Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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Solving for <math>r</math>, we obtain <math>r = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>r = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>r \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes | Solving for <math>r</math>, we obtain <math>r = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>r = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>r \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes | ||
− | <math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> | + | <math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. Then would make sure the above solution works. <math>\textbf{- Emathmaster}</math> |
==Video Solution== | ==Video Solution== |
Revision as of 13:52, 1 February 2020
Contents
[hide]Problem
A point is chosen at random within the square in the coordinate plane whose vertices are and . The probability that the point is within units of a lattice point is . (A point is a lattice point if and are both integers.) What is to the nearest tenth
Solution
We consider an individual one by one block.
If we draw a quarter of a circle from each corner, the area covered by the circles should be . Because of this, and the fact that there are four circles, we write
Solving for , we obtain , where with , we get , and from here, we simplify and see that ~Crypthes
To be more rigorous, note that since if then clearly the probability is greater than . Then would make sure the above solution works.
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.