Date: Mon, 3 Nov 1997 Subject: triangle problem To: Eugenio Calabi From: Steven Finch Dear Professor Calabi, We haven't met, but in the charming book J. H. Conway and R. K. Guy, The Book of Numbers, Springer-Verlag, 1996; page 206 it is said that you found a (unique) non-equilateral triangle containing three congruent largest squares. I am wondering if you have published this result. If not, is it available in preprint form or even in rough handwritten notes? Or has someone else published articles about this and related problems? With appreciation and very best wishes, Steve Finch ************************************************************************** Date: Tue, 18 Nov 1997 Subject: Re: triangle problem To: Steven Finch From: Eugenio Calabi I did in fact tell John Conway about the triangle problem, which I had originally meant to suggest as a Putnam competition problem. However, having used it since then several times in coaching Putnam contestants, the cat is out of the bag, but I never thought of publishing it for the glory of it. I am very pleased to send you an outline proof (minus elementary calculations).... Let us say that, given any triangle T, a square Q is wedged inside T if it cannot be expanded continuously (as a square) inside T. There are, in general, exactly three ways in which a square Q may be wedged inside any given triangle T, the exceptional cases occurring when T is a right triangle. To show this, any square Q that is not wedged in T can be rigidly moved until one side of Q is contained in (or Q lies flat on) one side, say s, of T; the square Q may then be expanded, while lying flat on the side s, as far possible until it becomes wedged in T; the configuration then falls into one of exactly three possibilities: (i) if both of the angles of T at either end of the side s are acute, then the two corners of Q that are not on s fall each in one of the remaining two sides of T; (ii) if one of the angles, say at the vertex A, is a right angle, then one vertex of Q lies on A, so that there are simultaneously two adjacent sides of Q flat on two corresponding sides of T (including s) and the remaining vertex of Q lies in the interior of the hypotenuse; (iii) if the angle at A is obtuse, then one vertex of Q lies on A with one adjacent side, say AP, lying in s, the vertex of Q opposite to A touches the long side of T and the remaining vertex of Q, opposite to P, remains in the interior of T. 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 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) We see that, 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\ = (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\ = (1 + cos\mu\ - sin\mu\) / (cos\mu\ + sin\mu\). This equation reduces to a cubic equation with three distinct real solutions for \mu\ (modulo 2\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 = 0.631109489049... , cos\mu\ is the smaller of the two positive roots of the cubic polynomial 8x^3 - 4x^2 - 3x + 1 , namely x = 0.775693762274... , tan\mu\ is the one positive root of the cubic polynomial x^3 + 4x^2 + x - 4 , namely x = 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 = 0.355415726777... . As a result we have (3) \mu\ ~ 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. Sincerely yours, Eugenio Calabi
Copyright (c) 2002 Steven R. Finch.
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