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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 ==
+
== Solutions ==
−
We consider an individual one by one block.
+
==== 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.
 +
 
 +
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 \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>
 +
 
 +
=== 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]]
 +
 
 +
=== 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>.
 +
 
 +
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>.
  
−
If we draw a quarter of a circle from each corner, 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
+
-Solution by Joeya
  
−
<cmath>4 * \frac{1}{4} * \pi r^2 = \frac{1}{2}</cmath>
+
=== 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>.
  
−
Solving for <math>r</math>, we obtain <math>r = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>r = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>r \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes
+
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
 +
 
 +
~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.

Problem

A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth$?$

$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

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 $2020 \times 2020$ 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 $d$, the area covered by the circles should be $0.5$. Because of this, and the fact that there are four circles, we write

\[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\]

Solving for $d$, we obtain $d = \frac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}$, and from here, we see that $d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4}.$

~Crypthes

~ Minor Edits by BakedPotato66

$\textbf{Note:}$ To be more rigorous, note that $d<0.5$ since if $d\geq0.5$ then clearly the probability is greater than $\frac{1}{2}$. This would make sure the above solution works, as if $d\geq0.5$ there is overlap with the quartercircles. $\textbf{- Emathmaster}$

Solution 2

As in the previous solution, we obtain the equation $4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}$, which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$. Since $\pi$ is slightly more than $3$, $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$. We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$, so $d$ is roughly $\boxed{\textbf{(B) } 0.4}.$ ~emerald_block

Solution 3 (Estimating)

As above, we find that we need to estimate $d = \frac{1}{\sqrt{2\pi}}$.

Note that we can approximate $2\pi \approx 6.28318 \approx 6.25$ and so $\frac{1}{\sqrt{2\pi}}$ $\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4$.

And so our answer is $\boxed{\textbf{(B) } 0.4}$.

~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 $2020\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\frac{1}{2}$, we can find which answer will come the closest to covering $\frac{1}{2}$ of the area.

Since the closest is $0.4$ which turns out to be $(0.4)^2\times \pi = 0.16 \times \pi$ which is about $0.502$, we find that the answer rounded to the nearest tenth is $0.4$ or $\boxed{\textbf{(B)}}$.

~RuiyangWu

Solution 5 (Estimating but differently again)

As per the above diagram, realize that $\pi d^2 = \frac{1}{2}$, so $d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}$.

$\sqrt{2} \approx 1.4 = \frac{7}{5}$.

$\sqrt{\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$, so we can say $\sqrt{\pi} \approx 1.75 = \frac{7}{4}$.

So $d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}$. This is slightly above $\boxed{\textbf{(B) } 0.4}$, since $\frac{20}{49} \approx \frac{2}{5}$.

-Solution by Joeya

Solution 6 (Estimating but differently again, again)

As above, we have the equation $\pi d^2 = \frac{1}{2}$, and we want to find the most accurate value of $d$. We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\pi$.

Starting off in the middle, we try option C with $d=0.5$. Plugging this in, we get $\pi \left(\frac{1}{2}\right)^2 = \frac{1}{2},$ and after simplifying we get $\pi = \frac{1}{2} \cdot 4 = 2.$ That's not very good. We know $\pi \approx 3.14.$

Let's see if we can do better. Trying option A with $d = 0.3,$ we get $\pi = \frac{1}{2} \cdot \frac{100}{9} = \frac{50}{9} = 5 \frac{5}{9}.$

Hm, let's try option B with $d = 0.4.$ We get $\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}$. This is very close to $\pi$ and is the best estimate for $\pi$ of the 5 options.

Therefore, the answer is $\boxed{\textbf{(B) } 0.4}.$ ~ epiconan

Solution 7 (Sol. 1, but rigorous (and excessive))

PLEASE NOTE: Solution 1 IS rigorous. Say there are $k$ unit squares (it doesn't matter how many). There is a $\frac{1}k$ probability the point is in some unit square. There is a $\pi d^2$ probability the point is in the shaded region. So, there is a $\frac{1}k \pi d^2 \cdot k = \pi d^2$ probability the point is in any shaded region (since there are $k$ unit squares).

Let $n$ be the side length of a square. When $n=1$, 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 $n=2$: [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 $n=3$:

[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 $n$. There are $(n-1)^2$ full circles on the inside, $4(n-1)$ semicircles on the sides, and $4$ 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 $\dfrac14$ of a circle. Thus, weighing our sum gives $(n-1)^2+\dfrac{4(n-1)}2+\dfrac44=n^2-2n+1+2(n-1)+1=n^2-2n+2n+2-2=n^2.$ Thus, there is $n^2\cdot\pi r^2$ worth of the shaded area for any $n$, and since the area of each circle is $\pi r^2$ if $r$ is the radius of each.

We want the ratio of this shaded area to the entire to be $\dfrac12$. The area of the entire square is $n^2$, so dividing, we see that $\dfrac{n^2\cdot\pi r^2}{n^2}=\pi r^2=\dfrac12$.

The rest is the same as solution $1$.

Video Solutions

Video Solution 1

Education, The Study of Everything

https://youtu.be/napCkujyrac

Video Solution 2

https://youtu.be/RKlG6oZq9so

~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

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