Is it possible to choose six triples of lengths from $1,2,3,\ldots, 20$ to form six triangles with equal perimeters?

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Rachael has 20 thin rods whose lengths, in centimeters, are $1, 2, 3, \ldots, 20$. Any two rods can be connected at their ends. Rachael selects three rods to make a triangle, then three other rods to make a second triangle, and so on.

However, Rachael wants to use 18 rods to form a set of six triangles with equal perimeters. Either find such a set, or explain why no such set exists.

So here are my thoughts on this problem:
I know that the $1\mathrm cm$ rod is unusable because of triangle inequality. Now I am remained with $2 \mathrm cm, 3 \mathrm cm, 4 \mathrm cm, \ldots, 20 \mathrm cm$ rods, but this is still 19 rods. So my initial thought is to go through all the different possibilities and see if such set exist but I feel like this will be too time consuming. I am just wondering if there are a more time efficient method to determining if such set exist.

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The sum of the integers from $1$ to $20$ is $210$, and $210/6=35$. Since we only use $18$ of the $20$ rods, the perimeter of the triangles must be less than $35$, so it's $34$ at most. We discard $2$ of the rods, but we must use one of length at least $18$, so there is a triangle whose longest side is at least $18$ and whose perimeter is at most $34$. This is absurd, so the problem is impossible.

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The sum of the 19 rod lengths is 209. Even if we throw out the rod of length 20 (to make a set of 18 rods), the perimeter must be at least 2(19)+1 = 39. Six triangles with that perimeter would require a total rod length of 6(3) = 234. The total length of rods is insufficient to make the required triangles.