Difference between revisions of "2020 AMC 10A Problems/Problem 13"
(→Problem) |
m (numbering of solutions was off) |
||
| Line 7: | Line 7: | ||
<math>\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78</math> | <math>\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78</math> | ||
| − | ==Solution | + | ==Solution 1== |
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>. | If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>. | ||
- Lingjun | - Lingjun | ||
| − | ==Solution | + | ==Solution 2 (Complete States)== |
Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>. | Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>. | ||
Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: | Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: | ||
| Line 32: | Line 32: | ||
-mathisawesome2169 | -mathisawesome2169 | ||
| − | ==Solution | + | ==Solution 3 (States)== |
this is basically another version of solution 4; shoutout to mathisawesome2169 :D | this is basically another version of solution 4; shoutout to mathisawesome2169 :D | ||
Revision as of 21:16, 4 November 2024
- The following problem is from both the 2020 AMC 12A #11 and 2020 AMC 10A #13, so both problems redirect to this page.
Contents
Problem
A frog sitting at the point 
begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length 
, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices 
and 
. What is the probability that the sequence of jumps ends on a vertical side of the square?
Solution 1
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes 
. Since it starts on 
, there is a 
chance (up, down, or right) it will reach a diagonal on the first jump and 
chance (left) it will reach the vertical side. The probablity of landing on a vertical is 
.
- Lingjun
Solution 2 (Complete States)
Let 
denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at 
. Note that 
by reflective symmetry over the line 
. Similarly, 
, and 
.
Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point:
![\[P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\]](http://latex.artofproblemsolving.com/b/e/f/befd78b0b1bbec210c23d337155eed6c7b6ceccc.png)
![\[P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\]](http://latex.artofproblemsolving.com/e/d/0/ed070fe0984fb02949c5b4e8c2d92b94ff731da2.png)
![\[P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\]](http://latex.artofproblemsolving.com/3/c/8/3c8078b6b38cd9f42f490fc7624b8cb5029abef0.png)
![\[P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\]](http://latex.artofproblemsolving.com/a/1/0/a101a1795f4bf5aed1dc96e592eb709c68efeb46.png)
We have a system of 
equations in variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives
![\[P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)\]](http://latex.artofproblemsolving.com/d/3/6/d36c919abef162c5e773b7feac15fbe463f48490.png)
![\[P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}\]](http://latex.artofproblemsolving.com/7/6/9/7696ecfc2ad73aa1cdaa896476f0ec0fad3abef4.png)
Plugging in the third equation into this gives
![\[P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}\]](http://latex.artofproblemsolving.com/b/b/6/bb613413c794a72bc75f637709a9973cd312976a.png)
![\[P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}\]](http://latex.artofproblemsolving.com/9/1/8/918d0301c95978a28f4ee5f8f52b6236e9158af9.png)
Next, plugging in the second and third equation into the first equation yields
![\[P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)\]](http://latex.artofproblemsolving.com/f/c/4/fc4431a4578fc22f24ec3a6de3ca03d8c4a85f08.png)
![\[P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\]](http://latex.artofproblemsolving.com/4/7/e/47e0f7c9e215880a3a2148a8d363a02210812754.png)
Now plugging in (*) into this, we get
![\[P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)\]](http://latex.artofproblemsolving.com/a/6/a/a6a08ad179679de998e79a297b4e4822b6b01400.png)
![\[P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}\]](http://latex.artofproblemsolving.com/1/d/4/1d4333f9c11e369df593a7f047f49a1dbed27864.png)
-mathisawesome2169
Solution 3 (States)
this is basically another version of solution 4; shoutout to mathisawesome2169 :D
First, we note the different places the frog can go at certain locations in the square:
If the frog is at a border vertical point (
), it moves with probability to a vertical side of the square, probability to the center of the square, and probability to a corner square.
If the frog is at a border horizontal point (
), it moves with probability to a horizontal side of the square, probability to the center of the square, and probability to a corner square.
If the frog is at a center square (
), it moves with probability to a border horizontal point and probability to a border vertical point.
If the frog is at a corner (
), it moves with probability to a vertical side of the square, probability to a horizontal side, probability to a border horizontal point, and probability to a border vertical point.
Now, let 
denote the probability of the frog reaching a vertical side when it is at a border vertical point. Similarly, let 
denote the probability of the frog reaching a vertical side when it is at a border horizontal point.
Now, the probability of the frog reaching a vertical side of the square at any location inside the square can be expressed in terms of and .
First, the two easier ones: 
, and 
.
Now, we can write and in terms of and , allowing us to solve a system of two variables:
![\[x=\frac{1}{4}+\frac{1}{4}P_{center}+\frac{1}{2}P_{corner}=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}y\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y\right)\]](http://latex.artofproblemsolving.com/b/1/b/b1b8cbc816a39dd7ef95f94a77fb9b6257aa9615.png)
and ![\[y=\frac{1}{4}P_{center}+\frac{1}{2}P_{corner}=\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}y\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y\right).\]](http://latex.artofproblemsolving.com/b/b/6/bb659dab4585d011dba2a7b890f5847920cb1123.png)
From these two equations, it is apparent that 
. We can then substitute this value for back into any of the two equations above to get ![\[x=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}\left(x-\frac{1}{4}\right)\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}\left(x-\frac{1}{4}\right)\right).\]](http://latex.artofproblemsolving.com/5/9/f/59fd07cad979e2a52656a1be70e4bde8033113ef.png)
Although this certainly looks intimidating, we can expand the parentheses and multiply both sides by 16 to eliminate the fractions, which upon simplification yields the equation ![\[16x=5+8x,\]](http://latex.artofproblemsolving.com/9/f/5/9f55638381a61d103dbdb3e0a136e22fb18ac513.png)
giving us the desired probability 
. The answer is then 
.
- curiousmind888 & TGSN
Video Solutions
Video Solution 1
IceMatrix's Solution (Starts at 4:40)
Video Solution 2 (Simple & Quick)
Video Solution 3
On The Spot STEM
Video Solution 4
~savannahsolver
Video Solution 5
https://youtu.be/IRyWOZQMTV8?t=5173
~ pi_is_3.14
Video Solution 6
https://www.youtube.com/watch?v=R220vbM_my8
~ amritvignesh0719062.0
Video Solution 7
https://www.youtube.com/watch?v=TvYoiU_zct8
~ mathgenius2012
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| 2020 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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. 
