Difference between revisions of "2023 AMC 12A Problems/Problem 15"

Revision as of 17:46, 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.)

$[asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); [/asy]$

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

Solution 1

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

$[asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); dot((5,45)); dot((32.5,75)); dot((50,94.09090909090909)); draw((-22.5,15)--(50,94.09090909090909)); draw((50,-4.09090909090909)--(50,94.09090909090909)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); label("$Q'$",(5,35)); label("$R'$",(32.5,85)); label("$S'$",(58,94.09090909090909)); [/asy]$

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

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

~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 $DP = \sqrt{x^2 - 900}$ . 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

Solution 3 (No Trig)

Let $x$ be the length of $DP$ . Apply the Pythagoras theorem on $\triangle{ADP}$ to get $AP = \sqrt{900 + x^2}$ , which is also the length of every zigzag segment.

There are $\frac{100}{x}$ such segments. Thus the total length formed by the zigzags is $\frac{100}{x} \times \sqrt{900+x^2} = 120$ $\sqrt{900+x^2} = \frac{6}{5}x$ $900 + x^2 = \frac{36}{25}x^2$ $x = \frac{150}{\sqrt{11}} = DP$ $AP = \sqrt{900 + x^2} = \frac{180}{\sqrt{11}}$ $\cos\theta = \frac{DP}{AP} = \frac{5}{6}$ $\theta = \arccos\frac{5}{6}$

(note that $\frac{100}{x}$ is not an integer, but it doesn't matter because of similar triangles. The length of the incomplete segment is always proportionate proportionate to the length of the incomplete base)

~dwarf_marshmallow

Video Solution 1 by OmegaLearn

https://youtu.be/NhUI-BNCIUE

@@ Line 113: / Line 113: @@
 <cmath>x = \frac{150}{\sqrt{11}} = DP</cmath>
 <cmath>AP = \sqrt{900 + x^2} = \frac{180}{\sqrt{11}}</cmath>
-<cmath>cos\theta = \frac{DP}{AP} = \frac{5}{6}</cmath>
+<cmath>\cos\theta = \frac{DP}{AP} = \frac{5}{6}</cmath>
-<cmath>\theta = arccos\frac{5}{6}</cmath>
+<cmath>\theta = \arccos\frac{5}{6}</cmath>
 (note that <math>\frac{100}{x}</math> is not an integer, but it doesn't matter because of similar triangles. The length of the incomplete segment is always proportionate proportionate to the length of the incomplete base)

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