Difference between revisions of "2015 AMC 10A Problems/Problem 25"
Cocohearts (talk | contribs) m (→Solution 2) |
m (→Solution 3) |
||
(11 intermediate revisions by 7 users not shown) | |||
Line 4: | Line 4: | ||
<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 1== | + | ==Solution 1 (Calculus)== |
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 | + | 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> | ||
Line 28: | Line 28: | ||
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> | <cmath> | ||
− | \frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \frac{1}{4} =\frac{26-\pi}{32}. | + | \frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}\left(1 - \frac{\pi}{16}\right) + \frac{1}{4} =\frac{26-\pi}{32}. |
</cmath> | </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> | 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> | ||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc10a/399 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}} | {{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}} | {{MAA Notice}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 11:13, 29 November 2020
Contents
Problem 25
Let be a square of side length . Two points are chosen independently at random on the sides of . The probability that the straight-line distance between the points is at least is , where , , and are positive integers with . What is ?
Solution 1 (Calculus)
Divide the boundary of the square into halves, thereby forming segments. Without loss of generality, let the first point be in the bottom-left segment. Then, it is easy to see that any point in the segments not bordering the bottom-left segment will be distance at least apart from . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least apart from is because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
If the second point is on the left-bottom segment, then if is distance away from the left-bottom vertex, then must be up to away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is
(Alternatively, one can equate the problem to finding all valid with such that , i.e. is outside the unit circle with radius )
Thus, averaging the probabilities gives
Our answer is .
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 , on an adjacent side is , and on the opposite side is . We discuss these three cases.
Case 1: Two points are on the same side. Let the first point be and the second point be in the -axis with . Consider a point on the unit square on the Cartesian plane. The region has the area of . Therefore, the probability that is .
Case 2: Two points are on two adjacent sides. Let the two sides be on the x-axis and on the y-axis and let one point be and the other point be . Then and the distance between the two points is . As in Case 1, is a point on the unit square . The area of the region is and the area of its complementary set inside the square (i.e. ) is . Therefore, the probability that the distance between and is at least is .
Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least is obviously .
Thus the probability that the probability that the distance between the two points is at least is given by Therefore , , and . Thus, and the answer is
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 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 to and and be the values representing the positions of Point A and Point B respectively. Our problem now asks for the probability that Graphing the inequality on a coordinate plane with and as the and -axes gives us a "good area" of out of a "total area" of Hence, the probability the inequality is satisfied is
Case 2: Point B is on a side adjacent to the side with Point A: This setup occurs with probability This is a slight deviation of the same geometric probability principle. This time, let the common vertex of the two sides be and the sides have side length Again, let and be the values representing the positions of Point A and Point B respectively, so the Pythagorean Theorem yields graphing into a quarter circle of radius and a total area that is Hence, our probability is
Case 3: Point B is on a side opposite to the side with Point A: This setup occurs with probability
Clearly, the distance between Point A and Point B are at least 1, so it must be at least The probability in this case is
Now taking the probabilities of the setups into account, our final probability is Thus so
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/399
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.