Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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Diagram by [[User:Mathandski|MathandSki]] Using Asymptote | Diagram by [[User:Mathandski|MathandSki]] Using Asymptote | ||
− | Note: The diagram represents each unit square of the given <math>2020 | + | Note: The diagram represents each unit square of the given <math>2020 \times 2020</math> square. |
===Solution=== | ===Solution=== | ||
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If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write | If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write | ||
− | <cmath>4 | + | <cmath>4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}</cmath> |
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes | Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes | ||
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== Solution 2 == | == Solution 2 == | ||
− | As in the previous solution, we obtain the equation <math>4 | + | As in the previous solution, we obtain the equation <math>4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]] |
== Solution 3 (Estimating) == | == Solution 3 (Estimating) == |
Revision as of 16:42, 17 January 2021
- 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
Solution 4 (Estimating but a bit different)
We only need to figure out the probability for a unit square, as it will scale up to the square. Since we want to find the probability that a point inside a unit square that is units away from a lattice point (a corner of the square) is , we can find which answer will come the closest to covering of the area.
Since the closest is which turns out to be which is about , we find that the answer rounded to the nearest tenth is or .
~RuiyangWu
Solution 5 (Estimating but differently again)
As per the above diagram, realize that , so .
.
is between and and , so we can say .
So . This is slightly above , since .
-Solution by Joeya
Video Solution
Education, The Study of Everything
~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.