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 == | == 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 | + | 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>. | ||
− | ==Video Solution== | + | <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>. |
+ | |||
+ | 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 | ||
+ | |||
+ | === Solution 6 (Estimating but differently again, again)=== | ||
+ | As above, we have the equation <math>\pi d^2 = \frac{1}{2}</math>, and we want to find the most accurate value of <math>d</math>. We resort to the answer choices and can plug those values of <math>d</math> in and see which value of <math>d</math> will lead to the most accurate value of <math>\pi</math>. | ||
+ | |||
+ | Starting off in the middle, we try option C with <math>d=0.5</math>. Plugging this in, we get <math>\pi \left(\frac{1}{2}\right)^2 = \frac{1}{2},</math> and after simplifying we get <math>\pi = \frac{1}{2} \cdot 4 = 2.</math> | ||
+ | That's not very good. We know <math>\pi \approx 3.14.</math> | ||
+ | |||
+ | Let's see if we can do better. Trying option A with <math>d = 0.3,</math> we get <math>\pi = \frac{1}{2} \cdot \frac{100}{9} = \frac{50}{9} = 5 \frac{5}{9}.</math> | ||
+ | |||
+ | Hm, let's try option B with <math>d = 0.4.</math> We get <math>\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}</math>. This is very close to <math>\pi</math> and is the best estimate for <math>\pi</math> of the 5 options. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) } 0.4}.</math> | ||
+ | ~ epiconan | ||
+ | |||
+ | == Solution 7 (Sol. 1, but rigorous (and excessive)) == | ||
+ | |||
+ | PLEASE NOTE: Solution 1 IS rigorous. Say there are <math>k</math> unit squares (it doesn't matter how many). There is a <math>\frac{1}k</math> probability the point is in some unit square. There is a <math>\pi d^2</math> probability the point is in the shaded region. So, there is a <math>\frac{1}k \pi d^2 \cdot k = \pi d^2</math> probability the point is in any shaded region (since there are <math>k</math> unit squares). | ||
+ | |||
+ | Let <math>n</math> be the side length of a square. When <math>n=1</math>, the shaded areas represent half of the total area: | ||
+ | <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> | ||
+ | |||
+ | When <math>n=2</math>: | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | filldraw((arc((0,0), 0.1994, 0, 90))--(0,0)--cycle, gray); | ||
+ | draw(arc((1,0), 0.1994, 90, 180)); | ||
+ | filldraw((arc((1,0), 0.1994, 90, 180))--(1,0)--cycle, gray); | ||
+ | draw(arc((1,1), 0.1994, 180, 270)); | ||
+ | filldraw((arc((1,1), 0.1994, 180, 270))--(1,1)--cycle, gray); | ||
+ | draw(arc((0,1), 0.1994, 270, 360)); | ||
+ | filldraw(arc((0,1), 0.1994, 270, 360)--(0,1)--cycle, gray); | ||
+ | draw(arc((0.5,0.5), 0.1994,0,360)); | ||
+ | filldraw(arc((0.5,0.5), 0.1994,0,360)--(0.5,0.5)--cycle, gray); | ||
+ | draw(arc((0.5,0), 0.1994,0,180)); | ||
+ | filldraw(arc((0.5,0), 0.1994,0,180)--(0.5,0)--cycle, gray); | ||
+ | draw(arc((0,0.5), 0.1994,-90,90)); | ||
+ | filldraw(arc((0,0.5), 0.1994,-90,90)--(0,0.5)--cycle, gray); | ||
+ | filldraw(arc((1,0.5), 0.1994,90,270)--(1,0.5)--cycle, gray); | ||
+ | filldraw(arc((0.5,1), 0.1994,0,-180)--(0.5,1)--cycle, gray); | ||
+ | |||
+ | draw((0,0)--(0.5,0)--(0.5,1)--(0,1)--(0,0)--(1,0)--(1,1)--(0,1)--(0,0.5)--(1,0.5)); | ||
+ | </asy> | ||
+ | |||
+ | For <math>n=3</math>: | ||
+ | |||
+ | <asy>size(10cm); | ||
+ | |||
+ | filldraw(arc((0,0),0.1330,0,90)--(0,0)--cycle, gray); | ||
+ | filldraw(arc((0,1),0.1330,-90,0)--(0,1)--cycle, gray); | ||
+ | filldraw(arc((1,0),0.1330,90,180)--(1,0)--cycle, gray); | ||
+ | filldraw(arc((1,1),0.1330,-180,-90)--(1,1)--cycle, gray); | ||
+ | |||
+ | filldraw(arc((0.333,0.333),0.133,0,360)--(0.333,0.333)--cycle, gray); | ||
+ | filldraw(arc((0.667,0.333),0.133,0,360)--(0.667,0.333)--cycle, gray); | ||
+ | filldraw(arc((0.333,0.667),0.133,0,360)--(0.333,0.667)--cycle, gray); | ||
+ | filldraw(arc((0.667,0.667),0.133,0,360)--(0.667,0.667)--cycle, gray); | ||
+ | |||
+ | filldraw(arc((0.333,0),0.133,0,180)--(0.333,0)--cycle, gray); | ||
+ | filldraw(arc((0.667,0),0.133,0,180)--(0.667,0)--cycle, gray); | ||
+ | filldraw(arc((0.333,1),0.133,-180,0)--(0.333,1)--cycle, gray); | ||
+ | filldraw(arc((0.666,1),0.133,-180,0)--(0.666,1)--cycle, gray); | ||
+ | filldraw(arc((0,0.333),0.133,-90,90)--(0,0.333)--cycle, gray); | ||
+ | filldraw(arc((0,0.667),0.133,-90,90)--(0,0.667)--cycle, gray); | ||
+ | filldraw(arc((1,0.333),0.133,90,270)--(1,0.333)--cycle, gray); | ||
+ | filldraw(arc((1,0.667),0.133,90,270)--(1,0.667)--cycle, gray); | ||
+ | |||
+ | draw((0,0)--(0,1)--(0.333,1)--(0.333,0)--(0.667,0)--(0.667,1)--(1,1)--(1,0)--(0,0)--(0,0.333)--(1,0.333)--(1,0.667)--(0,0.667)); | ||
+ | draw((0.333,1)--(0.667,1));</asy> | ||
+ | |||
+ | We can calculate the total number of shaded circles given some <math>n</math>. There are <math>(n-1)^2</math> full circles on the inside, <math>4(n-1)</math> semicircles on the sides, and <math>4</math> quarter circles for the corners. | ||
+ | |||
+ | Full circles are, of course, worth one circle. Semicircles are worth half a circle each, and quarter circles are worth <math>\dfrac14</math> of a circle. Thus, weighing our sum gives <math>(n-1)^2+\dfrac{4(n-1)}2+\dfrac44=n^2-2n+1+2(n-1)+1=n^2-2n+2n+2-2=n^2.</math> Thus, there is <math>n^2\cdot\pi r^2</math> worth of the shaded area for any <math>n</math>, and since the area of each circle is <math>\pi r^2</math> if <math>r</math> is the radius of each. | ||
+ | |||
+ | We want the ratio of this shaded area to the entire to be <math>\dfrac12</math>. The area of the entire square is <math>n^2</math>, so dividing, we see that <math>\dfrac{n^2\cdot\pi r^2}{n^2}=\pi r^2=\dfrac12</math>. | ||
+ | |||
+ | The rest is the same as solution <math>1</math>. | ||
+ | |||
+ | == 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== | ||
{{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}} |
Latest revision as of 21:29, 4 November 2024
- The following problem is from both the 2020 AMC 12A #16 and 2020 AMC 10A #16, so both problems redirect to this page.
Contents
[hide]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
Solution 6 (Estimating but differently again, again)
As above, we have the equation , and we want to find the most accurate value of . We resort to the answer choices and can plug those values of in and see which value of will lead to the most accurate value of .
Starting off in the middle, we try option C with . Plugging this in, we get and after simplifying we get That's not very good. We know
Let's see if we can do better. Trying option A with we get
Hm, let's try option B with We get . This is very close to and is the best estimate for of the 5 options.
Therefore, the answer is ~ epiconan
Solution 7 (Sol. 1, but rigorous (and excessive))
PLEASE NOTE: Solution 1 IS rigorous. Say there are unit squares (it doesn't matter how many). There is a probability the point is in some unit square. There is a probability the point is in the shaded region. So, there is a probability the point is in any shaded region (since there are unit squares).
Let be the side length of a square. When , the shaded areas represent half of the total area:
When :
For :
We can calculate the total number of shaded circles given some . There are full circles on the inside, semicircles on the sides, and quarter circles for the corners.
Full circles are, of course, worth one circle. Semicircles are worth half a circle each, and quarter circles are worth of a circle. Thus, weighing our sum gives Thus, there is worth of the shaded area for any , and since the area of each circle is if is the radius of each.
We want the ratio of this shaded area to the entire to be . The area of the entire square is , so dividing, we see that .
The rest is the same as solution .
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.