Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
m (→‎Solution 1)
Line 7: Line 7:
 
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.)
 
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 up to <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}{\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>
 
<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>
  

Revision as of 22:36, 14 February 2018

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

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 \[P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).\]

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

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 \[\frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \frac{1}{4} =\frac{26-\pi}{32}.\] Therefore $a=26$, $b=1$, and $c=32$. Thus, $a+b+c=59$ and the answer is $\textbf{(A).}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_25&oldid=91227"