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What's the minimal number of distinct right angled triangles needed to tile a unit square? I suspect it cannot be done in less than four, thus here are the solutions for $4$ triangles so far:

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Define a unit square $ABCD$; then the three distinct constructions so far are:

  1. Connect two opposite corners ($\overline{AC}$), and place a fifth point ($E$) anywhere on some side ($\overline{BC}$) of the square. Connect the fifth point with the remaining available point ($A$) and draw a perpendicular line ($\overline{EF}$) from it to the diagonal line. You now have $4$ distinct triangles tiling the square.

  2. Place a fifth point ($E$) anywhere on some side ($\overline{BC}$) of the square except on the middle point ($P$) of that side. Connect it with the two points of the opposing side ($\overline{DA}$) and draw a perpendicular line ($\overline{DF}$ or $\overline{AF}$) from any of those two points to the opposing line segment. You now have $4$ distinct triangles tiling the square.

  3. Oscar Lanzi's construction; From $A$ draw a line segment to a point $E$ anywhere on side $\overline{BC}$. Construct the perpendicular to $\overline{AE}$ through $E$ which intersects side $\overline{CD}$ at $F$. Finish the division by drawing $AF$. You now have $4$ distinct triangles tiling the square.


Do these constructions contain all valid solutions to tile the unit square with $4$ distinct right angle triangles? Or are there some other ways missing?

The goal is to find all valid solutions for this simple problem, and prove no more exist.

But I'm not sure how can one tell whether we have all the solutions contained in these constructions so far or not?


Follow up question; is it possible that there exists a solution among the solutions such that all four triangles have all side lengths rational?

Since I stumbled upon a tweet showing $5$ Pythagorean triples packed in a square; If that square were scaled down to a unit square, we would have a solution with $5$ triangles that have all sides rational.

Is it possible to do it in $4$?

Pack four rational distinct triangles in a unit square?

I suppose this can be shown false if all constructions were found and shown to not contain such a solution. So far, these three constructions seem to not contain such a solution, if I'm not mistaken.

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Another solution with four triangles later incorporated into the question. Start with square $ABCD$. From$A$ draw a line segment to a point $E$ on side $BC$. Construct the perpendicular to $AE$ through $E$ which intersects side $CD$ at $F$. Finish the division by drawing $AF$. This must be distinct from the cases given above because only one vertex of the square is used for triangle vertices with acute angles.

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    $\begingroup$ Added to the constructions. $\endgroup$ – Vepir Jun 7 '17 at 9:21

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