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I looked through this site and the internet for a while and found some related problems, but not with my particular constraints. The problem is:

Given a $n*m$ rectangle, cut four congruent right triangles using the corners as the right angles. What is the maximum area that each triangle can have?

Let me explain a bit further:
This is an example illustration to help with the visual side of things
enter image description here

You can see that the four lines have created four (theoretically) identical right triangles each with the same area. What I am wondering is how we could maximize this area, while still having each triangle be congruent.

My attempts:
I thought that if we placed the triangles' vertices halfway across each side, such that the $4$ hypotenuses created a rectangle with the empty space. It would look like this:
enter image description here

This would mean that each triangle has area $\frac{1}{8}nm$

For some reason, half of me feels as if this is the solution while the other half feels as if there is something I am overlooking. If anyone could provide some aid as to the correctness of my solution. Many thanks.

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  • $\begingroup$ Do all the triangle have to be congruent? $\endgroup$ – Ahmed S. Attaalla Feb 23 '17 at 2:19
  • $\begingroup$ Yes, they all have to be congruent, otherwise you cut just cut one huge triangle with the diagonal as the hypotenuse. $\endgroup$ – Harnoor Lal Feb 23 '17 at 2:22
  • $\begingroup$ Of course if you change the corner requirements slightly to say that each corner of the rectangle is a vertex of exactly one of the congruent triangles, you can get arbitrarily close to using the whole rectangle. $\endgroup$ – Joffan Feb 23 '17 at 3:35
  • $\begingroup$ FYI: Regarding "if we placed the triangles' vertices halfway across each side (of the m*n rectangle with $m \ne n$), such that the 4 hypotenuses created a rectangle with the empty space", that empty space is a rhombus. $\endgroup$ – Mick Feb 23 '17 at 6:13
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As far as I am able to understand your line of thinking is correct. If you call the two sides of the triangle which aren't the hypotenuse $a$ and $b$. Then the area of each triangle is given by $\frac{ab}{2}$. Increasing a, or increasing b will increase the area of the triangle.

Therefore we seek the maximum value of $a$ and $b$. The only limiting factor is that (without loss of generality) $2a\leqslant n$ and $2b \leqslant m$. Therefore we take $a = n/2$ and $b = m/2$. This gives $area = \frac{mn}{8}$ as you said.

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  • $\begingroup$ Or the condition that $a+b=S \leq L$ for a square. But this gives the same result. That maximum occurs occurs when $a=b=S/2$, and taking $S=L$ gives the max out of all these possibilities. $\endgroup$ – Ahmed S. Attaalla Feb 23 '17 at 2:36

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