- Following is a message (edited with some corrections) from an

old thread in sci.math newsgroup.

[I had used the pseudonym Athanatos Karamelopoulos and the e-mail

address joyce@..., which explains that name:

AK is a Greek hero of Joyce _Ulysses_.

Ilias is the Greek-American mathematician Ilias Kastanas]

APH

--------------------------------------------------------------------

From: Antreas P. Hatzipolakis (xpolakis@...)

Subject: Re: Perpendicular bisestors

Newsgroups: sci.math

Date: 3 April 1999

In article <7e5i8f$e5n$1@...>, ikastan@... (ilias kastanas

08-14-90) wrote:

> In article <joyce-0104992223300001@...>,

And neither this one, I guess, Ilias, since it is long before your (and my!)

> Athanatos Karamelopulos <joyce@...> wrote:

> @In article <4rddgj8f7g0y@...>, joyce@...

> @(Athanatos Karamelopulos) wrote:

> @

> @> We are given 4 lines which are the

> @> perpendicular bisectors of a quadrilateral.

> @>

> @> Construct the quadrilateral.

> @

> @

> @The story of the problem:

> @

> @It was proposed in the student mathematical journal EUKLEIDHS (=Euclid)

> @of the Greek Mathematical Society.

> @Although at that period (70's) there was a number of students contributed

> @nice solutions (among them was Ilias Kastanas), no solution

> @appeared.

>

>

> And I never wrote "Age 11" either!

>

>

> (Ehmm, Antrea... stop revealing my secret sources!!...) :-))>

>

> Heh, heh... it was before my time... -- but _NOW_(!!) at last...

> the moment has COME!! (_Nobody_ expects the Spanish Inquisition!! etc etc)

>

>

>

> But seriously, no -- I've never seen this before. Thanks to Antrea

time:

To Construct a quadrilateral when are given the points P,R,S,T which are

the orthogonal projections of O: the diagonals intersection point,

on the sides of the quadrilateral.

[SBMS, May-July 1956, pp. 146-157]

> for bringing it up; and it seems he received some rather unwarranted

A nice analysis, indeed, Ilias!

> criticism, too. E.g. he never claimed uniqueness.

>

>

> Solution by geometry is nice and pleasing. Say, the bisectors of

> AB, BC have intersection M (at angle 2*pi - B) and the other two N;

> then MA has known direction, as its angle with MN is B. Likewise with

> NA. This constructs A. OPER EDEI PRAXAI.

>

The editors' solution was as follows:

Let ABCD be the quadrilateral in question, with x = the perp. bisector of AB,

y of BC, w of CD and z of DA.

I will use the following symbolism:

Sx(A) = B : the symmetrical point of A is B with axis of symmetry x.

Now, let H be any point on the plane, and consider the points sequence:

H, Sw(D), Sy(Sw(D), Sx(Sy(Sw(D)), Sz(Sx(Sy(Sw(D))) =: K

K, Sw(K), Sy(Sw(K), Sx(Sy(Sw(K)), Sz(Sx(Sy(Sw(K))) =: L

We have that DH = DK = DL.

Therefore the vertex D is the center of the circle whose the circumference

passes through the known points H,K,L.

>

And now let's use "a bazooka to kill a fly".

>

> Another approach, less elegant but making degeneracies and extremes

> easy to handle, is: on the complex plane let P be the projection of the

> origin O on a line; represent the line by the complex number a = OP.

> Call arg(a) "theta". Plainly, the point symmetric to z w.r.t. the

> line is: -exp(i 2theta) z* +2a. Iterate four times (a_1,..., a_4 of the

> bisectors), set the result = z, and you get a linear equation

>

> (exp(2i [theta_1 - theta_2 + theta_3 - theta_4]) -1) z = (stuff)

>

>

> (1) If [theta_1 - theta_2 + theta_3 - theta_4] != 0,

> (or ditto for some permutation), a solution z is immediate.

>

> (2) Otherwise (that is, theta's all equal -- 4 parallel lines):

> A solution still exists iff distance between two lines = distance

> between the other two (0 included).

>

>

>

>

> Uh... what about "3 bisectors with a common point not on the 4th"??

> It's under (1) ?!... Yes, it is. Just use a triangle with a vertex on the

> 4th bisector! (For that matter, if all 4 have a common point, one of the

> solution quadrilaterals is that lone point, natch).

>

> Come to think of it, the no-solution case (subcase of (2)) _is_

> 4 lines concurrent at a "point at infinity"... All right, all right --

> that's stretching it. Such a "quadrilateral" would have surely made

> Euclid raise his eyebrows!

>

>

>

>

> Ilias

There is an interesting theorem (with an interesting history; see APPENDIX)

namely:

The perpendicular bisectors of the sides of quadrilateral ABCD

form a quadrilateral Q (unless ABCD is a rectangle), and the

perpendicular bisectors of the sides of Q form a quadrilateral

Q'. Show that Q' is similar to Q.

Now, let Q_0 be the original quadrilateral ABCD, and Q_1 the

quadrilateral formed by the perp. bisectors of Q_0,.....; Q_n+1 the

one formed by the p.b. of Q_n ......

We have Q_0 ~ Q_2 ~ Q_4 ~ .......

Q_1 ~ Q_3 ~ Q_5 ~ .......

Now, Q_2,Q_4 are known. So we can find the sides of Q_0 etc, etc

Antreas

----------------------------------------------------

APPENDIX:

From: schattdo@... (Doris Schattschneider)

Subject: The "Quad Perp" problem

Date: Mon, 31 Oct 1994 17:37:57 -0500

Newsgroups: geometry-announcements

[...]

THE PROBLEM: The perpendicular bisectors of the sides of quadrilateral ABCD

form a quadrilateral Q (unless ABCD is a rectangle), and the perpendicular

bisectors of the sides of Q form a quadrilateral Q'. Show that Q' is

similar to Q.

More precisely, show that Q' is homothetic to Q, that is, there is a dilation

(with positive or negative scale factor) that maps Q onto Q'. Also, find the

scale factor of the dilation.

This problem was proposed by Josef Langr in the American Mathematical Monthly

in 1953 [Problem E 1050, volume 60, p. 551], and brought to our attention by

Branko Grunbaum in an article "Quadrangles, Pentagons, and Computers," in

GEOMBINATORICS 3(1993) 4-9. At that time, no solution to the problem had been

found. Grunbaum had investigated many cases with Mathematica, and all

experimental evidence confirmed the claim of similarity.

This is a great problem for investigation with the Geometer's Sketchpad, and

Jim King, Dan Bennett, and I (Doris Schattschneider) all played around with

it-- Sketchpad certainly confirmed it was true. But proof? Dan Bennett and

Jim King each found a proof that Q' is similar to Q that uses only elementary

geometry, but this proof does not provide the description of the dilation that

sends Q to Q'. Geoffrey Shephard did find a description of the dilation scale

factor, but through a complicated trigonometric argument. An elementary

argument, or an argument that lends geometric insight into how the shape of

ABCD affects the scale factor still is needed.

I (Schattschneider) looked at some special cases and discovered with Sketchpad

(and subsequently proved) some special results:

(1) When two sides of the quadrilateral are parallel, then the first derived

quadrilateral Q is similar to the original ABCD by a spiral similarity with a

90 degree turn and scale factor( cot D - cot A)/2, where A and D are adjacent

angles in ABCD that are not supplementary.

[This scale factor is the same as [sin(D-A)]/(2 sin D sin A).]

So for this special case, the homothety that relates ABCD to Q' is the

product of the spiral symmetry with itself.

(2) When the quadrilateral is a parallelogram, all adjacent angles are

supplementary and the scale factor of the spiral similarity that relates

ABCD to Q is cot D = -cot A.

(3) Given a triangle ABC (with AB not equal to AC), take the perpendicular

bisectors of sides AB and AC and the (extended) altitude from A to the

remaining side BC. The triangle formed by the intersections of these

three lines is similar to the original one by a spiral similarity with

scale factor sin(B-C)/(2 sinB sinC).

This result came about from the special quadrilateral case (1) by

observing that you could move one of the sides of the non-parallel

pair in ABCD parallel to itself, changing the size of the ABCD, but

not its angles, and the scale factor of the spiral similarity that

related ABCD to Q remained fixed. So in the extreme, ABCD is reduced

to triangle ABC, and the perpendicular bisector of the side that is

reduced to a point becomes the altitude of the triangle formed.

Grunbaum challenges us with these other problems to be solved:

(a) determine for what other quadrilaterals ABCD the quadrilateral

Q is similar to ABCD;

(b) investigate what happens with pentagons and with hexagons if a

similar construction is carried out.

Doris Schattschneider

-----------------------

Grunbaum's Exploration

The perpendicular bisectors of the sides of quadrilateral ABCD

form a quadrilateral Q (unless ABCD is a rectangle), and the

perpendicular bisectors of the sides of Q form a quadrilateral

Q'. Show that Q' is similar to Q.

This problem was proposed by Josef Langr in the American

Mathematical Monthly in 1953 [Problem E 1050, volume 60, p.

551], and brought to our attention by Branko Grunbaum in an

article "Quadrangles, Pentagons, and Computers," in

Geombinatorics 3 (1993) 4-9. No solution to the problem has

ever been published.

The sketch shows ABCD with Q in outline and Q' shaded. Lines

have been constructed through what appear to be

corresponding vertices of ABCD and Q'. The lines are

concurrent at point P, and with Sketchpad, you can perform a

dilation with center P and ratio ±(edge of ABCD/corresponding

edge of Q') that sends Q' onto ABCD. (To mark the ratio, first

choose an edge of Q', then the corresponding edge of ABCD.)

The ratio is negative when ABCD is convex, and positive when

ABCD is nonconvex. Remember that a dilation with a negative

ratio is the composite of a halfturn through P followed by the

dilation through P with the corresponding positive ratio. By

dragging any vertex of ABCD, the similarity of Q' to ABCD is

maintained - even in the self-intersecting cases. So

Sketchpad "proves" the theorem - we still await a

conventional proof.

Grunbaum challenges us to find a full proof of Langr's problem

and determine the ratio of the dilation which somehow

depends on the shape of ABCD. Other problems to be solved: (a)

determine for what quadrilaterals ABCD the quadrilateral Q is

similar to ABCD; (b) investigate what happens with pentagons

and with hexagons.

Do we call Langr's problem a theorem? Is the strong evidence

provided by Sketchpad enough to declare that there is a

theorem, even though we don't have a conventional proof?

Notes by Jim King on Grunbaum's exploration

Branko Grunbaum has recently written about a problem of

Langr which is easy to explore with Sketchpad.

Begin with a quadrilateral Q. Define a new quadrilateral Q1

whose sides are the perpendicular bisectors of the sides of Q,

and then define Q2 to be the quadrilateral whose sides are the

perpendicular bisectors of Q1.

It is easy to do this with Sketchpad and to observe visually

that Q2 is always a dilation of Q. However, the proof of the

theorem is another matter. No published elementary proof is

known, although this result is listed by Chou in his book as

one of the theorems proved by an automatic theorem-proving

program (S-C Chou, Mechanical Geometry Theorem Proving). No

published proof of the ratio of dilation is known.

In addition, Grunbaum has observed in computer examples that

the analogous construction for pentagons, beginning with P

and then constructing P1, P2, P3, ... by taking perpendicular

bisectors of the sides, always constructs P3 a dilation of P1

(though not P2 a dilation of P). Schattschneider points out

that this raises an interesting question: if a geometric result

of this nature is established in enough examples to convince

mathematicians that it must be true, then is it a theorem or

must one await a traditional proof?

(Note: Motivated by Schattschneider's talk, Dan Bennett and

Jim King independently found an elementary proof that Q2 is a

dilation or translation of Q, but with no information about the

ratio. G.C. Shephard, in a preprint called "The Perpendicular

Bisector Construction," proves by trigonometry that the

second perpendicular quad Q2 is homothetic (a dilation) of Q

and also gives a formula in terms to trig functions for the

ratio of similitude. The formula is quite complicated. There is

no obvious geometric interpretation of the ratio; the formula

is not even obviously symmetric in the order of vertex angles.)

http://forum.swarthmore.edu/sketchpad/maa/doris.html

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-- - Dear Antreas and Co.,

You wrote and cited:

>> And now let's use "a bazooka to kill a fly".

^^^^^^^^^

>> THE PROBLEM: The perpendicular bisectors of the

>> sides of quadrilateral ABCD form a quadrilateral

>> Q (unless ABCD is a rectangle), and the

inscribed quadrilateral !!>> perpendicular bisectors of the sides of Q form a

[...]

>> quadrilateral Q'. Show that Q' is similar to

>> Q.

>>

>> More precisely, show that Q' is homothetic to Q,

>> no solution to the problem had been found.

I have not read through the complete message, but what I

see here and elsewhere makes me wonder how much computer

power is needed to prove a triviality.

Sincerely,

Darij Grinberg