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This may be an easy question, although it stumped some of us over lunch. Consider the unit square and cut it up in several triangles, such that all the triangles have the same area. Is it possible that one triangle has a vertex with irrational coordinates (at least the abscissa or the ordinate, not necessarily both)?

This question was prompted by the theorem that in the cutting above, the number of triangles has to be even. The proof (that was inspired by this talk by Jim Fowler) uses that there exists an extension of the 2-adic valuation on $\mathbb{R}$, which follows from the axiom of choice. So we're wondering if it's necessary.


marked as duplicate by Watson, Community Oct 30 '18 at 8:09

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  • $\begingroup$ Just one vertex? The rest have rational coordinates? $\endgroup$ – Yuriy S Oct 29 '18 at 20:56
  • $\begingroup$ @YuriyS At least one. If there's one there's probably more but I don't know. $\endgroup$ – Najib Idrissi Oct 29 '18 at 20:56

To avoid fractions, consider a square with side-length $3$. Doing so doesn't affect the rationality-irrationality of coordinates.

enter image description here

The image shows a dissection into six triangles of area $3/2$. The coordinates of the inner vertices are

$$\left(1, \phi^{-2}\right) \qquad \left(\phi^{2},1\right) \qquad \left(2,\phi^{2}\right) \qquad \left(\phi^{-2},2\right)$$

where $\phi := \frac12\left(1+\sqrt{5}\right)$, the Golden Ratio, is an irrational number. $\square$


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