AMC 12 B, 2024

Let the smallest angle of the $3$-$4$-$5$ triangle have measure $\alpha$, the smallest angle of the $5$-$12$-$13$ triangle have measure $\beta$, and the smallest angle of the third triangle have measure $\gamma$. It is given that $\alpha + \beta + \gamma = 90^\circ$, so $\cos(\alpha + \beta + \gamma) = 0$. Expanding gives
$$\begin{aligned} \cos(\alpha + \beta + \gamma) &= \cos(\alpha + \beta) \cos \gamma - \sin(\alpha + \beta) \sin \gamma \\ &= (\cos \alpha \cos \beta - \sin \alpha \sin \beta) \cos \gamma - (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \sin \gamma \\ &= \cos \alpha \cos \beta \cos \gamma - \sin \alpha \sin \beta \cos \gamma - \sin \alpha \cos \beta \sin \gamma - \cos \alpha \sin \beta \sin \gamma \\ &= 0. \end{aligned}$$
Because $\sin \alpha = \frac{3}{5}$, $\cos \alpha = \frac{4}{5}$, $\sin \beta = \frac{5}{13}$, and $\cos \beta = \frac{12}{13}$,
$$\begin{aligned} \cos(\alpha + \beta + \gamma) &= \frac{4}{5} \cdot \frac{12}{13} \cos \gamma - \frac{3}{5} \cdot \frac{5}{13} \cos \gamma - \frac{3}{5} \cdot \frac{12}{13} \sin \gamma - \frac{4}{5} \cdot \frac{5}{13} \sin \gamma \\ &= \frac{48}{65} \cos \gamma - \frac{15}{65} \cos \gamma - \frac{36}{65} \sin \gamma - \frac{20}{65} \sin \gamma \\ &= \frac{33}{65} \cos \gamma - \frac{56}{65} \sin \gamma \\ &= 0. \end{aligned}$$
Therefore $33 \cos \gamma = 56 \sin \gamma$, so $\tan \gamma = \frac{33}{56}$. Because $33$ and $56$ are relatively prime, perhaps they are the lengths of the two legs of the third triangle. Indeed, $33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2$, and the perimeter of the triangle is $33 + 56 + 65 = 154$.

Let $\alpha$, $\beta$, and $\gamma$ be as in the first solution. Using complex numbers in polar form, note that $4 + 3i = 5 \text{ cis } \alpha$. Similarly, $12 + 5i = 13 \text{ cis } \beta$. Multiplying these two quantities yields
$$(4 + 3i)(12 + 5i) = 33 + 56i = 65 \text{ cis } \left(\frac{\pi}{2} - \gamma\right),$$
where the triangle with sides of length $33$, $56$, and $65$ has angle $\gamma$ opposite the side of length $33$ and angle $\frac{\pi}{2} - \gamma$ opposite the side of length $56$. Notice that
$$\begin{aligned} (5 \operatorname{cis} \alpha) \cdot (13 \operatorname{cis} \beta) \cdot (65 \operatorname{cis} \gamma) &= (4 + 3i)(12 + 5i)(56 + 33i) \\ &= 65^2 i \\ &= 65^2 \operatorname{cis}(\alpha + \beta + \gamma) \\ &= 65^2 \operatorname{cis} \frac{\pi}{2}. \end{aligned}$$
The third triangle has sides of length $33$, $56$, and $65$, and $33 + 56 + 65 = 154$.