Find all values of $n > 1$ for which one can dissect a rectangle into $n$ right triangles.

Question

Find all values of $n > 1$ for which one can dissect a rectangle into $n$ right triangles.

For $n=2$, it is easy to show that it is possible (just insert the diagonal).

For $n=3$, it is not possible ( I tried on some examples but not able to do it).

For $n=4$, it is possible first make two rectangle from the original rectangle then use the case of $n=2$

So if $n=2k$ (means even) then it is always possible.

I am not able to generalize the result. I am trying induction on $n$, but not getting anything.

Answer 1

You can always dissect a rectangle into $3$ right triangles:

And you can cut any of these trianlges into two, and so on... Eventually you can dissect a rectangle into any number $n \ge 2$ right triangles.

Answer 2

I assume that you're disect only into right triangles (otherwise we can disect it into one right triangle and one quadrilateral).

If you're allowed to have cuts that ends within the rectangle you can simply disect a right triangle by the normal to the hypotenuse. This way you can alwas dissect it if $n\ge 2$.

If on the other hand you require the cuts to go all the way through the rectangle it depends on the aspect ratio of the rectangle. If the aspect ratio is at least $2:1$ you can disect it if $n\ge 2$ otherwise you can only disect it if $n=2$ or $n\ge 4$.

For some rectangles you can disect it into three by drawing a line from the longer edge to the opposite corners.

To see which rectangles you can disect into three this way you consider the right triangle formed by the point on the longer edge and the opposite corners. You have that the length $c$ of the longer edge is given by pythagoras $c^2=a^2+b^2$ (where $a$ and $b$ are the legs of the triangle). You also have the area of the triangle to be $ch=ab$ where $h$ is the shorter edge of the rectangle. This means we can write the aspect ratio as:

$${c\over h} = {c\over ab/c} = {c^2\over ab} = {a^2+b^2\over ab} = a/b+b/a$$

So the possible aspect ratios are given by the range of the function $f(\xi)=\xi+1/\xi$, $\xi>0$. Which can be seen is $f(\xi)\ge2$.

For higher $n$ you can obviously disect a rectangle into rectangles and you can always make at least one of them have enough aspect ratio. Then you can disect the one in $2$ or $3$ triangles and the rest in $2$ triangles each.

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To see that these are the only possibilities we use the assumption that all parts be right triangles. For $n=1$ it's obvious that the only part must be the rectangle which is not a triangle.

For $n=3$ we note that a line through a convex polygon splits it into two convex polygon. So one line through the rectangle splits it into two. Another line would at least pass through one of these polygon and split that into two (but it can also pass through another polygon). This means that if cut with $l$ lines we divide the rectangle into at least $l+1$ convex regions, but precisely $2$ if $l=1$. This means that to split the rectangle into three regions we need precisely two lines.

Then we just have to go through all cases:

If they intersect within the rectangle we see that they divide the rectangle in at least $4$ parts. On the other hand if they don't we see that the second line can not pass through both parts formed by the first and we see that it divides the rectangle into precisely three parts. If now they do not intersect on the edge of the rectangle they will cross the edge at precisely four points we also see that at most two of these can be at the corners of the rectangle so at least two is on the sides this means that they have at least $6$ sides from the rectangle and $2$ shared sides from the lines. This means that at least one of the regions must have at least $4$ sides.

The only case left is that the lines crosses precisely at the edge of the rectangle. Now either thats at the corner or a side. The case of a corner means that one line must be the diagonal (we divide two sides in two parts and get the same situation as before where at least one region has at least $4$ sides), we also have that since crossing at a corner it need to cross at non-right angle so the right angles of the triangles must be those of the rectangle of which we already disected two.

Finally if they cross at a side we must have that they also cross at a corner since otherwise by the same reasoning as above we have at least $5$ sides from the rectangle and $2$ shared from the lines meaning that at least one of the region would need to have at least $4$ sides.

Answer 3

You can always dissect a right triangle into two right triangles, each similar to the original. Just consider the altitude at the vertex of the right angle. As you can dissect a rectangle by its diagonal, one can always dissect a rectangle into $n\ge2$ similar right triangles.