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

(→‎Solution 3)
m (Formatted answers)
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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 checks verifies that <math>\theta=90^\circ</math> indeed works.  
Line 21: Line 21:
 
Help with the diagram please?
 
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*\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>.
+
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{\textbf{(C) }3072}</math>.
  
 
==Solution 3==
 
==Solution 3==
Line 27: Line 27:
 
It is quite clear that this is just a 30-60-90 triangle. Its ratio is <math>\frac{48}{\sqrt{3}}</math>, so <math>\overline{AB}=\frac{96}{\sqrt{3}}</math>.
 
It is quite clear that this is just a 30-60-90 triangle. 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{\text{(C) }3072}</math>
+
Its square is then <math>\frac{96^2}{3}=\boxed{\textbf{(C) }3072}</math>
  
 
~not_slay
 
~not_slay

Revision as of 21:18, 9 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 Chaing 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 checks verifies that $\theta=90^\circ$ indeed works.

~Technodoggo

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 $120$. 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 $r=16*\sqrt3$. Due to the triangle inequality, $\overline{AB}$ is maximized when B is on the diameter passing through A, giving a length of $32*\sqrt3$ and when squared gives $\boxed{\textbf{(C) }3072}$.

Solution 3

It is quite clear that this is just a 30-60-90 triangle. 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

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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