Difference between revisions of "2024 AMC 10B Problems/Problem 14"

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{{duplicate|[[2024 AMC 10B Problems/Problem 14|2024 AMC 10B #14]] and [[2024 AMC 12B Problems/Problem 9|2024 AMC 12B #9]]}}
  
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==Problem==
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A dartboard is the region B in the coordinate plane consisting of points <math>(x, y)</math> such that <math>|x| + |y| \le 8</math>. A target T is the region where <math>(x^2 + y^2 - 25)^2 \le 49</math>. A dart is thrown and lands at a random point in B. The probability that the dart lands in T can be expressed as <math>\frac{m}{n} \cdot \pi</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?
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<math>
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\textbf{(A) }39 \qquad
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\textbf{(B) }71 \qquad
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\textbf{(C) }73 \qquad
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\textbf{(D) }75 \qquad
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\textbf{(E) }135 \qquad
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</math>
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==Diagram==
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<asy>
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// By Elephant200
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// Feel free to adjust the code
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size(10cm);
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pair A = (8, 0);
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pair B = (0, 8);
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pair C = (-8, 0);
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pair D = (0, -8);
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draw(A--B--C--D--cycle, linewidth(1.5));
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label("$(8,0)$", A, NE);
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label("$(0,8)$", B, NE);
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label("$(-8,0)$", C, NW);
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label("$(0,-8)$", D, SE);
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filldraw(circle((0,0),4*sqrt(2)), gray, linewidth(1.5));
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filldraw(circle((0,0),3*sqrt(2)), white, linewidth(1.5));
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draw((-10, 0)--(10,0),EndArrow(5));
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draw((10, 0)--(-10,0),EndArrow(5));
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draw((0,-10)--(0,10), EndArrow(5));
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draw((0,10)--(0,-10),EndArrow(5));
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</asy>
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~Elephant200
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==Solution 1==
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Inequalities of the form <math>|x|+|y| \le 8</math> are well-known and correspond to a square in space with centre at origin and vertices at <math>(8, 0)</math>, <math>(-8, 0)</math>, <math>(0, 8)</math>, <math>(0, -8)</math>.
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The diagonal length of this square is clearly <math>16</math>, so it has an area of
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<cmath>\frac{1}{2} \cdot 16 \cdot 16 = 128</cmath>
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Now,
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<cmath>(x^2 + y^2 - 25)^2 \le 49</cmath>
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Converting to polar form,
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<cmath>r^2 - 25 \le 7 \implies r \le \sqrt{32},</cmath>
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and
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<cmath>r^2 - 25 \ge -7\implies r\ge \sqrt{18}.</cmath>
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The intersection of these inequalities is the circular region <math>T</math> for which every circle in <math>T</math> has a radius between <math>\sqrt{18}</math> and <math>\sqrt{32}</math>, inclusive. The area of such a region is thus <math>\pi(32-18)=14\pi.</math> The requested probability is therefore <math>\frac{14\pi}{128} = \frac{7\pi}{64},</math> yielding <math>(m,n)=(7,64).</math> We have <math>m+n=7+64=\boxed{\textbf{(B)}\ 71}.</math>
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-AbhiSood1234, countmath1
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==Solution 2 (Calculus)==
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Expressing the Area of Region B
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Region B consists of points where |x|+|y|8
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In each quadrant, this can be expressed by the following functions:
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First quadrant: y=8x
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Second quadrant: y=8+x
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Third quadrant: y=8x
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Fourth quadrant: y=8+x
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In the first quadrant, x ranges from 0 to 8, and y ranges from 0 to 8x. Thus, the area in the first quadrant is:
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<cmath>
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\text{Area of first quadrant} = \int_0^8 \int_0^{8 - x} \, dy \, dx
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</cmath>
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<cmath>
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= \int_0^8 [y]_{y=0}^{y=8-x} \, dx = \int_0^8 (8 - x) \, dx
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</cmath>
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<cmath>
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= \left[ 8x - \frac{x^2}{2} \right]_0^8 = 64 - 32 = 32
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</cmath>
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The total area of region B is:
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<cmath>
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\text{Area of } B = 4 \times 32 = 128
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</cmath>
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Expressing the Area of Region T
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Region T is defined by the inequality (x2+y225)249, which can be rewritten as:
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<cmath>
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18 \le x^2 + y^2 \le 32
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</cmath>
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To find the area, we switch to polar coordinates with x=rcosθ and y=rsinθ, where x2+y2=r2. Here, r ranges from 18 to 32, and θ ranges from 0 to 2π.
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The area of T can then be found by:
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<cmath>
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\text{Area of } T = \int_0^{2\pi} \int_{\sqrt{18}}^{\sqrt{32}} r \, dr \, d\theta
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</cmath>
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<cmath>
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= \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_{r=\sqrt{18}}^{r=\sqrt{32}} \, d\theta = \int_0^{2\pi} \left( \frac{32}{2} - \frac{18}{2} \right) \, d\theta
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</cmath>
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<cmath>
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= \int_0^{2\pi} 7 \, d\theta = 14\pi
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</cmath>
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The probability P that a dart lands in region T is the area of T divided by the area of B:
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<cmath>
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P = \frac{\text{Area of } T}{\text{Area of } B} = \frac{14\pi}{128} = \frac{7\pi}{64}
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</cmath>
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So the probability is of the form mnπ, where m=7 and n=64, so m+n=7+64=71.
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<cmath>
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\boxed{\textbf{(B)}\ 71}
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</cmath>
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~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
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==Solution 3==
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[[Image: 2024_AMC_12B_P09.jpeg|thumb|center|600px|]]
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~Kathan
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/YqKmvSR1Ckk?feature=shared
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 +
~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=13|num-a=15}}
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{{AMC12 box|year=2024|ab=B|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 03:14, 30 November 2024

The following problem is from both the 2024 AMC 10B #14 and 2024 AMC 12B #9, so both problems redirect to this page.

Problem

A dartboard is the region B in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8$. A target T is the region where $(x^2 + y^2 - 25)^2 \le 49$. A dart is thrown and lands at a random point in B. The probability that the dart lands in T can be expressed as $\frac{m}{n} \cdot \pi$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

$\textbf{(A) }39 \qquad \textbf{(B) }71 \qquad \textbf{(C) }73 \qquad \textbf{(D) }75 \qquad \textbf{(E) }135 \qquad$

Diagram

[asy] // By Elephant200 // Feel free to adjust the code  size(10cm);  pair A = (8, 0); pair B = (0, 8); pair C = (-8, 0); pair D = (0, -8); draw(A--B--C--D--cycle, linewidth(1.5));  label("$(8,0)$", A, NE); label("$(0,8)$", B, NE); label("$(-8,0)$", C, NW); label("$(0,-8)$", D, SE);  filldraw(circle((0,0),4*sqrt(2)), gray, linewidth(1.5)); filldraw(circle((0,0),3*sqrt(2)), white, linewidth(1.5));  draw((-10, 0)--(10,0),EndArrow(5)); draw((10, 0)--(-10,0),EndArrow(5)); draw((0,-10)--(0,10), EndArrow(5)); draw((0,10)--(0,-10),EndArrow(5)); [/asy] ~Elephant200

Solution 1

Inequalities of the form $|x|+|y| \le 8$ are well-known and correspond to a square in space with centre at origin and vertices at $(8, 0)$, $(-8, 0)$, $(0, 8)$, $(0, -8)$. The diagonal length of this square is clearly $16$, so it has an area of \[\frac{1}{2} \cdot 16 \cdot 16 = 128\] Now, \[(x^2 + y^2 - 25)^2 \le 49\] Converting to polar form, \[r^2 - 25 \le 7 \implies r \le \sqrt{32},\] and \[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\]

The intersection of these inequalities is the circular region $T$ for which every circle in $T$ has a radius between $\sqrt{18}$ and $\sqrt{32}$, inclusive. The area of such a region is thus $\pi(32-18)=14\pi.$ The requested probability is therefore $\frac{14\pi}{128} = \frac{7\pi}{64},$ yielding $(m,n)=(7,64).$ We have $m+n=7+64=\boxed{\textbf{(B)}\ 71}.$

-AbhiSood1234, countmath1

Solution 2 (Calculus)

Expressing the Area of Region B

Region B consists of points where |x|+|y|8

In each quadrant, this can be expressed by the following functions:

First quadrant: y=8x Second quadrant: y=8+x Third quadrant: y=8x Fourth quadrant: y=8+x

In the first quadrant, x ranges from 0 to 8, and y ranges from 0 to 8x. Thus, the area in the first quadrant is: \[\text{Area of first quadrant} = \int_0^8 \int_0^{8 - x} \, dy \, dx\] \[= \int_0^8 [y]_{y=0}^{y=8-x} \, dx = \int_0^8 (8 - x) \, dx\] \[= \left[ 8x - \frac{x^2}{2} \right]_0^8 = 64 - 32 = 32\] The total area of region B is: \[\text{Area of } B = 4 \times 32 = 128\]

Expressing the Area of Region T Region T is defined by the inequality (x2+y225)249, which can be rewritten as: \[18 \le x^2 + y^2 \le 32\]

To find the area, we switch to polar coordinates with x=rcosθ and y=rsinθ, where x2+y2=r2. Here, r ranges from 18 to 32, and θ ranges from 0 to 2π.

The area of T can then be found by: \[\text{Area of } T = \int_0^{2\pi} \int_{\sqrt{18}}^{\sqrt{32}} r \, dr \, d\theta\] \[= \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_{r=\sqrt{18}}^{r=\sqrt{32}} \, d\theta = \int_0^{2\pi} \left( \frac{32}{2} - \frac{18}{2} \right) \, d\theta\] \[= \int_0^{2\pi} 7 \, d\theta = 14\pi\]

The probability P that a dart lands in region T is the area of T divided by the area of B: \[P = \frac{\text{Area of } T}{\text{Area of } B} = \frac{14\pi}{128} = \frac{7\pi}{64}\]

So the probability is of the form mnπ, where m=7 and n=64, so m+n=7+64=71.

\[\boxed{\textbf{(B)}\ 71}\]

~Athmyx

Solution 3

2024 AMC 12B P09.jpeg

~Kathan

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE


See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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
2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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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