Call a polygon side-rational if the lengths of all its sides are rational. Call a dissection of a polygon side-rational if all of the polygons within the dissection are side-rational. Then my question is as in the title:

Do any two rational triangles of the same area have a common side-rational dissection?

Of course it's a classical result that any two polygons of the same area have a common dissection; a standard proof goes through triangulating the polygons, using a triangle-to-rectangle dissection, partitioning the rectangles for both polygons into a set of common-area rectangles, and then using a rectangle-to-rectangle dissection. But many of the steps in this process will involve non-rational sides (for instance, the rectangle corresponding to a triangle of rational sides but irrational area certainly can't have rational sides), and it's not clear that such a dissection could be patched up.

This question was prompted by rediscovering my common dissection of two very special rational triangles, the $4-5-6$ triangle and the $3-8-10$ triangle (both of area $\frac{15}{4}\sqrt{7}$):

4-5-6 triangle 3-8-10 triangle

All the lengths here can be shown straightforwardly with the law of sines and law of cosines; these triangles are a somewhat special case because so many of the angles involved are multiples of each other, but it seems likely that any rational triangle will inherently be special in somewhat similar ways.

EDIT: the original version of this question asked about arbitrary polygons, but having thought about it more I would be amazed if that result is true. The area of a (side-)rational triangle is always the square root of a rational number (by Heron's formula, for instance), but already rational-sided quadrilaterals can have arbitrary area: just imagine 'flexing' a rhombus with all sides of unit length, from a square down to a trivial, collapsed shape: it will continuously take on all areas from $1$ to $0$. This implies, for instance, that there's a rhombus with all sides of unit length and area $\frac\pi4$; such a rhombus can't be dissected into rational triangles, and it's hard to imagine any shared dissection it would have with, e.g., the rhombus with all sides of length $2$ and area $\frac\pi4$. (Which admittedly isn't a proof that one doesn't exist!)

A fair bit of rummaging around the web has turned up nothing on the topic; if anyone has pointers to similar matters, they'd be very much appreciated.

  • 2
    $\begingroup$ Very nice problem! $\endgroup$ – André Nicolas Mar 19 '15 at 16:02
  • $\begingroup$ @AndréNicolas Thank you! I'm afraid my initial hope was much too optimistic; I've revised the Q after a bit of thinking, since the sheer freedom in choosing rational-sided polygons even for quadrilaterals makes things much less likely. I think the question is interesting even for triangles, though. $\endgroup$ – Steven Stadnicki Mar 19 '15 at 22:09
  • $\begingroup$ M is the midpoint of BC. If AM is rational, other triangle can be made with ABM and ACM. Your 6-5-4 triangle could be divided 3-4-5 and 3-4-4. $\endgroup$ – Takahiro Waki Jul 26 '16 at 21:38

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