Let \( AB = a - d, BC = a, CA = a + d \), then
according to the problem, we have \(\left\{\begin{array}{l}0a + b, \\ (a + d)^{2} > a^{2} + (a - d)^{2} .\end{array}\right.\)
Simplifying, we get \(\left\{\begin{array}{l}00) .
\[
(1) If \(4d - 1 \leq 50 - d\). That is, \(1 \leq d \leq 10\), then condition \((*)\) is
\[
2d + 1 \leq a \leq 4d - 1 (d > 0),
\]

For a fixed \(d\), \(a\) has \(2d - 1\) possible values.
(2) If \(4d - 1 > 50 - d\), that is, \(d \geq 11\), then condition \((*)\) is
\[
2d + 1 \leq a \leq 50 - d
\]

For a fixed \(d\), \(a\) has \(50 - 3d\) possible values.
Therefore, the total number of triangles is
\[
\begin{aligned}
& \sum_{d=1}^{10}(2d - 1) + \sum_{d=11}^{50-3d>0}(50 - 3d) = 100 + \sum_{d=11}^{16}(50 - 3d) \\
= & 100 + 50 \times 6 - 3 \times \frac{6(11 + 16)}{2} = 157 \text{ triangles.}
\end{aligned}
\]
When \(d = 10\), the perimeter of the triangle \(3a\) reaches its maximum value \(3 \times 39 = 117\).
At this time, the lengths of the three sides of the triangle are \(29, 39, 49\) or \(28, 39, 50\).
Four, it is known that the obtuse triangle $\triangle A B C$ satisfies the following conditions:
(1) The lengths of $A B, B C, C A$ are all positive integers;
(2) The lengths of $A B, B C, C A$ are all no more than 50;
(3) The lengths of $A B, B C, C A$ form an arithmetic sequence with a positive common difference.

Find the number of obtuse triangles that satisfy conditions (1), (2), and (3), and determine the side lengths of the triangle with the largest perimeter among them.