Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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== Problem == | == Problem == | ||
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A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math> | A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math> | ||
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math> | <math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math> | ||
− | == Solution 1 == | + | == Solutions == |
− | === Diagram === | + | === Solution 1 === |
+ | ==== Diagram ==== | ||
<asy> | <asy> | ||
size(10cm); | size(10cm); | ||
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Note: The diagram represents each unit square of the given <math>2020 \times 2020</math> square. | Note: The diagram represents each unit square of the given <math>2020 \times 2020</math> square. | ||
− | ===Solution=== | + | ==== Solution ==== |
− | |||
We consider an individual one-by-one block. | We consider an individual one-by-one block. | ||
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<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>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</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>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math> | ||
− | + | === Solution 2 === | |
− | == Solution 2 == | ||
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]] | 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) === |
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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>. | ||
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~Silverdragon | ~Silverdragon | ||
− | ==Solution 4 (Estimating but a bit different)== | + | === 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 <math>2020\times 2020</math> square. Since we want to find the probability that a point inside a unit square that is <math>d</math> units away from a lattice point (a corner of the square) is <math>\frac{1}{2}</math>, we can find which answer will come the closest to covering <math>\frac{1}{2}</math> of the area. | We only need to figure out the probability for a unit square, as it will scale up to the <math>2020\times 2020</math> square. Since we want to find the probability that a point inside a unit square that is <math>d</math> units away from a lattice point (a corner of the square) is <math>\frac{1}{2}</math>, we can find which answer will come the closest to covering <math>\frac{1}{2}</math> of the area. | ||
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~RuiyangWu | ~RuiyangWu | ||
− | ==Solution 5 (Estimating but differently again)== | + | === Solution 5 (Estimating but differently again) === |
As per the above diagram, realize that <math>\pi d^2 = \frac{1}{2}</math>, so <math>d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}</math>. | As per the above diagram, realize that <math>\pi d^2 = \frac{1}{2}</math>, so <math>d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}</math>. | ||
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-Solution by Joeya | -Solution by Joeya | ||
− | ==Video Solution== | + | == Video Solution == |
− | + | === Video Solution 1 === | |
Education, The Study of Everything | Education, The Study of Everything | ||
https://youtu.be/napCkujyrac | https://youtu.be/napCkujyrac | ||
+ | === Video Solution 2 === | ||
https://youtu.be/RKlG6oZq9so | https://youtu.be/RKlG6oZq9so | ||
Revision as of 13:32, 19 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
Solutions
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
Video Solution 1
Education, The Study of Everything
Video Solution 2
~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.