Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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-Solution by Joeya | -Solution by Joeya | ||
+ | |||
+ | == Solution 6 (Sol. 1, but rigorous (and excessive)) == | ||
+ | |||
+ | 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 Solutions == |
Revision as of 23:34, 2 November 2023
- 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
Solution 6 (Sol. 1, but rigorous (and excessive))
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.