To find the sum of all possible perimeters of a triangle

Question

Here's the exact question:

Consider a triangle $ACD$. Point $B$ is on $AC$ with $AB = 9$ and $BC = 21$. Also $AD = CD$ , and $AD$ and $BD$ are integers. Let $S$ be the sum of all possible perimeters of $\triangle$$ACD$. Find $S$.

So I tried making $BD$ and taking it as $x$ and then putting on inequalities related to the very famous, artistic statement:

The sum of any two sides of a triangle is greater than its third side

The only thing I found useful that they were integer sides...?
And I am kind of stuck after that...... I was told to use the Stewart's theorem, but I know it can be solved without using it too..... Anyways the answer is $380$ :)

Any suggestions will greatly help!! Also, any formatting edits/changes in my question are also welcomed!! Thank you

PS: I don't think this is a duplicate, I did not find any similar question to this

Answer 1

Hints for a solution without using Stewart's theorem. If you're stuck, state what you've tried:

Let $E$ be the foot of the perpendicular from $D$ to $AC$.
Find a formula relating $DB$ to $DA$.

Express $DE^2$ using Pythagoreas theorem in 2 ways.

Then, solve the diophantine equation (since all of these are integers).

Remember that $DA$ must be "large enough" ($>15$ as stated in the comments), so some cases might need to be rejected.

Hence conclude that the sum of perimeters is indeed 380.

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Note:

• Both approaches lead us to the same formula (which I did not state, and have repeatedly asked OP to state it).
  • Stewarts theorem gives the formula as a direct application knowing the various sides.
  • This approach is essentially follows one proof of Stewart's theorem using Pythagoreas theorem, so I don't consider them distinct approaches.