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Asked by SeemaAs we need to show the case to hold for all n>0, we will proceed by induction. To start we will need to show that the case holds when n=1 and when n=2. This can be done most easily by the use of drawings which are helpful for geometric proofs. (Make note here of the fact that after we have drawn all of our red lines our counters are back in the corners) [Now that we have those two cases we notice that for a triangle of side length n+2 we have an outlet layer of triangles of side length 1 formed around a triangle of side length n. This allows us to say that if we can find a solution for the triangle of length n then all that remains is to show that there is a way to solve the outermost layer of the triangle of n+2. This forms the basis for our proof.] Take the statement to hold for some k in the natural numbers not including zero and take this solution to be such that all counters lie on the outside corners at the end of the solution. Then consider the triangle of side length k+2. Begin by moving all counters from the outermost corners or the triangle of side length k+2 to the outermost corners of the triangle of side length n, this can be shown to take place in two moves. Using our inductive hypothesis we claim that this triangle of side length k has a solution in which all counters are placed on the outermost points at the end. And this we can cover all the edges covered in this triangle of side length k by the inductive hypothesis. It now remains to cover the remaining edges, which can easily be done by "zig zagging" along the edges swapping out counters on each corner for another counter. Then by covering the edges not covered by the previous counter to be in this position we find ourselves in a location with only 1 path of possible moves which completes the triangle. By the principle of mathematical induction we say that as the case holds when n=1 that case holds for all n = 1+2a for all positive integer values of a, and similarly for when n=2. This gives us that this can be done for all triangles of side length greater than or equal to 1. []

As we need to show the case to hold for all n>0, we will proceed by induction. To start we will need to show that the case holds when n=1 and when n=2. This can be done most easily by the use of drawings which are helpful for geometric proofs. (Make note here of the fact that after we have drawn all of our red lines our counters are back in the corners) [Now that we have those two cases we notice that for a triangle of side length n+2 we have an outlet layer of triangles of side length 1 formed around a triangle of side length n. This allows us to say that if we can find a solution for the triangle of length n then all that remains is to show that there is a way to solve the outermost layer of the triangle of n+2. This forms the basis for our proof.] Take the statement to hold for some k in the natural numbers not including zero and take this solution to be such that all counters lie on the outside corners at the end of the solution. Then consider the triangle of side length k+2. Begin by moving all counters from the outermost corners or the triangle of side length k+2 to the outermost corners of the triangle of side length n, this can be shown to take place in two moves. Using our inductive hypothesis we claim that this triangle of side length k has a solution in which all counters are placed on the outermost points at the end. And this we can cover all the edges covered in this triangle of side length k by the inductive hypothesis. It now remains to cover the remaining edges, which can easily be done by "zig zagging" along the edges swapping out counters on each corner for another counter. Then by covering the edges not covered by the previous counter to be in this position we find ourselves in a location with only 1 path of possible moves which completes the triangle. By the principle of mathematical induction we say that as the case holds when n=1 that case holds for all n = 1+2a for all positive integer values of a, and similarly for when n=2. This gives us that this can be done for all triangles of side length greater than or equal to 1. []

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