AMC12 2025 B

Let the angles between the $8$ and $9$ sides be $\theta_1$ and $\theta_2$. Relating the two triangles' areas, we get \[\frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_1} = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_2}\] \[\sin{\theta_1} = \sin{\theta_2}\] For this to be true, $\theta_1 + \theta_2 = 180.$ We will now apply Law of Cosines to get an expression for $s_1$ and $s_2$ in terms of $\theta_1$ and $\theta_2$. \[s_1^2 = 8^2 + 9^2 - (2)(8)(9)\cos(\theta_1)\] \[s_2^2 = 8^2 + 9^2 - (2)(8)(9)\cos(180-\theta_1).\] Noting that $\cos{\theta_1} = -\cos{(180 - \theta_1)}$, we can add the two equations to get that \[s_1^2 + s_2^2 = 290.\] We now need two perfect squares that add to $290$. After some analysis, we find $121 + 169 = 290$. Therefore, $s_1 = 11$ and $s_2 = 13$, so our answer is $11 + 13 = \boxed{24}.$ Note that the $\cos{\theta_1} = 1/6.$ Also, note that $289 + 1 = 290$, but $17$ and $1$ are not valid side lengths due to the Triangle Inequality.