User:William Ellison/sandbox

The Calibi triangle and the three inscribed squares of maximum area

The Calabi triangle is a curiosity from elementary geomety, discovered by Eugenio Calabi in 1996 and popularised by John H. Conway & Richard K. Guy in The Book of Numbers.

The problem considered is that of finding how many ways can a square of maximum area be inscribed in the interior of a given triangle.

In general, the largest square can be inscribed in the triangle in only one way. However, there are some isoceles triangles where two squares of maximum area can be inscribed and for an equilateral triangle there are THREE positions to inscribe the square of maximum area.

Calabi found that there is precisely one other type of triangle for which one can inscribe THREE equal squares of maximum area. It is an isosceles triangle with side lengths proportional to 1, 1, $x$ = 1.55138752455, where $x$ is the positive root of the equation

$\textstyle 2x^{3}-2x^{2}-3x+2=0$

The problem was initially intended to be a Putnam Competition problem (a mathematical contest in the U.S. for high-school students), but was never used.

Proof : For any given triangle T, we say that a square S is wedged inside T if it cannot be rotated, translated and then expanded so as to form a bigger square inside T.

Obviously a square with maximum area must be a wedged square, (because if not, we could move it and expand it to a bigger square !).

We first study wedged squares, and once we know all about them we can see which of them have a maximum area.

A square S that is not wedged in T can be rigidly moved until one side of S lies flat on a side s of T.

The square S may then be slid along s and expanded until it becomes wedged in T. The configuration then falls into one of exactly three possibilities:

1. If both of the angles of T at either end of the side s are acute, then the two corners of S that are not on s fall each in one of the remaining two sides of T;
2. If one of the angles, say at the vertex A, is a right angle, then one vertex of S lies on A, so that there are simultaneously two adjacent sides of S flat on two corresponding sides of T (including s) and the remaining vertex of S lies on the hypotenuse;
3. If the angle at A is obtuse, then one vertex of S lies on A with one adjacent side, say AB, lying in s, the vertex of S opposite to A touches the long side of T and the remaining vertex of S, is in the interior of T.

So, given any triangle we now know how to find all the wedged squares.

There are three wedged squares, except for the case of a right-angle triangle for which there are only two.

If the three sides lengths are different (scalene triangle), the wedged squares have different areas ; if two sides have equal lengths (isosceles triangle) at least two of the squares have equal area and if the sides have equel length (equilateral triange), the three wedged squares have equal area.

In the case of a right-angled triangle, the two wedged squares not have the same area. The square on the hypotenuses is strictly smaller than the square wedged in at the right-angle.

Since the square of largest possible area inside T is one of the wedged ones, there remain at most three possibilities, depending on the side of T on which the square lies flat; these three choices reduce to two if and only if T is a right triangle; in all cases one must compare the relative sizes of the wedged squares.

Obviously, if the triangle T is isosceles or equilateral, the two wedged squares lying flat on any two sides of T of equal length are of equal size, by symmetry.

If the triangle T is a right-angled triangle, the comparison of the size of the two possible wedged squares is easy: the one lying flat simultaneously on the two perpendicular sides of T is always larger than the one lying flat on the hypotenuse, so that the largest square is unique.

If all three angles of T are acute, then all three positions of the wedged squares have one side of the square and both of its end-points in the interior of one side of T, and each of the two other sides of T containing a vertex of the square Q in its interior. Comparing any two of the three positions of wedged squares in T, one sees that the shorter the side of T on which one whole side of Q rests, the larger is the size of the square. Hence the largest size square that fits is the one wedged with one side flat against the shortest side of T, and there are exactly one, two or three largest such squares, respectively, depending on whether the triangle is scalene, isosceles (not equilateral), or equilateral.

The situation is different if the triangle T = (ABC) has an obtuse angle, say at the vertex C. We compare first the sizes of the two wedged squares in T that lie flat along either of the sides AC and BC adjacent to the obtuse angle (both wedged squares in this case will have a vertex at the point C)

If the lengths of these two sides are unequal, let us say if |AC| > |BC|, then the square with the side flat along AC is larger than the other one. Hence, if the obtuse-angle triangle is scalene, then the shortest side can never be the one on which one side of the largest fitting square lies flat, contrary to the earlier case; therefore the contest for the largest fitting square is between the two wedged ones lying flat on either of the two longer sides.

Intuitively, one sees that, if the obtuse angle at C is very nearly a right angle, then the wedged square flat against the long side (AB) is smaller that the one lying flat against the middle-sized side of T;

On the other hand, if the obtuse angle is very nearly flat, then the wedged square with a side along the long side (AB) of T wins the day. So there must exist transitional shapes of triangles for which there are two equal-sized wedged-in triangles, exhibiting what R. Thom called a "catastrophe".

These "catastrophic" triangles may be constructed (in the Pythagorean sense) as follows:

Start with a right triangle APQ with the right-angle vertex at P, a longer "horizontal" side of length |AP| and a shorter "vertical" side of length |PQ| (roughly, say |PQ| < .82 |AP|, see below), and prolong the longer side AP beyond the right-angle vertex P to a length equal to at most 2|AP|+|PQ|, and the hypotenuse AQ beyond Q to the point C, where |AC| = |AP|+|PQ|.

On the half-line AP mark the point S at a distance |AP|+|PQ| from A, and on the half-line AQ (already prolonged to the segment AC) mark the point P' at a distance respectively equal to |AP| from A; then construct the points R and R' to constitute the vertices of two squares (of equal size) PQRS and P'CR'S' (symmetrically situated with respect to the bisector of the angle QAP).

Connect the points C and S by a line, intersecting the half-line AP at a point B, and look at the triangle ABC. This triangle, with an obtuse angle at C, contains the two squares just described wedged inside it in a non-symmetric way; by the earlier argument, these two squares are of the largest size that fits, provided that the side AC of the triangle ABC is not shorter than BC.

Note that, unless the slope of the hypotenuse AQ over AP is very close to the upper limit .82, the side BC of the triangle ABC will be shorter than AC.

Equivalently, the catastrophe occurs if the angle $\alpha =BAC$ is not larger than the angle $\beta =ABC$. The relation between these two angles is easily seen by taking the lengths $|AP|=|AP'|$ as unity and writing the lengths of the remaining relevant distances in terms of triginometric functions of $\alpha$ :

(1) $\tan \beta ={\frac {(1+\cos \alpha -\sin \alpha )}{(\cos \alpha +\sin \alpha )}}$.

Note that the relation (1) makes $\beta$ a monotone decreasing function of $\alpha$ . Thus, for the construction to work, we must have

                       $\beta >=\alpha$ ; 

this implies that the allowable values of $\alpha$ range from arbitrarily small positive values (as $\alpha$ decreases toward zero, $\beta$ increases to the limit value arctan(2)) to a maximum value $\mu$, which is the smallest positive value obtained by solving (1) when one sets \beta = $\alpha =\mu$ ;

this is the case where the catastrophic triangle becomes isosceles, and therefore describes the unique isosceles, non-equilateral triangle (up to similarity) in which the three wedged squares have all equal size.

Thus $\mu$ is the smallest positive angle satisfying the equation

  (2)  $\tan \mu ={\frac {(1+\cos \mu -\sin \mu )}{(\cos \mu +\sin \mu )}}$.

This equation reduces to a cubic equation with three distinct real solutions for $\mu (modulo2\pi$ radians);

of these three solutions only one is relevant to our problem; each of the other two describes the shape of an isosceles triangle that admits three congruent squares that are similarly adapted into the triangle, in the sense that the same incidence relations occur between sides and vertices of the squares and the sides of the triangles as for a wedged square, but not all three of the squares are contained in the triangle.

The angle $\mu$ is determined most easily by solving (more or less equivalently) any of the following equations:

  $\sin \mu$ is the smaller of the two positive roots of the cubic polynomial
               $8x^{3}-4x^{2}-7x+4$ ,  namely   $x\approx 0.631109489049...$ ,

  $\cos \mu$ is the smaller of the two positive roots of the cubic polynomial

               $8x^{3}-4x^{2}-3x+1$ ,  namely   $x\approx 0.775693762274...$ ,

  $\tan \mu$ is the one positive root of the cubic polynomial

               $x^{3}+4x^{2}+x-4$ ,    namely   $x\approx 0.813605502649...$ ,

  $\tan(\mu /2$) is the smaller of the two positive roots of the cubic polynomial,  

               $2x^{3}-3x^{2}-2x+1$ ,   namely   $x\approx 0.355415726777...$ .

As a result we have

  (3)  $\mu \approx 0.682982699161...rad$ ,  or  $39.1320261423...degrees$.

Thus the isosceles triangle with an acute base angle (2) is the unique (up to similarity) non-equilateral triangle in which all three of the wedged-in squares are of equal size.

Incidentally, of the "catastrophic" triangles, the only one I have found whose angles are elementary (meaning that they are all commensurate to right angles) is the triangle with angles measuring 30, 45 and 105 degrees; it may be seen as a triangle inscribed in a regular dodecagon with vertices denoted as on a watch dial; the triangle T is then the one with vertices, say, at 12- , 2-, and 10 o' clock.