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 | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/RKlG6oZq9so | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
Revision as of 22:20, 31 January 2020
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
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
![\[4 * \frac{1}{4} * \pi r^2 = \frac{1}{2}\]](http://latex.artofproblemsolving.com/d/6/c/d6c7a095fe1a98e416df416026b7e838984de7f0.png)
Solving for 
, we obtain 
, where with 
, we get 
, and from here, we simplify and see that 
~Crypthes
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. 
