We have three congruent, right-angled triangular pieces of paper. Any of these can be cut in half along the altitude. From the resulting triangles, we can again cut one of them in half along the altitude, and so on. What can the ratio of the legs be if we know that during the process, we can always find two congruent triangles?
We will prove that the ratio can only be $\lambda_{1}=1$ and $\lambda_{2}=\sqrt{\frac{\sqrt{5}+1}{2}}$. Here, $\lambda_{1}$ is the ratio of the legs of an isosceles right triangle, and $\lambda_{2}$ is the only one that, when cut as shown in Figure 1, results in the 1st and 3rd pieces being congruent. (This can be simply calculated.)

First, we will show that when the ratio is $\lambda_{1}$ or $\lambda_{2}$, there will always be two congruent pieces after every cut.

If the ratio is $\lambda_{1}$, then every cut divides the triangle into two congruent isosceles right triangles, so the final cut will always result in two congruent triangles.

If the ratio is $\lambda_{2}$, then suppose it is possible to cut in such a way that all pieces at the end are different. Choose the shortest (with the fewest cuts) such sequence of cuts! Note that the order of the cuts does not matter.

Of the three original triangles, two are definitely cut in half. Consider the two larger pieces formed by the two cuts. One of these must definitely be cut in half. Thus, there are three congruent, similar triangles to the original (Figure 2). Forget the other parts and continue the planned cuts only on these. According to our assumption, all pieces will be different at the end of the process. Therefore, fewer cuts were sufficient on the three smaller pieces, which is a contradiction, since we also assumed that we started from a minimal sequence of cuts.

Now we will show that if the ratio of the legs differs from $\lambda_{1}$ and $\lambda_{2}$, then among the eight similar triangles obtained after the cuts shown in Figure 3, there are no congruent ones. For brevity, in the following, $H_{i}$ will simultaneously denote the triangle numbered $i$ on the diagram, its area, and the set of points of the triangle.

If two triangles share a side on the diagram, their area ratio can be determined: the smaller one is where this common side is the hypotenuse, or where this common side is the larger leg, since our triangles are similar.

Thus, we get the following chain of inequalities:

$$
H_{1}>H_{2} \cong H_{4} \cup H_{5}>H_{4}>H_{5} \cong H_{7} \cup H_{8}>H_{8}>H_{7}
$$

that is,

$$
H_{1}>H_{2}>H_{4}>H_{5}>H_{8}>H_{7} .
$$

Only two triangles could not be classified, these are $H_{3}$ and $H_{6}$. Since $H_{2}>H_{3} \cong H_{6} \cup H_{7} \cup H_{8}>H_{5}$, it follows that $H_{2}>H_{3}>$ $H_{5}$ and similarly, $H_{5} \cong H_{7} \cup H_{8}>H_{6}>H_{7}$, so $H_{5}>H_{6}>H_{7}$.

According to inequality (1), only $H_{3} \cong H_{4}$ and $H_{6} \cong H_{8}$ would be possible, but both would mean that the ratio of the legs is $\lambda_{2}$.

This completes the proof.
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$a<b$
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