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

(Created page with "Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you ge...")
 
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Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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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?
  
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-Jonathan Yu
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<math>\textbf{(A) }\frac1728\qquad\textbf{(B) }2601\qquad\textbf{(C) }3072\qquad\textbf{(D) }4608\qquad\textbf{(E) }6912</math>
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==Solution 1==
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[[Image:2023_10a_13.png]]
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Let <math>\theta=\angle ACB</math> and <math>x=\overline{AB}</math>.
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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>.
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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>
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A quick checks verifies that <math>\theta=90^\circ</math> indeed works.
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~Technodoggo

Revision as of 20:47, 9 November 2023

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?

$\textbf{(A) }\frac1728\qquad\textbf{(B) }2601\qquad\textbf{(C) }3072\qquad\textbf{(D) }4608\qquad\textbf{(E) }6912$

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

A quick checks verifies that $\theta=90^\circ$ indeed works.

~Technodoggo

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