2023 AMC 12A Problems/Problem 15
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (also simple)
- 4 Solution 3 (Trig Bash)
- 5 Solution 4 (No Trig)
- 6 Solution 5 (Intuitive and Quick)
- 7 Solution 6
- 8 Video Solution (easy to digest) by Power Solve
- 9 Video Solution 1 by OmegaLearn
- 10 Video Solution (⚡Solved in 56 seconds⚡)
- 11 Video Solution
- 12 See also
Problem
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 (also simple)
Drop an altitude from to and let its base be . Note that if we repeat this for and , all four right triangles (including ) will have the same trig ratios. By proportion, the hypotenuse is , so .
~IbrahimNadeem
Solution 3 (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 4 (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 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 . Therefore . ~numerophile
Solution 6
Although the diagram is not fully accurate, we can use it to some extent.
It's given that the length of the rectangle is and the total length of the path Usain is taking is , so Usain will walk longer than he would if he were just to walk along the rectangle. Therefore for each point along his path, he will travel farther than he would if he were just to walk along the rectangle.
Drop a perpendicular from point to point on . Let : it follows that and By the Pythagorean Theorem,
so and .
Now, and
-Benedict T (countmath1)
Video Solution (easy to digest) by Power Solve
https://youtu.be/YXIH3UbLqK8?si=U1VEKC7S0PoUFjF5&t=2100
Video Solution 1 by OmegaLearn
Video Solution (⚡Solved in 56 seconds⚡)
https://youtu.be/jkujDM5aW3w ~Education, the Study of Everything
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.