Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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+ | {{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}} | ||
+ | |||
+ | == Problem == | ||
+ | |||
+ | 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> | ||
+ | |||
+ | == Solution 1 == | ||
+ | === Diagram === | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); | ||
+ | filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); | ||
+ | draw(arc((1,0), 0.3989, 90, 180)); | ||
+ | filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); | ||
+ | draw(arc((1,1), 0.3989, 180, 270)); | ||
+ | filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); | ||
+ | draw(arc((0,1), 0.3989, 270, 360)); | ||
+ | filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray); | ||
+ | </asy> | ||
+ | |||
+ | Diagram by [[User:Mathandski|MathandSki]] Using Asymptote | ||
+ | |||
+ | Note: The diagram represents each unit square of the given <math>2020 * 2020</math> 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 <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 * \frac{1}{4} * \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 | ||
+ | |||
+ | <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 == | ||
+ | As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \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) == | ||
+ | |||
+ | 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 6.28318 \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>. | ||
+ | |||
+ | ~Silverdragon | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/napCkujyrac | ||
+ | |||
+ | https://youtu.be/RKlG6oZq9so | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}} | {{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}} | ||
+ | {{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:52, 7 November 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
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.