Professor Halfbrain and the right-angled triangles

Question

Today I met professor Halfbrain at the tea house. The professor looked very tired, and apparently had not slept for the last couple of days. He told me that he had been spending his time with cutting right-angled triangles into smaller pieces.

• The professor had originally started out with six huge right-angled triangles: two congruent triangles with side length $9,12,15$, two congruent triangles with side length $12,16,20$, and two congruent triangles with side length $15,20,25$.
• In one cutting operation, the professor cut one of his current right-angled triangles into two smaller right-angled ones. The cut was always done along the altitude dropped onto the hypotenuse. (Note that all the resulting triangles are again right-angled.)

The professor was extremely happy to report that $-$ after many hours of cutting and cutting and cutting $-$ he had finally produced a triangle set in which there did not remain a single pair of congruent triangles.

Question: Is it indeed possible to generate a set of pairwise non-congruent triangles (according to the above rules), or has the professor once again made one of his mathematical blunders?

Answer 1

Professor Halfbrain

has made a blunder, it is not possible to generate a set of pairwise non-congruent triangles.

For integer $a$, $b\geq 0$, let $T_{a,b}$ denote the right triangle with side lengths $$ \frac{3^{a+1} 4^b}{5^{a+b-1}},\;\;\;\frac{3^a 4^{b+1}}{5^{a+b-1}}\;\;\;\frac{3^a 4^b}{5^{a+b-2}}. $$ Cutting triangle $T_{a,b}$ results in the triangles $T_{a+1,b}$ and $T_{a,b+1}$. The initial triangles are two copies each of $T_{1,0}$, $T_{0,1}$, and $T_{0,0}$, so every triangle ever produced will have the form $T_{a,b}$ for some integers $a$, $b\geq 0$.

To the triangle $T_{a,b}$, assign a weight $\frac{1}{2^{a+b}}$. Dissecting $T_{a,b}$ into $T_{a+1,b}$ and $T_{a,b+1}$ does not change the sum of the weights. At the start, the sum of the weights is $$ 2\cdot\frac{1}{2^1}+2\cdot\frac{1}{2^1}+2\cdot\frac{1}{2^0}=4. $$ If at any point the triangles are pairwise non-congruent, the sum of the weights must be less than $$ \sum_{a,b\geq 0}\frac{1}{2^{a+b}}=4. $$ So it is not possible that no pair of triangles is congruent.