Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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== Problem == | == 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> | 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 ==== | ||
+ | <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> | ||
+ | |||
+ | The diagram represents each unit square of the given <math>2020 \times 2020</math> square. | ||
+ | |||
+ | ==== Solution 1 ==== | ||
We consider an individual one-by-one block. | 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 | 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 \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}</math>, and from here, we see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4}.</math> | ||
− | + | ~Crypthes | |
+ | |||
+ | ~ Minor Edits by BakedPotato66 | ||
<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 | + | 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) === | ||
+ | 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 | ||
+ | |||
+ | === 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. | ||
+ | |||
+ | Since the closest is <math>0.4</math> which turns out to be <math>(0.4)^2\times \pi = 0.16 \times \pi</math> which is about <math>0.502</math>, we find that the answer rounded to the nearest tenth is <math>0.4</math> or <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | ~RuiyangWu | ||
+ | |||
+ | === 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>. | ||
+ | |||
+ | <math>\sqrt{2} \approx 1.4 = \frac{7}{5}</math>. | ||
+ | |||
+ | <math>\sqrt{\pi}</math> is between <math>1.7</math> and <math>1.8</math> <math>((1.7)^2 = 2.89</math> and <math>(1.8)^2 = 3.24)</math>, so we can say <math>\sqrt{\pi} \approx 1.75 = \frac{7}{4}</math>. | ||
− | ==Video Solution== | + | So <math>d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}</math>. This is slightly above <math>\boxed{\textbf{(B) } 0.4}</math>, since <math>\frac{20}{49} \approx \frac{2}{5}</math>. |
+ | |||
+ | -Solution by Joeya | ||
+ | |||
+ | == Video Solutions == | ||
+ | === Video Solution 1 === | ||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/napCkujyrac | ||
+ | |||
+ | === Video Solution 2 === | ||
https://youtu.be/RKlG6oZq9so | https://youtu.be/RKlG6oZq9so | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | === Video Solution 3 === | ||
+ | https://youtu.be/R220vbM_my8?t=238 | ||
+ | |||
+ | ~ amritvignesh0719062.0 | ||
==See Also== | ==See Also== | ||
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{{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}} | {{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:43, 8 November 2022
- 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
Diagram
The diagram represents each unit square of the given square.
Solution 1
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 see that
~Crypthes
~ Minor Edits by BakedPotato66
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 Solutions
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
https://youtu.be/R220vbM_my8?t=238
~ amritvignesh0719062.0
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.