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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?

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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 $15^2$.

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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.

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