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© Copyright 2002, Jim Loy
I received email asking me how to inscribe a square in a general quadrilateral (using compasses and straightedge). This is a report of my attempts to perform this construction.
We would seem to have several cases, some of which are shown in the diagrams above. In the first three, our inscribed square has two vertices on one side of the original quadrilateral. The first case is a trivial construction (but the square is not inscribed in the quadrilateral). The next two cases are the same as inscribing a square in a triangle (which we get by extending two of the sides of the quadrilateral). In the fourth case, we have each vertex of the square resting on a different side of the quadrilateral. This is not always possible. And when it is possible, it looks like it may be much more difficult than the other cases.
The first three cases show how we can place two
vertices of a square on each of the four sides of the quadrilateral. So, we can
inscribe four different squares in our quadrilateral, using the first three
orientations. We may want to do this in order to find the largest of these
squares. On the left, we see one way to inscribe a square in a triangle.
Instead of trying to inscribe the square directly, I have instead inscribed a
square into a right triangle of the same base and height, by drawing the 45
degree diagonal. This square is identical to the square I want to inscribe into
the original arbitrary triangle. From there, it is fairly easy to construct the
congruent square in the original triangle. That solves case 2, above. We can
probably do something similar for case 3.
Off on a tangent: So, it remains to construct case 4. If the original quadrilateral is a square, then infinitely many squares can be inscribed into it using the orientation of case 4. On the other hand, if our quadrilateral is very long and thin, then we cannot inscribe a square using case 4. Experimentation suggests that for many quadrilaterals we can inscribe one and only one square (maybe), using case 4.
On the right we see an interesting experiment, using the geometry
program Cinderella. I have chosen the
blue point on one side of the quadrilateral, and then let the red point move
along the next side (extending the side when necessary). These two points
define a rectangle. We have found an inscribed rectangle when the green point
and the blue point coincide.
As the red point moves, the green point traces out a complicated curve. Here we see that the blue and green points cannot coincide, and we must choose a different blue point. The complicated shape of the green curve would suggest that it is a graph of a fourth degree equation or higher. That would in turn suggest that we cannot place the blue point in its proper place using compasses and straightedge, as a general third degree or higher equation cannot be solved using compasses and straightedge. See addendum below.
If I move the blue point, the curve changes shape somewhat, as shown above left. Changing the shape of the original quadrilateral also changes the shape of the curve, as shown in the second diagram above.
And zooming out, we find that the curve is even more complicated than we thought (below), seeming to have three asymptotes. This experimental geometry suggests that, in general, we cannot construct the case 4 square.
Addendum (my email):
This solution was sent to me by Nikolaos
Dergiades. ABCD is our quadrilateral. From A draw l1 (AG) toward point C, such
that angle DAG = 45 degrees. From B draw l2 (BF) toward C, such that CAF = 45
degrees. G and F are not yet determined. Draw BE perpendicular to l1, and
meeting AD at some point E. Draw AH perpendicular to l2, meeting BC at H. Draw
EF parallel to l1, meeting L2 at F. Draw HG parallel to l2, meeting L1 at G.
Draw FG intersecting CD at J. Draw JI parallel to l2, meeting BC at I. Draw IL
perpendicular to l2, meeting AB at L. L and J are opposite corners of our
square. The perpendicular bisector of LJ meets AD at K and BC at M. Our square
is JKLM.
It doesn't seem to matter in what order we label our
original quadrilateral, we seem to get the same square, for most
quadrilaterals. So maybe it is unique, as I suggested above. I don't know how
Mr. Dergiades came up with this construction. He sent me a proof (below), which
I have not studied.
On the right is is construction applied to another fairly common-looking quadrilateral ABCD. Here the square is entirely outside our quadrilateral. I would have guessed that we can inscribe a square, maybe not. Or maybe his method provides only one square of many.
Here is Mr. Dergiades' proof: "Let P be an arbitrary point on BC. The perpendicular from P to LP meets the line IJ at Q. Since the points L, P, I, Q are concyclic (on the same circle) because angle LPQ = angle LIQ = 90 degrees, then angle PLQ = angle CIQ = 45 degrees and angle LQP = angle LIP = 45 degrees. This means that triangle PLQ is isosceles (PL = PQ) and if P moves to M then Q moves to J. Hence triangle LMJ is isosceles with angle JML = 90 degrees and similarly the triangle LKJ is isosceles with angle LKJ = 90 degrees or the quadrilateral is a square." He further comments: "It is obvious that if the line FG is parallel to CD then there is no construction."
Comments: I think that most quadrilaterals have only one
solution, maybe. Rectangles which are not squares seem to have no solutions,
even with sides extended. A square has infinitely many solutions. The
quadrilateral here (red) has two solutions, with sides extended. I would expect
the above construction to find both solutions, by relabeling the vertices of
the parallelogram.
Taking a clue from the above construction, it is useful to construct a square by starting with a pair of opposite vertices, and then constructing the diagonals, which are perpendicular and bisect each other.