# Four equilateral triangles equal to one equilateral triangle

Let ABC be an equilateral triangle with a height equal to a rational number. Is there a general method to obtain four unequal equilateral triangles with heights rational numbers and the sum of their surfaces equal to the surface of the ABC triangle?

There are many ways. Let the height of the given equilateral triangle be $r$, Then $4$ equilateral triangles with rational heights whose combined area is the same as the area of the given triangle have heights $$\frac{2r}{15}, \quad\frac{3r}{15},\quad\frac{4r}{15}, \quad\frac{14r}{15}.$$
There are many other ways to do it. The trick is to find $4$ distinct positive integers the sum of whose squares is a perfect square. One example of $4$ such integers is $2,3,4,14$. The sum of their squares is $15^2$.
The area of an equilateral triangle is proportional to its height, so your problem reduces to finding $a,b,c,d \in \mathbb{Q}$ such that $$a^2+b^2+c^2+d^2 = h^2$$ where $h$ is the height of the original triangle. Multiplying both sides by the denominator of $h^2$ shows that in particular Pythagorean 5-tuples solve the problem.