Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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As above, we find that we need to estimate <math>d = \frac{1}{\sqrt{2\pi}}</math>. | As above, we find that we need to estimate <math>d = \frac{1}{\sqrt{2\pi}}</math>. | ||
− | Note that we can approximate <math>2\pi \approx 6. | + | Note that we can approximate <math>2\pi \approx 6.28318 \approx 6.25</math> and so <math>\frac{1}{\sqrt{2\pi}}</math> <math>\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4</math>. |
And so our answer is <math>\boxed{\textbf{(B) } 0.4}</math>. | And so our answer is <math>\boxed{\textbf{(B) } 0.4}</math>. | ||
~Silverdragon | ~Silverdragon | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/RKlG6oZq9so | https://youtu.be/RKlG6oZq9so |
Revision as of 09:37, 19 March 2020
- The following problem is from both the 2020 AMC 12A #16 and 2020 AMC 10A #16, so both problems redirect to this page.
Contents
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 1
Diagram
Diagram by MathandSki Using Asymptote
Note: The diagram represents each unit square of the given square.
Solution
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius , 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 . This would make sure the above solution works, as if there is overlap with the quartercircles.
Solution 2
As in the previous solution, we obtain the equation , which simplifies to . Since is slightly more than , is slightly less than . We notice that is slightly more than , so is roughly ~emerald_block
Solution 3 (Estimating)
As above, we find that we need to estimate .
Note that we can approximate and so .
And so our answer is .
~Silverdragon
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 |
2020 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.