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Difference between revisions of "2023 AMC 12A Problems/Problem 15"
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==Video Solution 1 by OmegaLearn==
 
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https://youtu.be/NhUI-BNCIUE
 
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==Solution 2(Trig Bash)==
 
==Solution 2(Trig Bash)==
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==Video Solution 1 by OmegaLearn==
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https://youtu.be/NhUI-BNCIUE
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==See also==
 
==See also==

Revision as of 10:38, 10 November 2023

Question

Usain is walking for exercise by zigzagging across a $100$-meter by $30$-meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$. He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$. What angle $\theta$$\angle PAB=\angle QPC=\angle RQB=\cdots$ will produce in a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)

[Pls finish] [asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-11)); label("$\theta$", (-13,11)); [/asy]

$\textbf{(A)}~\arccos\frac{5}{6}\qquad\textbf{(B)}~\arccos\frac{4}{5}\qquad\textbf{(C)}~\arccos\frac{3}{10}\qquad\textbf{(D)}~\arcsin\frac{4}{5}\qquad\textbf{(E)}~\arcsin\frac{5}{6}$

Solution 1

By "unfolding" $APQRS$ into a straight line, we get a right angled triangle $ABS$.

$cos(\theta)=\frac{100}{120}$

$\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}$

~lptoggled

Solution 2(Trig Bash)

We can let $x$ be the length of one of the full segments of the zigzag. We can then notice that $\sin\theta = \frac{30}{x}$. By Pythagorean Theorem, we see that $BP = \sqrt{x^2 - 900}$. This implies that: \[RC = 100 - 3\sqrt{x^2 - 900}.\] We also realize that $RS = 120 - 3x$, so this means that: \[\cos\theta = \frac{100 - 3\sqrt{x^2 - 900}}{120 - 3x}.\] We can then substitute $x = \frac{30}{\sin\theta}$, so this gives: \begin{align*} \cos\theta &= \frac{100 - 3\sqrt{x^2 - 900}}{120 - 3x}\\ &= \frac{100 - 3\sqrt{\frac{900}{\sin^2\theta} - 900}}{120 - \frac{90}{\sin\theta}}\\ &= \frac{100 - 90\sqrt{\csc^2\theta - 1}}{120 - \frac{90}{\sin\theta}}\\ &= \frac{100 - \frac{90}{\tan\theta}}{120 - 90\sin\theta}\\ &= \frac{100\sin\theta - 90\cos\theta}{120\sin\theta - 90}\\ &= \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9}\\ \end{align*}

Now we have: \[\cos\theta = \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9},\] meaning that: \[12\sin\theta\cos\theta - 9\cos\theta = 10\sin\theta - 9\cos\theta \implies \cos\theta = \frac{10}{12} = \frac56.\] This means that $\theta = \arccos\left(\frac56\right)$, giving us $\boxed{\textbf{A}}$

~ap246

Video Solution 1 by OmegaLearn

https://youtu.be/NhUI-BNCIUE


See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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