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Difference between revisions of "2014 UMO Problems/Problem 6"

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== Solution ==
 
== Solution ==
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Consider just the equilateral triangle of length <math>n</math> made by cutting the parallelogram in half. In how many ways can you get from the SW vertex to any point on the opposite edge? Call this value <math>T_n</math> (in fact, it doesn't matter which edgepoint you go to - there are the same number of paths). Note that <math>T_{n} = 2n\cdot T_{n-1}</math>. Why? Consider the equilateral triangle of length <math>n-1</math>. Note that for each edgepoint in the <math>n</math> triangle, we can get there in <math>2</math> ways from each of the <math>n</math> edgepoints in the <math>n-1</math> triangle. Now it isn't hard to see that the general formula for <math>T</math> is <math>T_n = 2^nn!</math>.
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 +
Now to find the number of ways to get from <math>A</math> to <math>B</math>. It's the sum of the number of ways to get from <math>B</math> to each point along the diagonal, times <math>T_n</math>. But the former is just <math>2n\cdot T_{n-1} = T_n</math>. Therefore, the total number of paths is <math>T_n^2</math>, which is a square number.
  
 
== See Also ==
 
== See Also ==

Revision as of 16:04, 12 February 2020

Problem

Draw $n$ rows of $2n$ equilateral triangles each, stacked on top of each other in a diamond shape, as shown below when $n = 3$. Set point $A$ as the southwest corner and point $B$ as the northeast corner. A step consists of moving from one point to an adjacent point along a drawn line segment, in one of the four legal directions indicated. A path is a series of steps, starting at $A$ and ending at $B$, such that no line segment is used twice. One path is drawn below. Prove that for every positive integer $n$, the number of distinct paths is a perfect square. (Note: A perfect square is a number of the form $k^2$, where $k$ is an integer).

[asy] size(200); path p1=polygon(3),p2=shift(sqrt(3)/2,1/2)*rotate(180)*polygon(3); for(int k=0;k<3;++k) {for (int j=0;j<3;++j) {draw(shift(sqrt(3)*(k+j/2),1.5*j)*p1,black+linewidth(.4)); draw(shift(sqrt(3)*(k+j/2),1.5*j)*p2,black+linewidth(.4));} } pair O=(-sqrt(3)/2,-1/2),E=(sqrt(3),0),NW=(-sqrt(3)/2,3/2),NE=(sqrt(3)/2,3/2),SE=(sqrt(3)/2,-3/2); D(O--(O+E)--(O+E+NE)--(O+2E+NE)--(O+2E+NE+SE)--(O+2E+2NE+SE)--(O+2E+2NE+SE+NW)--(O+3E+2NE+SE+NW)--(O+3E+3NE+SE+NW),black+linewidth(2)); MP("A",O,SW); MP("B",(O+3E+3NE+SE+NW),NE); draw(shift(10,0)*p1,black+linewidth(.4)); draw(shift(10,sqrt(3))*p1,black+linewidth(.4)); draw(shift(12.5,0)*p1,black+linewidth(.4)); draw(shift(12.5,sqrt(3))*p1,black+linewidth(.4)); MP("Legal Steps",(11.4,4.2)); draw((10-sqrt(3)/2,-1/2)--(10,1.5-1/2),arrow=Arrow()); draw((12.5-sqrt(3)/2,-1/2)--(12.5+sqrt(3)/2,-1/2),arrow=Arrow()); draw((10,sqrt(3)+1)--(10+sqrt(3)/2,sqrt(3)-1/2),arrow=Arrow()); draw((12.5+sqrt(3)/2,sqrt(3)-1/2)--(12.5,sqrt(3)+1),arrow=Arrow()); [/asy]

Solution

Consider just the equilateral triangle of length $n$ made by cutting the parallelogram in half. In how many ways can you get from the SW vertex to any point on the opposite edge? Call this value $T_n$ (in fact, it doesn't matter which edgepoint you go to - there are the same number of paths). Note that $T_{n} = 2n\cdot T_{n-1}$. Why? Consider the equilateral triangle of length $n-1$. Note that for each edgepoint in the $n$ triangle, we can get there in $2$ ways from each of the $n$ edgepoints in the $n-1$ triangle. Now it isn't hard to see that the general formula for $T$ is $T_n = 2^nn!$.

Now to find the number of ways to get from $A$ to $B$. It's the sum of the number of ways to get from $B$ to each point along the diagonal, times $T_n$. But the former is just $2n\cdot T_{n-1} = T_n$. Therefore, the total number of paths is $T_n^2$, which is a square number.

See Also

2014 UMO (Problems • Answer Key • Resources)
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Problem 5 		Followed by
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All UMO Problems and Solutions
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