Square with the minimum possible area that satisfy the given conditions

Question

We're given a square with integer side lenght, it is divided into 4 triangles such that all the triangle's side lenghts are integers and that no two triangles are congruent, what is the minimum possible area of the square? (picture linked below)

I tried working a bit with the two right triangles, trying to find pythagorean triples that would do the trick (since I also tried drawing the altitude of the 2 "middle" triangles) but I'm not really sure how to tackle this type of problem...

Answer 1

Since somehow you already knew that the answer is $576$ (side length = $24$), let us see how we can find this answer.

Let us first look at the following graph:

By your conditions, we know that $(a_1, x, b_1), (a_2, x, b_2)$ and $(a_3, x, b_3)$ form Pythagorean triples, and $a_1 + a_2 + a_3 = x$. There is an additional condition that the four solid line triangles are pairwise non-congruent, but just forget it for now.

So the core of the problem is to find Pythagorean triangles with a common side $x$. Let us try a few ideas to do this:

First. most naively you can just write down all Pythagorean triangles, starting from smallest side lengths:

$$ (3,4,5), (6,8,10), (5,12,13), (9,12,15), (8,15,17), (12,16,20), (15,20,25), (20,21,29), \cdots $$

This is actually doable. Since we know that all Pythagorean triangles have the form $(a(m^2-n^2), a\cdot2mn, a(m^2+n^2))$, and indeed we know by cheating that the answer is at $24$, it will not take too much time to get to the right answer. If I were you I might consider writing a program for this, if the solution is actually large.

Second, here is a way to increase the speed for your searching. Let $a = \min\{a_1, a_2, a_3\}$. Then clearly $x > 3a_1$. This is indeed a nice restriction. From the general formula

$$ (a(m^2-n^2), a\cdot 2mn, a(m^2+n^2)), \text{ $\gcd(m,n) = 1$ and exactly one of $m$ and $n$ is even} $$

you need either (i) $m^2-n^2 > 6mn$; or (ii) $3(m^2-n^2) < 2mn$.

1. If $m^2-n^2 > 6mn$, then $m^2-6mn-n^2 > 0$ or $m > (3 + \sqrt{10})n \approx 6.16n$. The smallest Pythagorean triangles starts from $(16,65,67), (20,99,101),\cdots$.
2. If $3(m^2-n^2) < 2mn$, then $n < m < n(1+\sqrt{10})/3 \approx 1.42n$. The smallest Pythagorean triangles starts from $(7,24,25), (9,40,41), (14,48,50),\cdots$.

This should shrink our search space significantly. Anyway the minimal answer happens at $7+7+10=24$.