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

Revision as of 21:09, 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

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 $x=32\sqrt3\implies x^2=\boxed{\text{(C) }3072.}$

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 $\boxed{\text{(C) }3072}$ .

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

@@ Line 22: / Line 22: @@
 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>.
+==Solution 3==
+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>
 ==See Also==
 {{AMC10 box|year=2023|ab=A|num-b=12|num-a=14}}
 {{MAA Notice}}

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

Page

Toolbox

Search