Difference between revisions of "2015 AMC 10A Problems/Problem 25"
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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 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 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 | + | 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 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>. |
Revision as of 00:44, 9 January 2018
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
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 at least away from that same vertex. 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
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 |
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