Outline of Proof Regarding Squares Wedged in Triangle

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 

	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 

	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.
All rights reserved.