Difference between revisions of "2019 AMC 8 Problems/Problem 6"
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Divide the grid into 4 4x5 quadrants. Each row of 5 points has 1 point on a horizontal/vertical line of symmetry + 1 point on a diagonal line of symmetry: <math>\boxed{\textbf{(C)} \frac{2}{5}}</math>. | Divide the grid into 4 4x5 quadrants. Each row of 5 points has 1 point on a horizontal/vertical line of symmetry + 1 point on a diagonal line of symmetry: <math>\boxed{\textbf{(C)} \frac{2}{5}}</math>. | ||
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==Video Solution by Math-X (First fully understand the problem!!!)== | ==Video Solution by Math-X (First fully understand the problem!!!)== |
Latest revision as of 13:52, 13 January 2024
Contents
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.
.
Solution 2
Divide the grid into 4 4x5 quadrants. Each row of 5 points has 1 point on a horizontal/vertical line of symmetry + 1 point on a diagonal line of symmetry: .
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=AdzSEy4Ocrte4gEU&t=1650
~Math-X
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
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
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
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.