15 As shown in the figure, it is known that the diameter of the circumcircle of $\triangle ABC$ is 25, the lengths of the three sides are all positive integers, the distances from the circumcenter $O$ to sides $AB$ and $BC$ are also positive integers, and $AB > BC$. Find the lengths of the three sides of $\triangle ABC$.
15 As shown in the figure, draw perpendiculars from the center $O$ to $AB$ and $BC$, with the feet of the perpendiculars being $D$ and $E$. Let $AB=a$, $BC=b$, $OD=d$, and $OE=e$. Then $BD=\frac{a}{2}$, $BE=\frac{b}{2}$, where $a$, $b$, $d$, and $e$ are all positive integers, and $a > b$. Since $DB^2 + OD^2 = OB^2$, we have $a^2 + (2d)^2 = 25^2$. Similarly, we get $b^2 + (2e)^2 = 25^2$. Consider the indeterminate equation $x^2 + (2y)^2 = 25^2$, which yields two sets of positive integer solutions $(x, y) = (15, 10), (7, 12)$. Given $a > b$, we have $a = 15$, $b = 7$, so $AB = 15$, $BC = 7$, and $OD = 10$, $OE = 12$. Since $OD \perp AB$ and $OE \perp BC$, points $O$, $D$, $B$, and $E$ are concyclic. By Ptolemy's theorem, we know
$$
DE \cdot OB = OD \cdot BE + OE \cdot DB,
$$

which gives
$$
DE = \frac{OD \cdot BE + OE \cdot DB}{OB} = 10.
$$

Since $D$ and $E$ are the midpoints of $AB$ and $BC$ respectively, $DE$ is the midline of $\triangle ABC$, so $AC = 20$. Therefore, the lengths of the three sides of the triangle are $15$, $7$, and $20$.
(Problem 15)