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

Revision as of 20:56, 18 November 2023

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

It is quite clear that this is just 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 (Straightforward)

We can represent Abdul, Bharat and Chiang as $A$ , $B$ , and $C$ , respectively. Since we have $\angle ABC=60^\circ$ and $\angle BCA=90^\circ$ , this is obviously a $30-60-90$ triangle, and it would not matter where $B$ is. 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

Video Solution by MegaMath

https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s

~megahertz13

Video Solution 1 by OmegaLearn

https://youtu.be/mx2iDUeftJM

Video Solution

https://youtu.be/wuew6LaAM48

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

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/N2lyYRMuZuk?si=_Y5mdCFhG-XD7SaG&t=631

~Math-X

@@ Line 56: / Line 56: @@
 ~ESAOPS
+==Video Solution by MegaMath==
+https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s
+~megahertz13
 ==Video Solution 1 by OmegaLearn ==

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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