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In this question, I was able to show how some Pythagorean triples can be divided into two smaller Pythagorean triples by the perpendicular from the hypotenuse to the right angle... as shown in the example here.

enter image description here

I found a number of these $\,A,B,C,D,J,K\,$ sets generated by this formula and shown in the table below.

\begin{align*} &A=(2n-1+k)^2-k^2&&=(2n-1)^2+2(2n-1)k\\ &B=2(2n-1+k)k &&=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1+k)^2+k^2 &&=(2n-1)^2+2(2n-1)k+2k^2 \end{align*}

$$\begin{array}{c|c|c|c|c|c|c|c|c|} (n,k) & A & B & C & D & GCD & J & K \\ \hline (3,5) & 75 & 100 & 125 & 60 & 25 & 45 & 80 \\ \hline (7,15) & 175 & 600 & 625 & 168 & 25 & 49 & 576 \\ \hline (7,26) & 845 & 2028 & 2197 & 780 & 169 & 325 & 1872 \\ \hline (8,15) & 675 & 900 & 1125 & 549 & 225 & 405 & 720 \\ \hline (8,45) & 1575 & 5400 & 5625 & 1512 & 225 & 441 & 5185 \\ \hline (13,25) & 1875 & 2500 & 3125 & 1500 & 625 & 1125 & 2000 \\ \hline (13,75) & 4375 & 15000 & 15625 & 4200 & 625 & 1225 & 14400 \\ \hline (15,580) & 34481 & 706440 & 707281 & 34440 & 841 & 1681 & 705600 \\ \hline (18,35) & 3675 & 4900 & 6125 & 2940 & 1225 & 2205 & 3820 \\ \hline (18,105) & 8575 & 29400 & 30625 & 8232 & 1225 & 2401 & 28224 \\ \hline (20,78) & 7605 & 18252 & 19773 & 7020 & 1521 & 2925 & 16848 \\ \hline \end{array}$$

These appear to be very rare. Have I missed any for $\,n\le20,\,$ and is there a more efficient way to find them than a spreadsheet full of formulas that I used?

poetasis
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    Note that in dividing a right triangle by a perpendicular from the hypotenuse, each resulting triangle has one apex angle and a right angle in common with the original triangle, so all three are similar. Hence, their sides (if constrained to be integers) must be multiples of the sides of some basic right triangle. – Keith Backman Dec 11 '24 at 16:05

1 Answers1

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Here's a way to generate more examples, several of which are "small" by some metric.

Given any triple $(a, b, c)$, scaling it by $c$ produces the triple $(ca, cb, c^2)$, so the hypotenuse can be decomposed as $c^2 = a^2 + b^2$. The two subtriangles are now similar via scaling the original triangle by $a$ and $b$, respectively: $(a^2, ab, ac)$ and $(ba, b^2, bc)$. Notice that these triangles share the altitude of length $ba$.

Here's what this construction produces for some primitive triples that are smaller than your give examples: $$ \begin{array}{c|ccc} (a, b, c) & (\color{red}{a^2}, \color{orange}{ab}, \color{blue}{ac}) & (\color{orange}{ba}, \color{red}{b^2}, \color{green}{bc}) & (\color{blue}{ca}, \color{green}{cb}, \color{red}{c^2}) \\ \hline (3, 4, 5) & (9, 12, 15) & (12, 16, 20) & (15, 20, 25) \\ (5, 12, 13) & (25, 60, 65) & (60, 144, 156) & (65, 156, 169) \\ (8, 15, 17) & (64, 120, 136) & (120, 225, 255) & (136, 255, 289) \\ (7, 24, 25) & (49, 168, 175) & (168, 576, 600) & (175, 600, 625) \end{array} $$

This is not an exhaustive list, and you can see that the final example replicates one of yours.

Sammy Black
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  • I loved your analysis but I was wondering. Is there any way that two primitive triples could form a larger one? It doesn't look like it. – poetasis Dec 13 '24 at 00:28
  • Well, the three triples are generated by scaling, so by definition they cannot be primitive. However, we could ask if, for example, the big triangle $(15, 20, 25)$ is primitive in this divisible-into-two-subtriangles sense. It's pretty clear that this triple is; whereas, your the similar triple $(75, 100, 125)$ is not, since it is evidently $5$ times bigger than necessary. – Sammy Black Dec 13 '24 at 17:18
  • Neither $,(15,20,25),$ nor it's smaller triples are primitive because they all have $,GCD>1.,$ I am looking for an imprimitive made of $,2,$ primitives. – poetasis Dec 13 '24 at 18:23