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

enter image description here

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.

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    $\begingroup$ Ok, now this is harder. Note that again we only have to solve it for rational area. Since $\mathbb{Q}$ is closed under addition, the area of the equilateral triangle is rational. Hence $s^2\sqrt{3}$ is rational, for side length $s$. $\endgroup$ Commented Jan 1, 2023 at 2:40
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    $\begingroup$ Start with an isoceles right-angled triangle on $(-1,0),(0,1),(1,0)$. Get seven rational areas. Then stretch the $y$-axis by $\sqrt[4]3$ and shrink the $x$-axis by the same factor. The shoelace formula gives the same areas as before. $\endgroup$
    – Empy2
    Commented Jan 1, 2023 at 3:09

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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 $5$.

enter image description here

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 :

$$\text{triangle} \ (-\sqrt{7\sqrt{3}},0),(\sqrt{7\sqrt{3}},0),(0,\sqrt{21\sqrt{3}}) \ \ \text{which has area } \ \ 21.$$

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    $\begingroup$ This is really just an example of my construction $\endgroup$ Commented Jan 1, 2023 at 18:05
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    $\begingroup$ Thank you for the links, Jean Marie. The "one-seventh area triangle" is a nice triangle. I believe $21$ is the minimum area of a triangle that can be separated by three cevians into seven regions of integer area. $\endgroup$
    – Dan
    Commented Jan 1, 2023 at 23:03
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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.

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    $\begingroup$ In the Wikipedia page linked by Jean Marie, I find that this can also be proven with Routh’s Theorem $\endgroup$ Commented Jan 1, 2023 at 18:05

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