Difference between revisions of "2020 CIME I Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
We can draw lines using <math>(x+1,y+2)</math> and <math>(x+2,y+1)</math>. Calculating the lines, we see that they are from <math>y=2x</math> and <math>y=\frac{1}{2}x</math> respectively. The point <math>(15,15)</math> is on the line <math>y=x</math>, which is in the "middle" between both, since by multiplying or dividing the slope by <math>2</math> we can get the other two lines. This means to get to <math>(15,15)</math>, for every <math>(x+1,y+2)</math> we do, we do one <math>(x+2,y+1)</math> to balance it. Call this system of moves <math>x</math>, and by performing <math>x</math> once, we get to <math>(3,3)</math>. | We can draw lines using <math>(x+1,y+2)</math> and <math>(x+2,y+1)</math>. Calculating the lines, we see that they are from <math>y=2x</math> and <math>y=\frac{1}{2}x</math> respectively. The point <math>(15,15)</math> is on the line <math>y=x</math>, which is in the "middle" between both, since by multiplying or dividing the slope by <math>2</math> we can get the other two lines. This means to get to <math>(15,15)</math>, for every <math>(x+1,y+2)</math> we do, we do one <math>(x+2,y+1)</math> to balance it. Call this system of moves <math>x</math>, and by performing <math>x</math> once, we get to <math>(3,3)</math>. | ||
− | If we repeat <math>x</math> five more times, we get to <math>(15,15)</math>. Thus this is now a word arrangement problem where we have to arrange five <math> | + | If we repeat <math>x</math> five more times, we get to <math>(15,15)</math>. Thus this is now a word arrangement problem where we have to arrange five <math>A</math>s and <math>B</math>s which represent each move (letter name is arbitrary). We get <math>\frac{10!}{5!5!}=\boxed{252}.</math> |
~ neeyakkid23 | ~ neeyakkid23 |
Latest revision as of 16:40, 2 October 2024
Problem 1
A knight begins on the point in the coordinate plane. From any point the knight moves to either or . Find the number of ways the knight can reach the point .
Solution
Let denote a move of units north and unit east, and let denote a move of unit north and units east. To get to the point using only these moves, say moves in direction and moves in direction , we must have because both the - and -coordinates have increased by since the knight started. Solving this system of equations gives us . This means we need the knight to make moves, of which are headed in direction , and the remaining are headed in direction . Any combination of these moves work, so the answer is
Solution 2
We can draw lines using and . Calculating the lines, we see that they are from and respectively. The point is on the line , which is in the "middle" between both, since by multiplying or dividing the slope by we can get the other two lines. This means to get to , for every we do, we do one to balance it. Call this system of moves , and by performing once, we get to . If we repeat five more times, we get to . Thus this is now a word arrangement problem where we have to arrange five s and s which represent each move (letter name is arbitrary). We get
~ neeyakkid23
Video Solution
https://www.youtube.com/watch?v=SFVt0JYLkHY ~Shreyas S
See also
2020 CIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All CIME Problems and Solutions |
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.