2020 CIME I Problems/Problem 1

Problem 1

A knight begins on the point $(0,0)$ in the coordinate plane. From any point $(x,y)$ the knight moves to either $(x+2,y+1)$ or $(x+1,y+2)$ . Find the number of ways the knight can reach the point $(15,15)$ .

Solution

Let $A$ denote a move of $2$ units north and $1$ unit east, and let $B$ denote a move of $1$ unit north and $2$ units east. To get to the point $(15,15)$ using only these moves, say $a$ moves in direction $A$ and $b$ moves in direction $B$ , we must have $2a+1b=1a+2b=15$ because both the $x$ - and $y$ -coordinates have increased by $15$ since the knight started. Solving this system of equations gives us $a=b=5$ . This means we need the knight to make $10$ moves, $5$ of which are headed in direction $A$ , and the remaining $5$ are headed in direction $B$ . Any combination of these moves work, so the answer is $\binom{10}{5}=\boxed{252}.$

Solution 2

We can draw lines using $(x+1,y+2)$ and $(x+2,y+1)$ . Calculating the lines, we see that they are from $y=2x$ and $y=\frac{1}{2}x$ respectively. The point $(15,15)$ is on the line $y=x$ , which is in the "middle" between both, since by multiplying or dividing the slope by $2$ we can get the other two lines. This means to get to $(15,15)$ , for every $(x+1,y+2)$ we do, we do one $(x+2,y+1)$ to balance it. Call this system of moves $x$ , and by performing $x$ once, we get to $(3,3)$ . If we repeat $x$ five more times, we get to $(15,15)$ . Thus this is now a word arrangement problem where we have to arrange five $X$ s and $Y$ s (name is arbitrary). We get $\frac{10!}{5!5!}=\boxed{252}.$

~ neeyakkid23

Video Solution

https://www.youtube.com/watch?v=SFVt0JYLkHY ~Shreyas S

2020 CIME I (Problems • Answer Key • Resources)
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2020 CIME I Problems/Problem 1

Contents

Problem 1

Solution

Solution 2

Video Solution

See also