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

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==Solution 6 (Logic)==
 
==Solution 6 (Logic)==
  
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As in the previous solution, refer to Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively- we also have <math>\angle ABC=60^\circ</math>. Note that we actually can't change the lengths (and thus the positions) of <math>AB</math> and <math>BC</math>, because that would change the value of <math>\angle ABC</math>. We can change the position of <math>AC</math> to alter the values of <math>AC</math> and <math>BC</math>, but throughout all of these changes, <math>AB</math> remains unvaried. Therefore, we can let <math>\angle ACB = 90^\circ</math>.  
+
As in the previous solution, refer to Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively- we also have <math>\angle ABC=60^\circ</math>. Note that we actually can't change the lengths, and thus the positions, of <math>AB</math> and <math>BC</math>, because that would change the value of <math>\angle ABC</math> (if we extended either of these lengths, then we could simply draw <math>AC'</math> such that <math>BC'</math> is perpendicular to <math>AC'</math>, so <math>AB</math> is unchanged). We can change the position of <math>AC</math> to alter the values of <math>AC</math> and <math>BC</math>, but throughout all of these changes, <math>AB</math> remains unvaried. Therefore, we can let <math>\angle ACB = 90^\circ</math>.  
  
−
It follows thfat <math>\triangle ABC</math> is <math>30</math>-<math>60</math>-<math>90</math>, and <math>BC = \frac{48}{\sqrt{3}}</math>. <math>AB</math> is then <math>\frac{96}{\sqrt{3}},</math> and the square of <math>AB</math> is <math>\boxed{\textbf{(C) 3072}}</math>.
+
It follows that <math>\triangle ABC</math> is <math>30</math>-<math>60</math>-<math>90</math>, and <math>BC = \frac{48}{\sqrt{3}}</math>. <math>AB</math> is then <math>\frac{96}{\sqrt{3}},</math> and the square of <math>AB</math> is <math>\boxed{\textbf{(C) 3072}}</math>.
  
 
-Benedict T (countmath1)
 
-Benedict T (countmath1)

Revision as of 09:07, 27 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

2023 10a 13.png

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

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 $\angle ACB = 90^\circ$.

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)

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


See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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