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

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(→‎Solution 2)
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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 the opposite side is <math>\frac{1}{4}</math>. We discuss these three cases.  
 
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 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. Then <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 <math>(x, y)</math>-plane. The region <math>\{(a,b): |b-a|> 1/2\}</math> has the area of  <math>(1/2)^2</math>. Therefore, the probability that <math>|b-a|> 1/2</math> is <math>1/4</math>.  
+
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 <math>(x, y)</math>-plane. The region <math>\{(a,b): |b-a|> 1/2\}</math> has the area of  <math>(1/2)^2</math>. Therefore, the probability that <math>|b-a|> 1/2</math> is <math>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 1/2, 0\le a, b\le 1\}</math> is <math>\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-\pi/16</math>. . Therefore, the probability that the distance between <math>(a, 0)</math> and <math>(0, b)</math> is at least <math>1/2</math> is <math>1-\pi/16</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 1/2, 0\le a, b\le 1\}</math> is <math>\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-\pi/16</math>. . Therefore, the probability that the distance between <math>(a, 0)</math> and <math>(0, b)</math> is at least <math>1/2</math> is <math>1-\pi/16</math>.  
Line 27: Line 27:
  
 
Thus the probability that the probability that the distance between the two points is at least <math>1/2</math> is given by  
 
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} \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \frac{1}{4} =\frac{26-\pi}{32}.
 
\frac{1}{4} \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \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 (a).
 
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 (a).
  

Revision as of 21:46, 19 May 2015

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

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 \[\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.\]

(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 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 $(x, y)$-plane. The region $\{(a,b): |b-a|> 1/2\}$ has the area of $(1/2)^2$. Therefore, the probability that $|b-a|> 1/2$ is $1/4$.

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 $\{(a,b): \sqrt{a^2+b^2} \le 1/2, 0\le a, b\le 1\}$ is $\pi/16$ 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 $1-\pi/16$. . Therefore, the probability that the distance between $(a, 0)$ and $(0, b)$ is at least $1/2$ is $1-\pi/16$.

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

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
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