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

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~lptoggled
 
~lptoggled
  
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==Solution 2(Trig Bash)==
+
==Solution 2 (also simple)==
 +
 
 +
Drop an altitude from <math>P</math> to <math>AB</math> and let its base be <math>x</math>. Note that if we repeat this for <math>Q</math> and <math>R</math>, all four right triangles (including <math>\triangle RSC</math>) will have the same trig ratios. By proportion, the hypotenuse <math>AP</math> is <math>\frac{x}{100}(120) = \frac65 x</math>, so <math>\cos\theta = \frac{x}{(\frac65x)} = \frac56 \Rightarrow \theta = \boxed{\textbf{(A) }\arccos{\frac56}}</math>.
 +
 
 +
 
 +
==Solution 3(Trig Bash)==
  
 
We can let <math>x</math> be the length of one of the full segments of the zigzag. We can then notice that <math>\sin\theta = \frac{30}{x}</math>. By Pythagorean Theorem, we see that <math>DP = \sqrt{x^2 - 900}</math>. This implies that: <cmath>RC = 100 - 3\sqrt{x^2 - 900}.</cmath> We also realize that <math>RS = 120 - 3x</math>, so this means that: <cmath>\cos\theta = \frac{100 - 3\sqrt{x^2 - 900}}{120 - 3x}.</cmath> We can then substitute <math>x = \frac{30}{\sin\theta}</math>, so this gives:
 
We can let <math>x</math> be the length of one of the full segments of the zigzag. We can then notice that <math>\sin\theta = \frac{30}{x}</math>. By Pythagorean Theorem, we see that <math>DP = \sqrt{x^2 - 900}</math>. This implies that: <cmath>RC = 100 - 3\sqrt{x^2 - 900}.</cmath> We also realize that <math>RS = 120 - 3x</math>, so this means that: <cmath>\cos\theta = \frac{100 - 3\sqrt{x^2 - 900}}{120 - 3x}.</cmath> We can then substitute <math>x = \frac{30}{\sin\theta}</math>, so this gives:
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~ap246
 
~ap246
  
−
==Solution 3 (No Trig)==
+
==Solution 4 (No Trig)==
 
Let <math>x</math> be the length of <math>DP</math>. Apply the Pythagoras theorem on <math>\triangle{ADP}</math> to get <math>AP = \sqrt{900 + x^2}</math>, which is also the length of every zigzag segment.  
 
Let <math>x</math> be the length of <math>DP</math>. Apply the Pythagoras theorem on <math>\triangle{ADP}</math> to get <math>AP = \sqrt{900 + x^2}</math>, which is also the length of every zigzag segment.  
  
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==Solution 4 (Intuitive and Quick)==
+
==Solution 5 (Intuitive and Quick)==
 
Imagine that Usain walks at a constant speed. The horizontal component of Usain's velocity does not change. (Imagine a beam of light reflecting off of mirrors. A mirror only changes the velocity of light in the direction perpendicular to the mirror.) The horizontal component of Usain's velocity divided by his total velocity must be <math>\frac{100}{120}</math>. Therefore <cmath>\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}</cmath>.
 
Imagine that Usain walks at a constant speed. The horizontal component of Usain's velocity does not change. (Imagine a beam of light reflecting off of mirrors. A mirror only changes the velocity of light in the direction perpendicular to the mirror.) The horizontal component of Usain's velocity divided by his total velocity must be <math>\frac{100}{120}</math>. Therefore <cmath>\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}</cmath>.
 
~numerophile
 
~numerophile

Revision as of 23:45, 11 November 2023

Problem

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 (also simple)

Drop an altitude from $P$ to $AB$ and let its base be $x$. Note that if we repeat this for $Q$ and $R$, all four right triangles (including $\triangle RSC$) will have the same trig ratios. By proportion, the hypotenuse $AP$ is $\frac{x}{100}(120) = \frac65 x$, so $\cos\theta = \frac{x}{(\frac65x)} = \frac56 \Rightarrow \theta = \boxed{\textbf{(A) }\arccos{\frac56}}$.


Solution 3(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 4 (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 \[\frac{100}{x} \times \sqrt{900+x^2} = 120\] \[\sqrt{900+x^2} = \frac{6}{5}x\] \[900 + x^2 = \frac{36}{25}x^2\] \[x = \frac{150}{\sqrt{11}} = DP\] \[AP = \sqrt{900 + x^2} = \frac{180}{\sqrt{11}}\] \[\cos\theta = \frac{DP}{AP} = \frac{5}{6}\] \[\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}\]

(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 to the length of the incomplete base)

~dwarf_marshmallow


Solution 5 (Intuitive and Quick)

Imagine that Usain walks at a constant speed. The horizontal component of Usain's velocity does not change. (Imagine a beam of light reflecting off of mirrors. A mirror only changes the velocity of light in the direction perpendicular to the mirror.) The horizontal component of Usain's velocity divided by his total velocity must be $\frac{100}{120}$. Therefore \[\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}\]. ~numerophile

Video Solution 1 by OmegaLearn

https://youtu.be/NhUI-BNCIUE

Video Solution

https://youtu.be/S5H8JEImiA8

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

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

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