Difference between revisions of "2023 AMC 12B Problems/Problem 20"
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where the last step holds by the double angle formula. By now, it is clear that our answer is <math>(B)\frac{2\arcsin{\frac{1}{4}}}{\pi}</math>. | where the last step holds by the double angle formula. By now, it is clear that our answer is <math>(B)\frac{2\arcsin{\frac{1}{4}}}{\pi}</math>. | ||
~ddk001 | ~ddk001 | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>A</math> be your starting position. WLOG, assume that the frog jumps toward the right <math>2</math> units to a point <math>B</math>. | ||
+ | |||
+ | Now, the frog will jump <math>2</math> more units in an arbitary direction. | ||
+ | We can draw a circle of radius <math>2</math> around center <math>B</math>. | ||
+ | |||
+ | We can draw another circle of radius <math>1</math> around center <math>A</math> to denote "1 unit away from his starting position" | ||
+ | |||
+ | |||
+ | Now we have a straight line <math>AB</math> of length <math>2</math>, with a circle of radius <math>1</math> around center <math>A</math> and a circle of radius <math>2</math> around center <math>B</math>. Let circle <math>A</math> and circle <math>B</math> intersect at <math>X</math> and <math>Y</math>. | ||
+ | |||
+ | <b>The probability the frog lands less than <math>1</math> unit away from his starting position is the same as <math>\frac{\angle XBY}{2\pi}</math></b> | ||
+ | |||
+ | That is the set up, now we just find <math>\angle XBY</math> using basic right triangles and trig. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Triangles <math>ABX</math> and <math>ABY</math> are congruent (radii, SSS), so <math>\angle XBY = \angle ABX + \angle ABY = 2\angle ABX</math> | ||
+ | |||
+ | Triange <math>ABX</math> is an isosceles triangle with the side lengths <math>2,2,1</math> (due to radii), we can drop the altitude <math>BC</math> to split it into 2 congruent right triangles. Now <math>\angle ABX=2\angle CBX</math> and <math>\angle XBY =2\angle ABX=4\angle CBX</math> | ||
+ | |||
+ | |||
+ | We look at the answer choices and see that answers could be left in terms of inverse trig functions, so solving the rest is relatively simple. Right triangle <math>CBX</math> has side length <math>\frac12</math> opposite of <math>\angle CBX</math>, and a hypotenuse length of <math>2</math>, so | ||
+ | <cmath>\sin{\angle CBX}=\frac{\frac12}{2}=\frac14</cmath> | ||
+ | <cmath>\angle CBX=\arcsin{\frac14}</cmath> | ||
+ | <cmath>\angle XBY=4\angle CBX=4\arcsin{\frac14}</cmath> | ||
+ | |||
+ | And the answer is <cmath>\frac{\angle XBY}{2\pi}=\boxed{\frac{2\arcsin{\frac14}}{\pi}}</cmath> | ||
+ | |||
+ | ~ CherryBerry | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2023|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2023|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:56, 15 November 2023
Contents
[hide]Problem
Cyrus the frog jumps 2 units in a direction, then 2 more in another direction. What is the probability that he lands less than 1 unit away from his starting position?
Solution 1
WLOG, let the place Cyrus lands in his first jump be . From , Cyrus can reach all the points on . The probability that Cyrus will land less than unit away from his starting position 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
Let be your starting position. WLOG, assume that the frog jumps toward the right units to a point .
Now, the frog will jump more units in an arbitary direction. We can draw a circle of radius around center .
We can draw another circle of radius around center to denote "1 unit away from his starting position"
Now we have a straight line of length , with a circle of radius around center and a circle of radius around center . Let circle and circle intersect at and .
The probability the frog lands less than unit away from his starting position is the same as
That is the set up, now we just find using basic right triangles and trig.
Triangles and are congruent (radii, SSS), so
Triange is an isosceles triangle with the side lengths (due to radii), we can drop the altitude to split it into 2 congruent right triangles. Now and
We look at the answer choices and see that answers could be left in terms of inverse trig functions, so solving the rest is relatively simple. Right triangle has side length opposite of , and a hypotenuse length of , so
And the answer is
~ CherryBerry
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.