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Bonjour! Being the low level matematician that I am (unfortunately) I can't seem I find a way to solve this practical problem.

I have a triangle, which sides are 40 for the base and 27 for the two sides.

I want to find out which is the biggest rectangle that I could fit in that triangle (in any way, could be rotated, etc.).

I've approached it as an optimization problem but I can't seem to find a solution. Cheers!

Brick Top
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  • Do you mean isosceles triangle? – Calvin Lin Feb 06 '14 at 17:27
  • FYI - The general case isn't easy. Most people make the assumption that one side of the square must lie on a side of the triangle. This is not true, and there are triangles where a tilted square works much better – Calvin Lin Feb 06 '14 at 17:28
  • @CalvinLin Indeed, isosceles. Actually in an isosceles triangle I came to the conclusion that the best solution to the problem is for the shape to be a square, and to have one of the vortex's on the triangle's top vortex, and the opposite one on the triangle's base. Is that true? I'm wondering if there is a general solution, there should be, shouldn't it? ;-) – Brick Top Feb 06 '14 at 20:37
  • @CalvinLin Are you sure? Looks to me in any triangle the maximum is when all four vertices of the rectangle are on the triangle, which necessarily means some side of the rectangle is on some side of the triangle. – Macavity Feb 07 '14 at 09:23
  • @Macavity Yup, it surprised me the first time I saw it. Here's a Wolfram Demonstration. The main issue arises when there is an obtuse angle in the triangle, in which case constructing the square would not completely touch the sides, and hence you can tilt the square slightly to enlarge it. – Calvin Lin Feb 07 '14 at 12:26
  • @CalvinLin Nice demonstration. You are right for an inscribed square. However this is not the case for an inscribed rectangle - there you can always have the maximum on the triangle, I will try describing the process below. – Macavity Feb 08 '14 at 08:14
  • @Macavity Ah yes, i missed that. Rectangles are fine. – Calvin Lin Feb 08 '14 at 23:42
  • @CalvinLin this is driving me nuts, I do not see this at all for an inscribed square using the Wolfram demo and I'm wondering what the nature of your special triangle is (I tried a lot of obtuse ones, yet a side of the square seems to always be on a side of the triangle) – James May 28 '14 at 16:54
  • @James Look at the 3rd snapshot in the Wolfram demonstration. Most isosceles obtuse triangles work. – Calvin Lin May 29 '14 at 22:57
  • @CalvinLin I see the third snapshot, and one side of the square is still on a side of the triangle. I reckon that's always the case. However not all of the square's corners are touching the triangle. – James May 30 '14 at 04:49

1 Answers1

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I assume in this case you mean by inscribed, that all four vertices need not be on the triangle. We can consider this in several cases. Let the triangle be $PQGR$ and the quadrilateral be $ABCD$:

Case 1: All the vertices $A, B, C, D$ are in fact on $\triangle PQR$ - note that we always have this case in any triangle (for an obtuse triangle you can take the base to be the longest side). Necessarily this means two sides of the rectangle are parallel to one side, say $QR$. Let $h$ be the height of $\triangle PQR$ with $b = \lvert QR \rvert$ as base. Also WLOG let $AB$ lie on $QR$ and let $P$ be opposite $DC$. Then we have $$\triangle PQR \sim \triangle PDC \implies k = \dfrac{\lvert DC \rvert}{\lvert QR \rvert} = \dfrac{\text{height of } \triangle PDC}{\text{height of } \triangle PQR}$$

Hence we have $\lvert DC \rvert = kb$ as the base of the rectangle and the height of $\triangle PDC$ is $kh$, so we have the area of the rectangle as $kb(h-kh) = bh \cdot k(1-k)$. This clearly gets maximised when we choose $k=\frac12$, and we can thus get half the area of the triangle as the area of the maximal such rectangle.

Case 2: Exactly one vertex, say $D$ is not on $\triangle PQR$. Here, you can chop off parts of the triangle to get Case $1$ for a smaller triangle. There are many orientations, but in each case this is possible. In one particular case, you may need to draw a line parallel to $CD$ or $DA$ from one of the triangle vertices, which cuts the rectangle and triangle into two parts. One part has case $1$ applicable, and for the other part, we can chop off part of the triangle to again invoke case $1$. Hence necessarily this case generates only rectangles of area strictly less than $\frac12$ of the area of $\triangle PQR$.

Other cases - only two vertices on $\triangle PQR$ etc. are similar, we can chop off parts of the triangle to reduce them to Case $1$. Hence in all such cases we can never achieve $\frac12$ the area of $\triangle PQR$, hence Case $1$ is the optimal one.

Macavity
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