Difference between revisions of "2015 AMC 12A Problems/Problem 23"
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Our answer is <math>\textbf{(A)}</math>. | Our answer is <math>\textbf{(A)}</math>. | ||
+ | |||
+ | ==Case Solution== | ||
+ | Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability <math>0.25, 0.5, 0.25</math>, respectively. | ||
+ | |||
+ | Opposite side: Probability is obviously <math>1</math>, no matter what. | ||
+ | |||
+ | Same side: Pretend the points are on a line with coordinates <math>x</math> and <math>y</math>. If <math>|a-b|<=\frac{1}{2}</math>, drawing a graph will give probability <math>\frac{1}{4}</math>. | ||
+ | |||
+ | Peripheral side: superimpose a coordinate system over the points; call them <math>(0, x)</math> and <math>(0, y)</math>. WLOG set <math>x, y >= 0</math> and <math>x, y <= 1</math>. We need <math>0.25x^2+0.25y^2>0.25</math>, and drawing the coordinate system with bounds <math>(0, 0), (1, 0), (0, 1), (1, 1)</math> gives probability <math>1-\frac{\pi}{16}</math>. | ||
+ | |||
+ | Adding these up and finding the fraction gives us <math>\frac{1}{32} (26 - \pi)</math> for an answer of <math>\boxed{A}</math>. | ||
== 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}} | {{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}} |
Revision as of 21:40, 2 February 2018
Contents
Problem
Let be a square of side length 1. 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 and . What is ?
Solution
Divide the boundary of the square into halves, thereby forming 8 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 5 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 0.5 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. (x, y) is outside the unit circle with radius 0.5.)
Thus, averaging the probabilities gives
Our answer is .
Case Solution
Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability , respectively.
Opposite side: Probability is obviously , no matter what.
Same side: Pretend the points are on a line with coordinates and . If , drawing a graph will give probability .
Peripheral side: superimpose a coordinate system over the points; call them and . WLOG set and . We need , and drawing the coordinate system with bounds gives probability .
Adding these up and finding the fraction gives us for an answer of .
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 |