In an equilateral triangle, lines are drawn from each vertex to the opposite side. Can there be seven regions of integer area? In an equilateral triangle, lines are drawn from each vertex to the opposite side. Can there be seven regions of integer area?

If we did not require that the starting triangle is equilateral, then the answer would be yes, by assuming that the starting triangle's vertices, and the points where the lines meet the sides, all have rational coordinates, and then applying the shoelace formula.
But the starting triangle is equilateral, so its vertices cannot all have rational coordinates. We can still apply the shoelace formula, but it is not obvious whether the resulting areas can all be integers.
 A: It's possible again. Make the area of the equilateral triangle rational. It's possible to make each area a rational multiple of the total area. Call the triangle $\triangle ABC$, then draw segments $AD, BE, CF$ in the picture. If  $CF$ is any segment so that $AF/AB\in\mathbb{Q}$ and likewise with $AD$ and $BE$, then this is true. This can be shown by deleting lines and using mass points (I'll leave this as an exercise to the reader). Finally, enlarge the picture by the common denominator and you get the desired outcome.
A: Do you know the one-seventh area triangle ? By joining each vertex to the point on the opposite side situated at one-third of it, one gets a triangle whose area is $1/7$ of the area of the triangle.
It is a direct way to exhibit such a triangle, after a certain transformation for a triangle with area $21$ we have three types of shapes:

*

*a central triangle with area $3$,


*3 small identical triangles with common area $1$,


*3 identical quadrilaterals with common area $5$.

How can an equilateral triangle be with area $21$ ? Simply by taking the good magnifying factor. In particular, if the reference equilateral triangle has vertices'coordinates
$$(-1,0),(1,0),(0,\sqrt{3}) \ \ \text{with area } \ \ \sqrt{3},\tag{1}$$
as we must reach an area equal to $21$, the magnifying factor is $k= 7\sqrt{3}$, but, caveat !, the coordinates in (1) have to be multiplied by the square  root of $k$ i.e., $\sqrt{7\sqrt{3}}$, i.e, we end with :
$$\text{triangle} \ (-\sqrt{7\sqrt{3}},0),(\sqrt{7\sqrt{3}},0),(0,\sqrt{21\sqrt{3}}) \ \ \text{which has area } \ \ 21.$$
