# Let $t_n$ denote the no. of integral sided triangles with distinct sides chosen from $\{1,2,3,\ldots,n\}$. Then find $t_{20}-t_{19}$.

Let $t_n$ denote the no. of integral sided triangles with distinct sides chosen from $\{1,2,3,\ldots,n\}$. Then find $t_{20}-t_{19}$.

My Approach:
$t_{19}$ is the no. of triangles formed by sides from set $\{1,2,3,\ldots,19\}$. When 20 is added to this set, the no. of sides triangles formed by sides from set $\{1,2,3,\ldots,19,20\}$ is given by $t_{20}$. Now the expression $t_{20}-t_{19}$ denotes the increase in the no. of triangles which are basically the triangles containing 20 as one side and the other two sides are from the set $\{1,2,3,\ldots,19\}$. We also know that the sum of the two sides is always greater than the third side. Hence for two sides $a,b$ $a+b>20$.

Problem:
I cannot figure out how to calculate the no. of solutions to this inequality from the given set.

If the second side has length $19$, the third side must have length greater than $1$ but less than $19$, or $17$ different possibilities. Every time you decrease the length of the second largest side, the possibilities for the length of the third side decrease by $2$. Can you complete it from here?

• Yeah, I can proceed forward. Thank you! – Osheen Sachdev Oct 28 '16 at 6:14