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Difference between revisions of "2015 AMC 10A Problems/Problem 25"

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==Problem 25==
 
==Problem 25==
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Let <math>S</math> be a square of side length <math>1</math>.  Two points are chosen independently at random on the sides of <math>S</math>.  The probability that the straight-line distance between the points is at least <math>\tfrac12</math> is <math>\tfrac{a-b\pi}c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers with <math>\gcd(a,b,c)=1</math>.  What is <math>a+b+c</math>?
+
Let <math>S</math> be a square of side length <math>1</math>.  Two points are chosen independently at random on the sides of <math>S</math>.  The probability that the straight-line distance between the points is at least <math>\dfrac{1}{2}</math> is <math>\dfrac{a-b\pi}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers with <math>\gcd(a,b,c)=1</math>.  What is <math>a+b+c</math>?
  
 
<math>\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63</math>
 
<math>\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63</math>
  
−
==Solution==
+
==Solution 1 (Calculus)==
−
Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
+
Divide the boundary of the square into halves, thereby forming <math>8</math> segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the <math>5</math> segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least <math>0.5</math> apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
  
−
If the second point <math>B</math> is on the left-bottom segment, then if <math>A</math> is distance <math>x</math> away from the left-bottom vertex, then <math>B</math> must be at least <math>\dfrac{1}{2} - \sqrt{0.25 - x^2}</math> away from that same vertex. Thus, using an averaging argument we find that the probability in this case is
+
If the second point <math>B</math> is on the left-bottom segment, then if <math>A</math> is distance <math>x</math> away from the left-bottom vertex, then <math>B</math> must be up to <math>\dfrac{1}{2} - \sqrt{0.25 - x^2}</math> away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is
−
<cmath>\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4(\frac{1}{4} - \frac{\pi}{16}) = 1 - \frac{\pi}{4}.</cmath>
+
<cmath>\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.</cmath>
  
−
(Alternatively, one can equate the problem to finding all valid <math>(x, y)</math> with <math>0 < x, y < \dfrac{1}{2}</math>  such that <math>x^2 + y^2 \ge \dfrac{1}{4}</math>, i.e. (x, y) is outside the unit circle with radius 0.5.)
+
(Alternatively, one can equate the problem to finding all valid <math>(x, y)</math> with <math>0 < x, y < \dfrac{1}{2}</math>  such that <math>x^2 + y^2 \ge \dfrac{1}{4}</math>, i.e. <math>(x, y)</math> is outside the unit circle with radius <math>0.5.</math>)
  
 
Thus, averaging the probabilities gives
 
Thus, averaging the probabilities gives
−
<cmath>P = \frac{1}{8} (5 + \frac{1}{2} + 1 - \frac{\pi}{4}) = \frac{1}{32} (26 - \pi).</cmath>
+
<cmath>P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).</cmath>
  
−
Our answer is <math>\textbf{(A)}</math>.
+
Thus our answer is <math>\boxed{\textbf{(A) } 59}</math>.
 +
 
 +
~minor edit by [[User: Yiyj1|Yiyj1]]
 +
 
 +
== Solution 2 ==
 +
Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is <math>\frac{1}{4}</math>, on an adjacent side is <math>\frac{1}{2}</math>, and on the opposite side is <math>\frac{1}{4}</math>. We discuss these three cases.
 +
 
 +
Case 1: Two points are on the same side. Let the first point be <math>a</math> and the second point be <math>b</math> in the <math>x</math>-axis with <math>0\le a, b\le 1</math>. Consider <math>(a, b)</math> a point on the unit square <math>[0,1]\times [0,1]</math> on the Cartesian plane. The region <math>\{(a,b): |b-a|> \frac{1}{2}\}</math> has the area of  <math>(\frac{1}{2})^2</math>. Therefore, the probability that <math>|b-a|> \frac{1}{2}</math> is <math>\frac{1}{4}</math>.
 +
 
 +
Case 2: Two points are on two adjacent sides. Let the two sides be <math>[0,1]</math> on the x-axis and <math>[0,1]</math> on the y-axis and let one point be <math>(a, 0)</math> and the other point be <math>(0, b)</math>. Then <math>0\le a, b\le 1</math> and the distance between the two points is <math>\sqrt{a^2+b^2}</math>. As in Case 1, <math>(a, b)</math> is a point on the unit square <math>[0,1]\times [0,1]</math>. The area of the region <math>\{(a,b): \sqrt{a^2+b^2} \le \frac{1}{2}, 0\le a, b\le 1\}</math> is <math>\frac{\pi}{16}</math> and the area of its complementary set inside the square (i.e. <math>\{(a,b): \sqrt{a^2+b^2} > 1/2, 0\le a, b\le 1\}</math> ) is <math>1-\frac{\pi}{16}</math>. Therefore, the probability that the distance between <math>(a, 0)</math> and <math>(0, b)</math> is at least <math>\frac{1}{2}</math> is <math>1-\frac{\pi}{16}</math>.
 +
 
 +
Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least <math>1/2</math> is obviously <math>1</math>.
 +
 
 +
Thus the probability that the probability that the distance between the two points is at least <math>1/2</math> is given by
 +
<cmath>
 +
\frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}\left(1 - \frac{\pi}{16}\right) + \frac{1}{4} =\frac{26-\pi}{32}.
 +
</cmath>
 +
Therefore <math>a=26</math>, <math>b=1</math>, and <math>c=32</math>. Thus, <math>a+b+c=59</math> and the answer is <math>\textbf{(A).}</math>
 +
 
 +
==Solution 3==
 +
Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases:
 +
 
 +
Case 1: Point B is on the same side as Point A: This setup occurs with probability <math>\dfrac{1}{4}.</math>
 +
This is the standard geometric probability problem. Since Point A and Point B can be anywhere on the side, we can't really "count" all of the possibilities. Hence, we translate the problem to a problem involving areas, which can be a finite value while still "containing" infinitely many points. Let side of a square be a number line from <math>0</math> to <math>1,</math> and <math>a</math> and <math>b</math> be the values representing the positions of Point A and Point B respectively. Our problem now asks for the probability that <math>|a-b| \geq 1/2.</math> Graphing the inequality on a coordinate plane with <math>a</math> and <math>b</math> as the <math>x</math> and <math>y</math>-axes gives us a "good area" of <math>1/4</math> out of a "total area" of <math>1.</math> Hence, the probability the inequality is satisfied is <math>1/4 \div 1 = 1/4.</math>
 +
 
 +
Case 2: Point B is on a side adjacent to the side with Point A: This setup occurs with probability <math>\dfrac{1}{2}.</math>
 +
This is a slight deviation of the same geometric probability principle. This time, let the common vertex of the two sides be <math>0,</math> and the sides have side length <math>1.</math> Again, let <math>a</math> and <math>b</math> be the values representing the positions of Point A and Point B respectively, so the Pythagorean Theorem yields <math>\sqrt{a^2+b^2}\geq 1/2,</math> graphing into a quarter circle of radius <math>1/2,</math> and a total area that is <math>1.</math> Hence, our probability is <math>1-\frac{\pi(1/2)^2}{4} = \frac{16-\pi}{16}.</math>
 +
 
 +
 
 +
Case 3: Point B is on a side opposite to the side with Point A: This setup occurs with probability <math>\dfrac{1}{4}.</math>
 +
Clearly, the distance between Point A and Point B are at least 1, so it must be at least <math>1/2.</math> The probability in this case is <math>1.</math>
 +
 
 +
 
 +
Now taking the probabilities of the setups into account, our final probability is <cmath>\frac{1}{4}\cdot \left( \frac{1}{4} \right)+ \frac{1}{2}\cdot \left(\frac{16-\pi}{16}\right) + \frac{1}{4}\cdot \left(1\right)=\frac{26-\pi}{32}.</cmath>Thus <math>(a,b,c)=(26,1,32),</math> so <math>a+b+c= \boxed{\mathbf{(A)}\;59}</math>
 +
 
 +
==Geometric way to solve case 2==
 +
The probability of case 2 (if the two points fall on adjacent sides) can be evaluated geometrically. Let the square have vertices at <math>(0,0)</math>, <math>(1,0)</math>, <math>(0,1)</math> and <math>(1,1)</math>, and WLOG, let point <math>A</math> be on the <math>x</math> axis and let point <math>B</math> be on the <math>y</math> axis. Let the midpoint of <math>AB</math> be <math>M</math>. We can draw the following conclusions:
 +
 
 +
1. <math>M</math> must fall inside the square with vertices at <math>(0,0)</math>, <math>(0.5,0)</math> and <math>(0,0.5)</math> and <math>(0.5,0.5)</math>.
 +
 
 +
2. <math>M</math> will fall inside the square described above randomly, with a uniform probability of landing anywhere.
 +
 
 +
3. If and only if <math>M</math> is more than <math>0.25</math> units from the origin will <math>AB</math> be more than <math>0.5</math> units long.
 +
 
 +
The proof of each of these statements is left as an exercise to the reader.
 +
 
 +
Thus, the probability of case 2 can be mapped to the probability that a randomly chosen point inside the square with vertices at <math>(0,0)</math>, <math>(0.5,0)</math> and <math>(0,0.5)</math> and <math>(0.5,0.5)</math> is more than <math>0.25</math> units from the origin. By calculating areas, this is <math>\frac{16 - \pi}{16}</math>.
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://artofproblemsolving.com/videos/amc/2015amc10a/399
 +
 
 +
~naren_pr
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}}
 +
 
 +
{{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}}
 +
 
 +
{{MAA Notice}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]

Latest revision as of 12:00, 18 August 2024

Problem 25

Let $S$ be a square of side length $1$. Two points are chosen independently at random on the sides of $S$. The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$, where $a$, $b$, and $c$ are positive integers with $\gcd(a,b,c)=1$. What is $a+b+c$?

$\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

Solution 1 (Calculus)

Divide the boundary of the square into halves, thereby forming $8$ segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the $5$ segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least $0.5$ apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be up to $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is \[\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.\]

(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$, i.e. $(x, y)$ is outside the unit circle with radius $0.5.$)

Thus, averaging the probabilities gives \[P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).\]

Thus our answer is $\boxed{\textbf{(A) } 59}$.

~minor edit by Yiyj1

Solution 2

Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is $\frac{1}{4}$, on an adjacent side is $\frac{1}{2}$, and on the opposite side is $\frac{1}{4}$. We discuss these three cases.

Case 1: Two points are on the same side. Let the first point be $a$ and the second point be $b$ in the $x$-axis with $0\le a, b\le 1$. Consider $(a, b)$ a point on the unit square $[0,1]\times [0,1]$ on the Cartesian plane. The region $\{(a,b): |b-a|> \frac{1}{2}\}$ has the area of $(\frac{1}{2})^2$. Therefore, the probability that $|b-a|> \frac{1}{2}$ is $\frac{1}{4}$.

Case 2: Two points are on two adjacent sides. Let the two sides be $[0,1]$ on the x-axis and $[0,1]$ on the y-axis and let one point be $(a, 0)$ and the other point be $(0, b)$. Then $0\le a, b\le 1$ and the distance between the two points is $\sqrt{a^2+b^2}$. As in Case 1, $(a, b)$ is a point on the unit square $[0,1]\times [0,1]$. The area of the region $\{(a,b): \sqrt{a^2+b^2} \le \frac{1}{2}, 0\le a, b\le 1\}$ is $\frac{\pi}{16}$ and the area of its complementary set inside the square (i.e. $\{(a,b): \sqrt{a^2+b^2} > 1/2, 0\le a, b\le 1\}$ ) is $1-\frac{\pi}{16}$. Therefore, the probability that the distance between $(a, 0)$ and $(0, b)$ is at least $\frac{1}{2}$ is $1-\frac{\pi}{16}$.

Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least $1/2$ is obviously $1$.

Thus the probability that the probability that the distance between the two points is at least $1/2$ is given by \[\frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}\left(1 - \frac{\pi}{16}\right) + \frac{1}{4} =\frac{26-\pi}{32}.\] Therefore $a=26$, $b=1$, and $c=32$. Thus, $a+b+c=59$ and the answer is $\textbf{(A).}$

Solution 3

Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases:

Case 1: Point B is on the same side as Point A: This setup occurs with probability $\dfrac{1}{4}.$ This is the standard geometric probability problem. Since Point A and Point B can be anywhere on the side, we can't really "count" all of the possibilities. Hence, we translate the problem to a problem involving areas, which can be a finite value while still "containing" infinitely many points. Let side of a square be a number line from $0$ to $1,$ and $a$ and $b$ be the values representing the positions of Point A and Point B respectively. Our problem now asks for the probability that $|a-b| \geq 1/2.$ Graphing the inequality on a coordinate plane with $a$ and $b$ as the $x$ and $y$-axes gives us a "good area" of $1/4$ out of a "total area" of $1.$ Hence, the probability the inequality is satisfied is $1/4 \div 1 = 1/4.$

Case 2: Point B is on a side adjacent to the side with Point A: This setup occurs with probability $\dfrac{1}{2}.$ This is a slight deviation of the same geometric probability principle. This time, let the common vertex of the two sides be $0,$ and the sides have side length $1.$ Again, let $a$ and $b$ be the values representing the positions of Point A and Point B respectively, so the Pythagorean Theorem yields $\sqrt{a^2+b^2}\geq 1/2,$ graphing into a quarter circle of radius $1/2,$ and a total area that is $1.$ Hence, our probability is $1-\frac{\pi(1/2)^2}{4} = \frac{16-\pi}{16}.$


Case 3: Point B is on a side opposite to the side with Point A: This setup occurs with probability $\dfrac{1}{4}.$ Clearly, the distance between Point A and Point B are at least 1, so it must be at least $1/2.$ The probability in this case is $1.$


Now taking the probabilities of the setups into account, our final probability is \[\frac{1}{4}\cdot \left( \frac{1}{4} \right)+ \frac{1}{2}\cdot \left(\frac{16-\pi}{16}\right) + \frac{1}{4}\cdot \left(1\right)=\frac{26-\pi}{32}.\]Thus $(a,b,c)=(26,1,32),$ so $a+b+c= \boxed{\mathbf{(A)}\;59}$

Geometric way to solve case 2

The probability of case 2 (if the two points fall on adjacent sides) can be evaluated geometrically. Let the square have vertices at $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$, and WLOG, let point $A$ be on the $x$ axis and let point $B$ be on the $y$ axis. Let the midpoint of $AB$ be $M$. We can draw the following conclusions:

1. $M$ must fall inside the square with vertices at $(0,0)$, $(0.5,0)$ and $(0,0.5)$ and $(0.5,0.5)$.

2. $M$ will fall inside the square described above randomly, with a uniform probability of landing anywhere.

3. If and only if $M$ is more than $0.25$ units from the origin will $AB$ be more than $0.5$ units long.

The proof of each of these statements is left as an exercise to the reader.

Thus, the probability of case 2 can be mapped to the probability that a randomly chosen point inside the square with vertices at $(0,0)$, $(0.5,0)$ and $(0,0.5)$ and $(0.5,0.5)$ is more than $0.25$ units from the origin. By calculating areas, this is $\frac{16 - \pi}{16}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc10a/399

~naren_pr

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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
2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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