2023 AMC 12B Problems/Problem 20
Contents
Problem
Cyrus the frog jumps units in a direction, then more in another direction. What is the probability that he lands less than unit away from his starting position?
Solution 1
Let Cyrus's starting position be . WLOG, let the place Cyrus lands at for his first jump be . From , Cyrus can reach all the points on . The probability that Cyrus will land less than unit away from is .
Therefore, the answer is
Solution 2
Denote by the position after the th jump. Thus, to fall into the region centered at and with radius 1, .
Therefore, the probability is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3(coord bash)
Let the orgin be the starting point of frog. Then, WLOG assume that after the first jump, it is at the point (2,0). Then, the range of all possible places the frog can jump to at its second jump is the circle with equation .If it landed unit within its starting point (the orgin), then it is inside the circle . We clearly want the intersection point. So we're trying to solve the system of equations and . We have , so . Therefore, our final answer would be (the angle we want divided by ). But that is not one of our answer choices! Don't worry though, because
where the last step holds by the double angle formula. By now, it is clear that our answer is . ~Ddk001
Solution 4 - Law of Cosines and Double Angle Formula
Let be Cyrus's starting point, be the first point he jumps to (), and be the second point he jumps to (). Let angle be , such that . The probability of would therefore be (since could be on either side of so there are two possible areas of having ) which simplifies to . Converting to radians gives us . To find , we use the law of cosines.
The probability is
which is .
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
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