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Difference between revisions of "2023 AMC 12B Problems/Problem 20"

m (→‎Solution 3(coord bash))
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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

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?

$\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{1}{5}\qquad\textbf{(C)}~\frac{\sqrt{3}}{8}\qquad\textbf{(D)}~\frac{\arctan \frac{1}{2}}{\pi}\qquad\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}$

Solution 1

2023AMC12BP20.png

WLOG, let the place Cyrus lands in his first jump be $O$. From $O$, Cyrus can reach all the points on $\odot O$. The probability that Cyrus will land less than $1$ unit away from his starting position $S$ is $\frac{4 \alpha }{ 2 \pi}$.

\[\sin \alpha = \frac{ \frac12 }{2} = \frac14, \quad \alpha = \arcsin \frac14\]

Therefore, the answer is

\[\frac{4 \arcsin \frac14 }{ 2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}\]

~isabelchen

Solution 2

Denote by $A_i$ the position after the $i$th jump. Thus, to fall into the region centered at $A_0$ and with radius 1, $\angle A_2 A_1 A_0 < 2 \arcsin \frac{1/2}{2} = 2 \arcsin \frac{1}{4}$.

Therefore, the probability is \[ \frac{2 \cdot 2 \arcsin \frac{1}{4}}{2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}. \]

~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 $(x-2)^2+y^2=2^2$.If it landed $1$ unit within its starting point (the orgin), then it is inside the circle $x^2+y^2=1$. We clearly want the intersection point. So we're trying to solve the system of equations $x^2+y^2=1$ and $(x-2)^2+y^2=2^2$. We have $x=\frac{1}{4}$, so $y=\pm\frac{\sqrt{15}}{4}$. Therefore, our final answer would be $\frac{\arcsin{\frac{\sqrt{15}}{8}}}{\pi}$ (the angle we want divided by $2\pi$). But that is not one of our answer choices! Don't worry though, because

$\arcsin{\frac{\sqrt{15}}{8}}=\arccos{\frac{7}{8}}=\arccos{(1-2 \cdot (\frac{1}{4})^{2})}=2\arcsin{\frac{1}{4}}$

where the last step holds by the double angle formula. By now, it is clear that our answer is $(B)\frac{2\arcsin{\frac{1}{4}}}{\pi}$. ~ddk001

Solution 4

Let $A$ be your starting position. WLOG, assume that the frog jumps toward the right $2$ units to a point $B$.

Now, the frog will jump $2$ more units in an arbitary direction. We can draw a circle of radius $2$ around center $B$.

We can draw another circle of radius $1$ around center $A$ to denote "1 unit away from his starting position"


Now we have a straight line $AB$ of length $2$, with a circle of radius $1$ around center $A$ and a circle of radius $2$ around center $B$. Let circle $A$ and circle $B$ intersect at $X$ and $Y$.

The probability the frog lands less than $1$ unit away from his starting position is the same as $\frac{\angle XBY}{2\pi}$

That is the set up, now we just find $\angle XBY$ using basic right triangles and trig.



Triangles $ABX$ and $ABY$ are congruent (radii, SSS), so $\angle XBY = \angle ABX + \angle ABY = 2\angle ABX$

Triange $ABX$ is an isosceles triangle with the side lengths $2,2,1$ (due to radii), we can drop the altitude $BC$ to split it into 2 congruent right triangles. Now $\angle ABX=2\angle CBX$ and $\angle XBY =2\angle ABX=4\angle CBX$


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 $CBX$ has side length $\frac12$ opposite of $\angle CBX$, and a hypotenuse length of $2$, so \[\sin{\angle CBX}=\frac{\frac12}{2}=\frac14\] \[\angle CBX=\arcsin{\frac14}\] \[\angle XBY=4\angle CBX=4\arcsin{\frac14}\]

And the answer is \[\frac{\angle XBY}{2\pi}=\boxed{\frac{2\arcsin{\frac14}}{\pi}}\]

~ 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

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