Difference between revisions of "2018 AIME II Problems/Problem 8"
(→Solution 3 (General Case)) |
(→Solution 3 (General Case)) |
||
Line 67: | Line 67: | ||
==Solution 3 (General Case)== | ==Solution 3 (General Case)== | ||
− | Mark the total number of distinct sequences of jumps for the frog to reach the point <math>(x,y)</math> as <math>\varphi (x,y)</math>. Consider for each point <math>(x,y)</math> in the first quadrant, there are only <math>4</math> possible points in the first quadrant for frog to reach point <math>(x,y)</math>, and these 4 points are <cmath>(x-1,y); (x-2,y); (x,y-1); (x,y-2)</cmath>. As a result, the way to count <math>\varphi (x,y)</math> is <cmath>\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)</cmath> | + | Mark the total number of distinct sequences of jumps for the frog to reach the point <math>(x,y)</math> as <math>\varphi (x,y)</math>. Consider for each point <math>(x,y)</math> in the first quadrant, there are only <math>4</math> possible points in the first quadrant for frog to reach point <math>(x,y)</math>, and these <math>4</math> points are <cmath>(x-1,y); (x-2,y); (x,y-1); (x,y-2)</cmath>. As a result, the way to count <math>\varphi (x,y)</math> is <cmath>\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)</cmath> |
Also, for special cases, <cmath>\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)</cmath> | Also, for special cases, <cmath>\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)</cmath> |
Revision as of 20:30, 6 October 2018
Problem
A frog is positioned at the origin of the coordinate plane. From the point , the frog can jump to any of the points , , , or . Find the number of distinct sequences of jumps in which the frog begins at and ends at .
Solution 1
We solve this problem by working backwards. Notice, the only points the frog can be on to jump to in one move are and . This applies to any other point, thus we can work our way from to , recording down the number of ways to get to each point recursively.
, , ,
A diagram of the numbers:
5 - 20 - 71 - 207 -
3 - 10 - 32 - 84 - 207
2 - 5 - 14 - 32 - 71
1 - 2 - 5 - 10 - 20
1 - 1 - 2 - 3 - 5
Solution 2
We'll refer to the moves , , , and as , , , and , respectively. Then the possible sequences of moves that will take the frog from to are all the permutations of , , , , , , , , and . We can reduce the number of cases using symmetry.
Case 1:
There are possibilities for this case.
Case 2: or
There are possibilities for this case.
Case 3:
There are possibilities for this case.
Case 4: or
There are possibilities for this case.
Case 5: or
There are possibilities for this case.
Case 6:
There are possibilities for this case.
Adding up all these cases gives us ways.
Solution 3 (General Case)
Mark the total number of distinct sequences of jumps for the frog to reach the point as . Consider for each point in the first quadrant, there are only possible points in the first quadrant for frog to reach point , and these points are . As a result, the way to count is
Also, for special cases,
Start with , use this method and draw the figure below, we can finally get (In order to make the LaTeX thing more beautiful to look at, I put to make every number a -digits number)
So the total number of distinct sequences of jumps for the frog to reach is .
~Solution by (Frank FYC)
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.