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Solution for September 2010

 The Problem: A plumber with a pile of equal-length straight pipes and a box full of right-angled elbows realizes that he cannot use them to form a closed pentagon, although he can make a closed polygon with six or with seven pipes. Prove that plumber's observations are indeed correct. (For those mathematicians with no training in plumbing we should add that the pipes cannot be joined directly together -- neighbouring pipes can be joined only by using an elbow, thereby forming a right angle.)

Correct responses:

Solutions were submitted to us by

 Bojan Basic (Serbia) Verena Haider (Austria) Bernard Collignon (France) Benoît Humbert (France) Shai Covo (Israel) Matthew Lim (USA) Mei-Hui Fang (Austria) Patrick J. LoPresti (USA) Philippe Fondanaiche (France) John T. Robinson (USA) Bruce Golfman (Austria) Albert Stadler (Switzerland) Cornel Gruian (Romania)

The solution:

The pentagon.
We first look at the impossibility in three dimensions of a regular pentagon having five right angles.  Fang commented that our claim follows immediately from a 1970 theorem of B.L. van der Waerden: Any pentagon in three dimensions that is both equilateral and equiangular lies in a plane.  Van der Waerden's rather complicated argument [Ein Satz über räumliche Fünfecke. Elem. Math. 25 (1970), 73-78; addendum: Elem. Math 27 (1972), 63] has given way to several simpler ones.  (A proof is very easy if you know that a 3-dimensional isometry is either a rotation or the product of a rotation and at most two reflections, one in a plane perpendicular to the axis of rotation and one in a plane that contains the axis.  See, for example, Introduction to Geometry by H.S.M. Coxeter, Sections 7.1 to 7.4.)  We shall instead offer two proofs of our result using the direct approach taken by all but one of our correspondents.  The first closely follows Collignon; the sketch is based on Fondanaiche's submission.
Proof 1.  Observe that any three consecutive vertices of the pentagon ABCDE form an isosceles right triangle with sides 1, 1, and √2.  We introduce coordinates with A, B, C in the xy-plane: A = (0, 1, 0), B = (0, 0, 0), C = (1, 0, 0).  Because CD BC, D must lie in the plane x = 1; because D is equidistant from A and B, it must lie in the plane y = ½.  Also, if we denote the foot of the perpendicular from D to the xy-plane by D', then CD'D

is a 30-60 right triangle with sides 1, 1/2, and √3/2. We may as well take D above the xy-plane, so that

D = (1, 1/2, √3/2).

Similar reasoning (interchanging the roles of x and y) gives us two possibilities for E:

E1 = (1/2, 1, √3/2) or E2 = (1/2, 1, -√3/2).

In the first case the distance from D to E is ; in the second, . In either case, the distance from D to E is not 1.

Proof 2.  Robinson showed somewhat more with his argument: An equilateral pentagon ABCDE in 3-dimensional space with unit sides and right angles at A, B, C, and D must have an angle at E of either 60° or cos-1(-5/√14) (which is about 111°); in particular, ∠E cannot be 90°. We will modify his notation to fit with the previous paragraph.  Thus the coordinates for E are still (1/2, 1, √3/2) but since there is an unknown angle at E, we no longer have the length of DA to work with; we know only that D is in the plane x = 1, and that DC = 1, whence D = (1, cos δ, sin δ) for some angle δ. To find cos δ we use ED = 1 to get

which simplifies to

.                                   (1)

We wish to solve (1) for cos δ; Let c = cos δ and replace sin δ  by ;

then (1) becomes

,

so that (by squaring)

.

We solve this quadratic equation for c and find that

c = cos δ  = 1 or 1/7.                                    (2)

Finally, the cosine of the angle we seek is then the scalar product of unit vectors,

which, from (1), equals cos δ– ½; Equation (2) now tells us that the cos ∠E = 1– ½ = ½, or cos ∠E =1/7 - 1/2 = -5/14, as claimed.

Further comments.  LoPresti suggested we call for a four-dimensional plumber — a pentagon with unit sides and five right angles is easily constructed in four dimensions!  One way to see this is to put it in the 4-dimensional subspace u + v + x + y + z = √2 of 5-dimensional space.  One then easily checks that the pentagon whose vertices are

(1/√2, 1/√2, 0, 0, 0), (0, 1/√2, 1/√2, 0, 0), (0, 0, 1/√2, 1/√2, 0),
(0, 0, 0, 1/√2, 1/√2), (1/√2, 0, 0, 0, 1/√2),

has edges of length 1 and diagonals of length √2 (which implies that the angles between consecutive edges are 90°).  LoPresti says nothing about what 4-dimensional plumbers charge these days; one can only imagine that the cost would be out of this world.

Haider's solution was completely different.  She considered the five vectors that join each vertex of the pentagon to the next, and determined that any four of them must be linearly independent; consequently the smallest space in which the pentagon could exist would necessarily have four dimensions.

Hexagons, heptagons, and other n-gons with n > 5.
People say that seeing is believing:

Mathematicians, however, say that one shouldn't believe everything one sees:

Penrose Triangle in a park in Gotschuchen, Austria
Picture reproduced from the Wikipedia files.

Because the angles of a triangle sum to 180°, no triangle can have more than one right angle.  The illusion of a triangle with three right angles was, according to Wikipedia, first devised in 1934 by the Swedish artist Oscar Reutersvärd; it was independently discovered and popularized by the mathematician Roger Penrose and his father Lionel Penrose, the distinguished British psychiatrist and geneticist.
It is easy to prove the existence of right-angled hexagons without resorting to coordinates: use the edges surrounding the region formed by two adjacent faces of a cube (as in the picture using pipes above, and in the diagram below on the left).  Lim and Golfman both described a second example using the edges of a cube: take edges along a path for which every two consecutive sides, but no three, belong to a face of the cube (as below on the right).

Here are the paths shown in the figures (starting with (0, 0, 0) in the back at the lower left):

For the heptagon, it might help to imagine the figure laid out on a flight of stairs.  The portion on the bottom stair consists of three sides of a square.  Consider the vertical

central portion that consists of two sides of an isosceles trapezoid ABCD with AB = BC = CD = 1, and AD || BC with AD = √2.  As shown in the diagram below, one can think of the isosceles right triangle ADP as a flap that rotates about a hinge AD.  When the flap

is "open" (with P outside the trapezoid, but in the same vertical plane), BAP >∠P'AP = 90°; with the flap "closed" (in the position AP'D), ∠BAP' < ∠BAD < 90°.  Somewhere between these two positions ∠BAP = ∠CDP = 90°.  There will also be a symmetric position of the flap on the other side of the vertical plane where the angle between adjacent sides is 90°.  Although half the correspondents used a similar dynamic argument to prove the existence of a right-angled heptagon, everybody supplied coordinates for good measure.  Here are the coordinates that go up the stairs on the right (with the positive x, y, and z axes pointing to the right, in, and up, respectively):

.

The three vertices on the left in the stair figure have negative x-coordinates, but the same values for y and z.  One checks easily that the distance between consecutive vertices is 1, and the angle between consecutive sides is 90°.