Difference between revisions of "2023 AMC 12B Problems/Problem 20"
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which is <math>E</math>. | which is <math>E</math>. | ||
+ | |||
+ | ==Solution 5 - Complex Number and Double Angle Formula== | ||
+ | |||
+ | The frog jumps 2 units in one direction, which we can represent as <math> z_1 = 2e^{ia} </math>, where <math> a </math> is the angle in radians. Then, the frog jumps another 2 units in a different direction, represented as <math> z_2 = 2e^{ib} </math>, where <math> b </math> is the angle in radians. | ||
+ | |||
+ | The total displacement from the starting position is <math> z3 = z_1 + z_2 = 2e^{ia} + 2e^{ib} </math>. | ||
+ | |||
+ | |||
+ | We need to find the condition under which <math> |z3| <= 1 </math>. This translates to: | ||
+ | <cmath> |2e^{ia} + 2e^{ib}| <= 1 </cmath> | ||
+ | |||
+ | <cmath> |e^{ia} + e^{ib}| <= \frac{1}{2} </cmath> | ||
+ | |||
+ | <cmath> e^{ia} + e^{ib} = \cos(a) + i\sin(a) + \cos(b) + i\sin(b) </cmath> | ||
+ | <cmath> = (\cos(a) + \cos(b)) + i(\sin(a) + \sin(b)) </cmath> | ||
+ | |||
+ | The magnitude squared of this sum is: | ||
+ | <cmath> |e^{ia} + e^{ib}|^2 = (\cos(a) + \cos(b))^2 + (\sin(a) + \sin(b))^2 </cmath> | ||
+ | <cmath> = \cos^2(a) + 2\cos(a)\cos(b) + \cos^2(b) + \sin^2(a) + 2\sin(a)\sin(b) + \sin^2(b) </cmath> | ||
+ | <cmath> = 2 + 2(\cos(a)\cos(b) + \sin(a)\sin(b)) </cmath> | ||
+ | <cmath> = 2 + 2\cos(a - b) </cmath> | ||
+ | |||
+ | Thus: | ||
+ | <cmath> 2 + 2\cos(a - b) <= \frac{1}{4} </cmath> | ||
+ | <cmath> 2\cos(a - b) <= -\frac{7}{4} </cmath> | ||
+ | <cmath> \cos(a - b) <= -\frac{7}{8} </cmath> | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 00:44, 3 August 2024
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 desired answer would be
(the angle we want divided by
). Since
where the last step holds by the double angle formula, 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 .
Solution 5 - Complex Number and Double Angle Formula
The frog jumps 2 units in one direction, which we can represent as , where
is the angle in radians. Then, the frog jumps another 2 units in a different direction, represented as
, where
is the angle in radians.
The total displacement from the starting position is .
We need to find the condition under which . This translates to:
The magnitude squared of this sum is:
Thus:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.