Difference between revisions of "2019 AMC 8 Problems/Problem 6"
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Lines of symmetry go through point <math>P</math>, and there are <math>8</math> directions the lines could go, and there are <math>4</math> dots at each direction.<math>\frac{4\times8}{80}=\boxed{\textbf{(C)} \frac{2}{5}}</math>. | Lines of symmetry go through point <math>P</math>, and there are <math>8</math> directions the lines could go, and there are <math>4</math> dots at each direction.<math>\frac{4\times8}{80}=\boxed{\textbf{(C)} \frac{2}{5}}</math>. | ||
− | == | + | == Video Solution == |
+ | The Learning Royal : https://youtu.be/8njQzoztDGc | ||
+ | |||
+ | == Video Solution 2 == | ||
+ | |||
+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=4L95z9DwlhI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=7 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/TAKmC11vitM | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See also== | ||
{{AMC8 box|year=2019|num-b=5|num-a=7}} | {{AMC8 box|year=2019|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:28, 7 February 2022
Problem 6
There are grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point
is in the center of the square. Given that point
is randomly chosen among the other
points, what is the probability that the line
is a line of symmetry for the square?
Solution 1
Lines of symmetry go through point
, and there are
directions the lines could go, and there are
dots at each direction.
.
Video Solution
The Learning Royal : https://youtu.be/8njQzoztDGc
Video Solution 2
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=4L95z9DwlhI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=7
Video Solution 3
~savannahsolver
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.