Solution, but not sure!

Bugi, triangles are scalene (raznostranicni), right?

Sorry!

Let $ K$ have side lengths $ a \ge b \ge c$; then $ K'$ has side lengths $ d \ge a \ge b$ and the similarity condition implies $ \frac {a}{d} = \frac {b}{a} = \frac {c}{b} < 1$ which implies $ d, a, b, c$ is a geometric progression. Write its common ratio as $ \frac {r}{s}, (r, s) = 1$; then we can write $ (a, b, c) = (ks^2, ksr, kr^2)$ for integers $ k, s, r$. (Note that $ d = \frac {ks^3}{r}$ need not be an integer.) a) The integral disguisable triangle of minimal perimeter occurs when $ k = 1, r = 2, s = 3$; this gives $ 4 + 6 + 9 = 19$. (This is the valid common ratio of minimal height; $ r = 1, s = 2$ violates the triangle inequality.) b) A primitive integral disguisable triangle must have $ k = 1$ and is therefore of the form $ (s^2, sr, r^2)$.

Call a scalene triangle K disguisable if there exists a triangle $K'$ similar to $K$ with two shorter sides precisely as long as the two longer sides of $K$, respectively. Call a disguisable triangle integral if the lengths of all its sides are integers. (a) Find the side lengths of the integral disguisable triangle with the smallest possible perimeter. (b) Let $K$ be an arbitrary integral disguisable triangle for which no smaller integral disguisable triangle similar to it exists. Prove that at least two side lengths of $K$ are perfect squares.