Difference between revisions of "2023 AMC 10A Problems/Problem 13"

Revision as of 06:35, 6 September 2024

1 Problem
2 Solution 1
3 Solution 2 (no law of sines)
4 Solution 3 (Guessing)
5 Solution 4
6 Solution 5
7 Solution 6 (Logic)
8 Solution 7
9 Video Solution by Math-X
10 Video Solution 🚀 Under 2 min 🚀
11 Video Solution by Power Solve
12 Video Solution by SpreadTheMathLove
13 Video Solution 1 by OmegaLearn
14 Video Solution by CosineMethod
15 Video Solution
16 Video Solution by MegaMath
17 See Also

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 by Math-X

https://youtu.be/GP-DYudh5qU?si=unB-KAz2AXgMuLSS&t=3337

~Math-X

Video Solution 🚀 Under 2 min 🚀

https://youtu.be/d5XeBKZvTGQ

~Education, the Study of Everything

Video Solution by Power Solve

https://www.youtube.com/watch?v=jkfsBYzBJbQ

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=nmVZxartc-o

Video Solution 1 by OmegaLearn

https://youtu.be/mx2iDUeftJM

Video Solution by CosineMethod

https://www.youtube.com/watch?v=BJKHsHQyoTg

Video Solution

https://youtu.be/wuew6LaAM48