Difference between revisions of "2023 AMC 12A Problems/Problem 15"
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https://youtu.be/NhUI-BNCIUE | https://youtu.be/NhUI-BNCIUE | ||
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+ | ==Solution 2(Trig Bash)== | ||
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+ | We can let <math>x</math> be the length of one of the full segements of the zigzag. We can then notice that <math>\sin\theta = \frac{30}{x}</math>. By Pythagorean Theorem, we see that <math>BP = \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: | ||
+ | <cmath>\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*}</cmath> | ||
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+ | Now we have: <cmath>\cos\theta = \frac{10\sin\theta - 9\cos\theta}{12\sin\theta - 9},</cmath> meaning that: <cmath>12\sin\theta\cos\theta - 9\cos\theta = 10\sin\theta - 9\cos\theta \implies \cos\theta = \frac{10}{12} = \frac56.</cmath> | ||
+ | This means that <math>\theta = \arccos\left(\frac56\right)</math>, giving us <math>\boxed{\textbf{A}}</math> | ||
==See also== | ==See also== |
Revision as of 10:32, 10 November 2023
Contents
[hide]Question
Usain is walking for exercise by zigzagging across a -meter by -meter rectangular field, beginning at point and ending on the segment . He wants to increase the distance walked by zigzagging as shown in the figure below . What angle will produce in a length that is 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.)
[someone add diagram]
Solution 1
By "unfolding" into a straight line, we get a right angled triangle .
~lptoggled
Video Solution 1 by OmegaLearn
Solution 2(Trig Bash)
We can let be the length of one of the full segements of the zigzag. We can then notice that . By Pythagorean Theorem, we see that . This implies that: We also realize that , so this means that: We can then substitute , so this gives:
Now we have: meaning that: This means that , giving us
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.