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"

(→‎Solution 2 (no law of sines))
m (→‎Solution 7)
 
(10 intermediate revisions by 7 users not shown)
Line 4: Line 4:
 
<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>
  
−
==Solution 1==
+
==Solution 1 (Using Trigonometry)==
  
 
[[Image:2023_10a_13.png]]
 
[[Image:2023_10a_13.png]]
Line 18: Line 18:
 
~Technodoggo
 
~Technodoggo
 
~(minor grammar edits by vadava_lx)
 
~(minor grammar edits by vadava_lx)
 +
 +
==Solution 2 (Inscribed Angles)==
 +
We can draw a circle such that the chord AC inscribes an arc of 120 degrees. This way, any point B on the circle not in the inscribed arc will form an angle of 60 degrees with <math>\angle{ABC}</math>. To maximize the distance between A and B, they must be opposite each other. So, the problem is now finding the length of the diameter of the circle. We know AOC is 120 degrees, so dropping a perpendicular form O to AC gives us the radius as <math>16\sqrt{3}</math>. So, the diameter is <math>32\sqrt{3}</math> which gives us the answer <math>\boxed{\textbf{(C) }3072}</math>
 +
 +
~AwesomeParrot
  
 
==Solution 3 (Guessing)==
 
==Solution 3 (Guessing)==
Line 24: Line 29:
  
 
Its square is then <math>\frac{96^2}{3}=\boxed{\textbf{(C) }3072}</math>
 
Its square is then <math>\frac{96^2}{3}=\boxed{\textbf{(C) }3072}</math>
 +
 +
Note: The distance between Abdul and Chiang is constant, so let that be represented as <math>{x}</math>. If we were dealing with an equilateral triangle, the height would be <math>{{x\sqrt3}/2}</math>, and if we were dealing with a 30-60-90 triangle, the height would be <math>{x\sqrt3}</math>, which is greater than <math>{{x\sqrt3}/2}</math>.
  
 
~not_slay
 
~not_slay
Line 54: Line 61:
 
~ESAOPS
 
~ESAOPS
  
−
==Solution 6 (Logic)==
+
==Solution 6 ==
−
 
 
−
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>.
 
−
 
 
−
(What is the justification for all of these assumptions??)
 
−
 
 
−
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 ==
 
 
<math>\angle BAC = 90^\circ - 60^\circ = 30^\circ</math> (why?)
 
<math>\angle BAC = 90^\circ - 60^\circ = 30^\circ</math> (why?)
  
Line 75: Line 71:
 
-vvsss
 
-vvsss
  
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=ISeW3ruGd-iLhQZi&t=2819
 +
~little-fermat
 
==Video Solution by Math-X ==
 
==Video Solution by Math-X ==
 
https://youtu.be/GP-DYudh5qU?si=unB-KAz2AXgMuLSS&t=3337
 
https://youtu.be/GP-DYudh5qU?si=unB-KAz2AXgMuLSS&t=3337
 
   
 
   
 
~Math-X
 
~Math-X
 +
 
==Video Solution 🚀 Under 2 min 🚀==
 
==Video Solution 🚀 Under 2 min 🚀==
  

Latest revision as of 17:22, 19 December 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?

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

Solution 1 (Using Trigonometry)

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 (Inscribed Angles)

We can draw a circle such that the chord AC inscribes an arc of 120 degrees. This way, any point B on the circle not in the inscribed arc will form an angle of 60 degrees with $\angle{ABC}$. To maximize the distance between A and B, they must be opposite each other. So, the problem is now finding the length of the diameter of the circle. We know AOC is 120 degrees, so dropping a perpendicular form O to AC gives us the radius as $16\sqrt{3}$. So, the diameter is $32\sqrt{3}$ which gives us the answer $\boxed{\textbf{(C) }3072}$

~AwesomeParrot

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

Note: The distance between Abdul and Chiang is constant, so let that be represented as ${x}$. If we were dealing with an equilateral triangle, the height would be ${{x\sqrt3}/2}$, and if we were dealing with a 30-60-90 triangle, the height would be ${x\sqrt3}$, which is greater than ${{x\sqrt3}/2}$.

~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

$\angle BAC = 90^\circ - 60^\circ = 30^\circ$ (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 Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=ISeW3ruGd-iLhQZi&t=2819 ~little-fermat

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

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

Video Solution by MegaMath

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

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=237307"