Difference between revisions of "2023 AMC 12A Problems/Problem 15"
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+ | ==Solution 4 (Intuitive/Time Crunch)== | ||
+ | Note that if Usain walks at a constant speed, then 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 | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 14:58, 11 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.)
Solution 1
By "unfolding" into a straight line, we get a right angled triangle .
~lptoggled
Solution 2(Trig Bash)
We can let be the length of one of the full segments 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
~ap246
Solution 3 (No Trig)
Let be the length of . Apply the Pythagoras theorem on to get , which is also the length of every zigzag segment.
There are such segments. Thus the total length formed by the zigzags is
(note that 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 4 (Intuitive/Time Crunch)
Note that if Usain walks at a constant speed, then 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 . Therefore . ~numerophile
Video Solution 1 by OmegaLearn
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.