(C) 3072}}$.

-vvsss

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~Math-X