Difference between revisions of "2023 AMC 12A Problems/Problem 15"
(→Solution 2(Trig Bash)) |
(→Solution 1) |
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By "unfolding" <math>APQRS</math> into a straight line, we get a right angled triangle <math>ABS</math>. | By "unfolding" <math>APQRS</math> into a straight line, we get a right angled triangle <math>ABS</math>. | ||
+ | |||
+ | <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)); | ||
+ | label("$P$",(-22.5,23)); | ||
+ | label("$Q$",(5,-23)); | ||
+ | label("$R$",(32.5,23)); | ||
+ | </asy> | ||
<math>cos(\theta)=\frac{100}{120}</math> | <math>cos(\theta)=\frac{100}{120}</math> |
Revision as of 15:42, 10 November 2023
Contents
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
Video Solution 1 by OmegaLearn
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.