Triangle inside a unit square.

Question

Consider a square of size 1 and a triangle of sides $a$, $b$ and $c$.

Is there a way to know if the triangle can fit inside the square?

Clearly, if $a>\sqrt{2}$ or $b>\sqrt{2}$ or $c>\sqrt{2}$, it is not possible, but a triangle of sides $\frac{7}{5}$, $\frac{7}{5}$ and $\frac{1}{4}$ does not fit.

My first idea was to rotate the triangle and see if the maximum distance between points in each axis is less than 1...

The biggest triangle that fits inside a 1-sided square has an area 1/2. Anything with a bigger area cannot. There may be other necessary conditions for lower area triangles. — Bruno Peixoto, Oct 21 '23 at 10:53
This is essentially the same as asking, What is the smallest square containing a given triangle? — Dan, Oct 21 '23 at 13:22
It seems to me that this is tantamount to requiring that the circumcircle radius, $r\le\sqrt{2}/2$, For the equations of the circumcenter, see, for example, https://en.wikipedia.org/wiki/Circumcircle. — Cye Waldman, Oct 24 '23 at 22:43

KDP · Answer 1 · 2023-10-27T06:03:29.707

4

I attacked this question from various angles, with little success at first. Treating it as linear programming minimization problem turned out to be hideously complicated and maybe impossible. In the end the solution turned out to be relatively trivial.

Use this equation (that was obtained by trigonometry and the law of cosines)

$x = c\cdot \cos\left(\frac{\pi}{2} - \arccos \left(\frac{a^2+c^2-b^2}{2 a c} \right) - \arccos \left(\frac 1 a \right)\right)$

where a is the length of the longest side of a triangle with area $\leq 1/2$ and b is the length of the shortest side of the triangle.

If $x\leq 1$ the given triangle will fit.

For an arbitrary square with side lengths (s), replace $\frac 1 a$ with $\frac s a$ and use:

$y = c \cdot \cos\left(\frac{\pi}{2} - \arccos \left(\frac{a^2+c^2-b^2}{2 a c} \right) - \arccos \left(\frac s a \right)\right)$

and the triangle will fit if $y \leq s$

edited Oct 27 '23 at 06:03

answered Oct 24 '23 at 21:05

KDP

1,079

Gracias. What are $x$ and $y$ exactly? I mean what properties of a triangle are we evaluating? – Agent Smith Oct 27 '23 at 04:19
1

For a triangle of area less than or equal to $1/2 $, x is the shortest possible length of a side that is orthogonal to a side of unit length of an enclosing rectangle. – KDP Oct 27 '23 at 05:59
Way to go! Muchas gracias. – Agent Smith Oct 27 '23 at 06:59

score 0 · Answer 2 · answered Oct 25 '23 at 00:19

I believe that this is essentially a question of showing that the radius of the circumcircle is less than $\sqrt{2}/2$ because the largest triangle that can fit in the unit square is the right-triangle with unit sides, hypotenuse of $\sqrt{2}$ and circumradius of $\sqrt{2}/2$. Let's demonstrate how easy it is to calculate this in the complex plane. Given three random points, say $z_{1,2,3}$, we can determine the circumcenter, $z_c$ and radius, $r$ as shown here.

$$ z_c\ \;=\;\frac{\left| \begin{array}{ccc} z_1 & |z_1|^2 & 1 \\ z_2 & |z_2|^2 & 1 \\ z_3 & |z_3|^2 & 1 \end{array} \right|} {\left| \begin{array}{ccc} z_1\ & {z_1}^*\ & 1 \\ z_2\ & {z_2}^*\ & 1 \\ z_3\ & {z_3}^*\ & 1 \end{array} \right|} $$

and

$$ r=|z_c-z_1|=|z_c-z_2|=|z_c-z_3| $$

These calculations can probably be calculated by hand, but most likely will be done on a computer. I did just that in a matter of just six lines of code (in Matlab). You can then go on to plot the triangle, circumcircle, and overlay a unit square (though it may not be in the correct orientation).

score 0 · Answer 3 · answered Oct 27 '23 at 04:51

For a $\triangle ABC$, suppose the longest side = $AB$. Then

$AB > \sqrt 2$. $\triangle ABC$ will not fit.
$AB = \sqrt 2$. $\triangle ABC$ will fit IFF $\angle ABC = 45$ degrees $\vee$ $\angle BAC = 45$ degrees.
$AB < \sqrt 2$. It seems necessary to set down some conditions for $1 < AB < \sqrt 2$ especially so if $AB \to \sqrt 2$

Triangle inside a unit square.

3 Answers3

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