Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Fixed minor typo and added See Also)
(28 intermediate revisions by 17 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Abdul and Chiang are standing <math>48</math> 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 <math>60^\circ</math>. What is the square of the distance (in feet) between Abdul and Bharat?  
+
Abdul and Chiang are standing <math>48</math> 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 <math>60^\circ</math>. What is the square of the distance (in feet) between Abdul and Bharat?  
  
−
<math>\textbf{(A) }1728\qquad\textbf{(B) }2601\qquad\textbf{(C) }3072\qquad\textbf{(D) }4608\qquad\textbf{(E) }6912</math>
+
<math>\textbf{(A) } 1728 \qquad \textbf{(B) } 2601 \qquad \textbf{(C) } 3072 \qquad \textbf{(D) } 4608 \qquad \textbf{(E) } 6912</math>
 +
==Video Solution by MegaMath==
 +
 
 +
https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s
  
 
==Solution 1==
 
==Solution 1==
Line 12: Line 15:
 
By the Law of Sines, we know that <math>\dfrac{\sin\theta}x=\dfrac{\sin60^\circ}{48}=\dfrac{\sqrt3}{96}</math>. Rearranging, we get that <math>x=\dfrac{\sin\theta}{\frac{\sqrt3}{96}}=32\sqrt3\sin\theta</math> where <math>x</math> is a function of <math>\theta</math>. We want to maximize <math>x</math>.  
 
By the Law of Sines, we know that <math>\dfrac{\sin\theta}x=\dfrac{\sin60^\circ}{48}=\dfrac{\sqrt3}{96}</math>. Rearranging, we get that <math>x=\dfrac{\sin\theta}{\frac{\sqrt3}{96}}=32\sqrt3\sin\theta</math> where <math>x</math> is a function of <math>\theta</math>. We want to maximize <math>x</math>.  
  
−
We know that the maximum value of <math>\sin\theta=1</math>, so this yields <math>x=32\sqrt3\implies x^2=\boxed{\text{(C) }3072.}</math>  
+
We know that the maximum value of <math>\sin\theta=1</math>, so this yields <math>x=32\sqrt3\implies x^2=\boxed{\textbf{(C) }3072.}</math>  
  
−
A quick checks verifies that <math>\theta=90^\circ</math> indeed works.  
+
A quick check verifies that <math>\theta=90^\circ</math> indeed works.  
  
 
~Technodoggo
 
~Technodoggo
 +
~(minor grammar edits by vadava_lx)
  
 
==Solution 2 (no law of sines)==
 
==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 <math>120</math>. 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 <math>\overline{AC}</math>. We will find that <math>r=16\times\sqrt3</math>. Due to the triangle inequality, <math>\overline{AB}</math> is maximized when B is on the diameter passing through A, giving a length of <math>32\times\sqrt3</math> and when squared gives <math>\boxed{\textbf{(C) }3072}</math>.
 +
 
 +
==Solution 3==
 +
 
 +
It is quite clear that this is just a 30-60-90 triangle as an equilateral triangle gives an answer of <math>48^2=2304</math>, which is not on the answer choices. Its ratio is <math>\frac{48}{\sqrt{3}}</math>, so <math>\overline{AB}=\frac{96}{\sqrt{3}}</math>.
 +
 
 +
Its square is then <math>\frac{96^2}{3}=\boxed{\textbf{(C) }3072}</math>
 +
 
 +
~not_slay
 +
 
 +
~wangzrpi
 +
 
 +
==Solution 4==
 +
 
 +
We use <math>A</math>, <math>B</math>, <math>C</math> to refer to Abdul, Bharat and Chiang, respectively.
 +
We draw a circle that passes through <math>A</math> and <math>C</math> and has the central angle <math>\angle AOC = 60^\circ \cdot 2</math>.
 +
Thus, <math>B</math> is on this circle.
 +
Thus, the longest distance between <math>A</math> and <math>B</math> is the diameter of this circle.
 +
Following from the law of sines, the square of this diameter is
 +
<cmath>
 +
\[
 +
\left( \frac{48}{\sin 60^\circ} \right)^2
 +
= \boxed{\textbf{(C) 3072}}.
 +
\]
 +
</cmath>
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Solution 5 (Straightforward)==
 +
 
 +
We can represent Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively.
 +
Since we have <math>\angle ABC=60^\circ</math> and <math>\angle BCA=90^\circ</math>, this is obviously a <math>30-60-90</math> triangle, and it would not matter where <math>B</math> is.
 +
By the side ratios of a <math>30-60-90</math> triangle, we can infer that <math>AB=\frac{48\times 2}{\sqrt{3}}</math>.
 +
Squaring AB we get <math>\boxed{\textbf{(C) 3072}}</math>.
 +
 
 +
~ESAOPS
 +
 
 +
==Solution 6 (Logic)==
 +
 
 +
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 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)
 +
 
 +
==Solution 7 (Simple)==
 +
<cmath>\angle BAC = 90^\circ - 60^\circ = 30^\circ \implies BC = \frac {AB}{2} \implies AB^2 = BC^2+ AC^2 \implies</cmath>
 +
<cmath>AB^2 = \frac {4}{3} \cdot 48^2 = 4 \cdot 48 \cdot 16 \approx 200 \cdot 16 = 3200.</cmath>
 +
We look at the answers and decide: the square of <math>AB</math> is <math>\boxed{\textbf{(C) 3072}}</math>.
 +
 
 +
-vvsss
 +
 
 +
==Video Solution by Power Solve (easy to digest!)==
 +
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 [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=BJKHsHQyoTg
 +
 
 +
==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
  
−
Let us begin by circumscribing the two points A and C so that the arc it determines has measure <math>120</math>. 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 <math>\overline{AC}</math>. We will find that <math>r=16*\sqrt3</math>. Due to the triangle inequality, <math>\overline{AB}</math> is maximized when B is on the diameter passing through A, giving a length of <math>32*\sqrt3</math> and when squared gives <math>\boxed{\text{(C) }3072}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2023|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2023|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:56, 19 May 2024

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?

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

Video Solution by MegaMath

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

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)

Solution 7 (Simple)

\[\angle BAC = 90^\circ - 60^\circ = 30^\circ \implies BC = \frac {AB}{2} \implies AB^2 = BC^2+ AC^2 \implies\] \[AB^2 = \frac {4}{3} \cdot 48^2 = 4 \cdot 48 \cdot 16 \approx 200 \cdot 16 = 3200.\] We look at the answers and decide: the square of $AB$ is $\boxed{\textbf{(C) 3072}}$.

-vvsss

Video Solution by Power Solve (easy to digest!)

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 [🔥Fast and Easy🔥]

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_13&oldid=219511"