Let $ABC$ be a arbitrary triangle with integral sides such that the perimeter is $2006$. And one of the side $16$ times the other side. How many such triangles exist?


enter image description here

  • 2
    $\begingroup$ Note that the length of the longest side of a triangle must be less than the sum of the lengths of the two shorter sides. This gives the additional requirements $15 x < y < 17 x$. $\endgroup$ Jan 15 '15 at 2:15
  • $\begingroup$ Note that $2006=(118)(17)$, So all integer solutions of $17x+y=2006$ have shape $y=17t$, $x=118-t$. Now we need to pick out the $t$ that give us a real triangle. $\endgroup$ Jan 15 '15 at 2:31
  • $\begingroup$ Be careful, a straight line is not a triangle :-) $\endgroup$
    – Joffan
    Jan 15 '15 at 2:44

I'll offer some hints.

You've got the basic relationship:

$$17x + y = 2006,$$

where $x$ is the shortest side and $y$ is the side not constrained to be $16$ times the shortest side, $16x$.

Now consider this. What other relation constrains the sides? Hint: $(1, 16, 1989)$ aren't valid sides for a triangle, even though they add up to $2006$ and have one side sixteen times another. Why not?

This should make the number of triples manageable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.