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Quadrilaterals and Perpendicular bisectors

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  • Antreas P. Hatzipolakis
    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
    Message 1 of 2 , Jan 11, 2004
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      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@...>,
      > 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

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

      A nice analysis, indeed, Ilias!

      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.


      >
      >
      > 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

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

      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

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

      --
    • Darij Grinberg
      Dear Antreas and Co., ... ^^^^^^^^^ inscribed quadrilateral !! ... [...] ... I have not read through the complete message, but what I see here and elsewhere
      Message 2 of 2 , Jan 11, 2004
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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
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