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be the set of all triples 
of positive integers for which there exist triangles with side lengths 
Express ![\[\sum_{(a,b,c)\in T}\frac{2^a}{3^b5^c}\]](http://latex.artofproblemsolving.com/9/f/6/9f6e0cee87e5b6958e143ff23b887bcd876f6cba.png)
as a rational number in lowest terms.
from somewhere (so final answer of 
). Is that more of a -1 or a -9 level mistake?
noting that 
But I don't really trust my arithmetic.
.
, and consider the formal power series![\[
f(x,y,z)=\sum_{(a,b,c)\in T} x^ay^bz^c
\]](http://latex.artofproblemsolving.com/6/8/8/688643e6a301ac5934426ddc847a6093a14d0a20.png)
(so we're basically throwing convergence issues out the window.) Let 
be the set of positive triples with 
, and define 
similarly. Then
where the last equality follows after sufficient expansion. Plugging in 
gives us the answer.
, 
Everything works out absurdly nicely.
![\[\sum_{(a,b,c)\in T} \frac{3^a}{2^b 5^c}\]](http://latex.artofproblemsolving.com/5/8/c/58c693319031dbd70e41f21ebffc356c77e28929.png)
If we repeat the computations we will get that the intermediate sum 
diverges, and so the sum is going to be infinite.
can be any positive integers. The third side 
then runs between 
and 
. This lets us write the sum
.
and another with 
:
side for later, thus not seeing the two sums that can be merged. That's where factors of 
in the denominator come in; evaluating those sums without the merge leads to geometric series with ratio 
. A 
in the denominator indicates a geometric series with ratio 
somewhere, and that must be a mistake.
:clown:
for now, and we will sum it over the possible values of .
. Using the necessary triangle inequality conditions and letting 
, we get 
First let's factor out the denominator and simplify a bit as it only depends on 
which are defined before the third summation. We get 
Now notice how the inner sum is just a geometric series. We can sum it using the finite geometric series formula. The number of terms is 
. The common ratio is 
. So we can sum it to get 
. Plug this back into the summation to get 
Again, we factor out stuff to make it easier and we get 
Now this is just an infinite geometric series with first term 
and common ratio 
, which has sum 
. We can factor this out as its just a constant. So our sum is currently 
Now this is a simple sum which we can break apart to solve. Doing that, we get 
The first sum is an infinite geo series with first term 
and common ratio 
, so it has sum 
. The second sums to 
. Now multiplying and adding, we get 
. Using the necessary triangle inequality conditions and letting 
, we get 
First let's factor out the denominator and simplify a bit as it only depends on 
which are defined before the third summation. We get 
Now notice how the inner sum is just a geometric series. We can sum it using the finite geometric series formula. The number of terms is 
. The common ratio is . So we can sum it to get 
. Plug this back into the summation to get 
Again, we factor out stuff to make it easier and we get 
Now this is just an infinite geometric series with first term 
and common ratio 
, which has sum . We can factor this out as its just a constant. So our sum is currently 
Now this is a simple sum which we can break apart to solve. Doing that, we get 
The first sum is an infinite geo series with first term and common ratio , so it has sum . The second sums to . Now multiplying and adding, we get 
. Now the easy thing about this case is that there is no minimum. But there is still a maximum for . We can also just consider one variable 
which is 
and both. So our sum is a simple 
Using a similar method, this reduces to 
which by finite geo series is 
We break this apart to get 
Using the infinite geometric series for each sum, we get 
.
are all integers or they are rational numbers with denominator 2 in lowest terms. For the first case, for instance, we just want 
Evaluate the sum layer by layer for both cases to get 
is the answer.
and , the triple is in if and only if satisfies 
(triangle inequality). So we can rewrite the sum as![\[S=\sum_{(a,b,c)\in T}\frac{2^a}{3^b5^c}=\sum_{b\geq 1}\sum_{c\geq 1}\left(\frac{1}{3^b 5^c} \sum_{\substack{a \leq b+c-1 \\ a\geq |b-c|+1}} 2^a\right)=\sum_{b\geq 1}\sum_{c\geq 1}\left(\frac{1}{3^b 5^c}\left(2^{b+c}-2^{|b-c|+1}\right)\right).\]](http://latex.artofproblemsolving.com/0/d/d/0dd0adf3439b6bab00aa48226c59b20979293e26.png)
We can split this up:![\[S=\sum_{b\geq 1}\sum_{c\geq 1}\left(\frac23\right)^b\left(\frac25\right)^c-\sum_{b\geq 1}\sum_{c\geq 1}\frac{2^{|b-c|+1}}{3^b5^c}.\]](http://latex.artofproblemsolving.com/7/f/6/7f68c3a6d75788690f5e6c6b09a19742f34bc4ed.png)
The first term on the RHS can be written as 
.
or 
:![\[\sum_{b\geq 1}\sum_{c\geq 1}\frac{2^{|b-c|+1}}{3^b5^c}=-2\sum_{b\geq 1}\left(3^{-b}\left(\sum_{c\geq b}\frac{2^{c-b}}{5^c}+\sum_{c=1}^{b-1}\frac{2^{b-c}}{5^c}\right)\right).\]](http://latex.artofproblemsolving.com/1/1/9/119a0ab49ea7446fbacc65c5c9902255d19a31f9.png)
If we set 
, then 
. Similarly, the second term 
is a geometric series that evaluates to 
. Substituting these both in, we are left with a summation in just one variable . Evaluating the resulting combination of geometric series, we find that it is 
.
.
for 
" if 
.
are not dependent on each other..so we can split them like we do for finite sums...it is a different point that the splitting is valid as every infinite GP converges...
, where 
, such that 
. Note that are either all odd or all even.
.
are the side lengths of a triangle, 
where we set 
and 
respectively.
constant. Then we have
Hold 
constant. Then we have
Which is equal to
we decompose into
which is just a product of easily-computable geometric series. I really don't know how I missed this T.T
and cyclic permutations, for positive integers 
be the sum![\[\sum_{x,y,z}\frac{2^{\frac{x+y}2}}{3^{\frac{y+z}2}5^{\frac{z+x}2}}\]](http://latex.artofproblemsolving.com/c/5/5/c5562040f54468530c329a1123b673811cdc2a28.png)
be the same sum when ![\[S_1= \left(\sum_{x\text{ odd}} \sqrt{\frac25}^x\right)\left(\sum_{y\text{ odd}} \sqrt{\frac23}^y\right)\left(\sum_{z\text{ odd}} \sqrt{\frac1{15}}^z\right)=\left(\frac{\sqrt{\frac25}}{1-\frac25}\right)\left(\frac{\sqrt{\frac23}}{1-\frac23}\right)\left(\frac{\sqrt{\frac1{15}}}{1-\frac1{15}}\right)=\frac57\]](http://latex.artofproblemsolving.com/a/9/3/a930b16d75caeb22425c2b61384e5f378e0d3cb6.png)
![\[S_2=\left(\sum_{x\text{ even}} \sqrt{\frac25}^x\right)\left(\sum_{y\text{ even}} \sqrt{\frac23}^y\right)\left(\sum_{z\text{ even}} \sqrt{\frac1{15}}^z\right)=\left(\frac{\frac25}{1-\frac25}\right)\left(\frac{\frac23}{1-\frac23}\right)\left(\frac{\frac1{15}}{1-\frac1{15}}\right)=\frac2{21}\]](http://latex.artofproblemsolving.com/1/0/d/10d2bfebc21fda10d87c9693c8d833d6f7111493.png)
![\[S_1 + S_2 = \boxed{\frac{17}{21}}.\]](http://latex.artofproblemsolving.com/1/7/2/172ad25844853d410db62a6708683bc7b5879cc9.png)
,
.
.
and splitting into our two cases, we find that the total sum is equal to
![\[\displaystyle\sum_{x, y, z}{\frac{2^{x+y}}{3^{y+z}5^{x+z}}} = \displaystyle\sum_{x,y,z}{\biggl(\frac{2}{5}\biggr)^{x}\biggl(\frac{2}{3}\biggr)^{y}\biggl(\frac{1}{15}\biggr)^{z}}\]](http://latex.artofproblemsolving.com/0/5/e/05e27bcd1cfda8ba1a643807a97d59181578999c.png)
We have that 
or 
.![\[\frac{\frac{2}{5}\cdot\frac{2}{3}\cdot\frac{1}{15}}{(1-\frac{2}{5})(1-\frac{2}{3})(1-\frac{1}{15})} = \frac{2}{21}\]](http://latex.artofproblemsolving.com/9/5/0/9509520f5e9f4e1f5794f48cf89270054b768ef2.png)
If ![\[\frac{\sqrt{\frac{4}{225}}}{(1-\frac{2}{5})(1-\frac{2}{3})(1-\frac{1}{15})} = \frac{5}{7}\]](http://latex.artofproblemsolving.com/3/9/a/39a9d25b9d06712e723723bc30cc99f0d57dbe9c.png)
Adding these up, we get 
.