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 12A Problems/Problem 23"

m (Parentheses latex formatting)
(→‎See Also)
Line 20: Line 20:
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}}
 +
{{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}}

Revision as of 21:55, 28 January 2018

Problem

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 $\frac12$ is $\frac{a-b\pi}{c}$, where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$. 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

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 at least $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from that same vertex. Thus, using an averaging argument we find that the probability in this case is \[\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left(\frac{1}{4} - \frac{\pi}{16}\right) = 1 - \frac{\pi}{4}.\]

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

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

See Also

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
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_23&oldid=90043"