triangle with minial perimeter

Discussion in 'Recreational Math' started by Gunnar G, Mar 14, 2005.

  1. Gunnar G

    Gunnar G Guest

    The problem is as follow:
    Give an acute triangle ABC, find a triangle DEF with D on BC, E on AC and F
    on AB such that DEF has a minimal perimeter.

    I solved it by taking D as the shortest distance from BC to A, and then
    reflect the point D in AB and AC to get points G and H, draw the lines GD
    and HD to find the intersections with AB and AC respectivly, there you have
    the points E and F. I left out some of the reasoning for why this is

    Now, that took me a few days and many triangles to solve. I also tried to
    find this analyticaly. I managed to find the solution for where to put the
    points when I have a isosceles triangle, (at height 2*a^2*L/(a^2+L^2) from
    the odd side, where L is the common side, a is the half of the third side)
    but the question is: how can I find an analytical solution to this problem
    for the general acute triangle? There are three variables I can play with,
    and maple didn't like it any more than I do. Equations of the 6th grade...
    If you have a solution, I'm interested of seeing this.
    Gunnar G, Mar 14, 2005
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  2. Gunnar G

    Bob Pease Guest

    I would appeciate seeing the proof of the first part.

    If it is minimal, is it unique?

    I'll work on an analytic solution..

    BOb Pease
    Bob Pease, Mar 15, 2005
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  3. Gunnar G

    Ken Pledger Guest

    See "Schwarz's triangle problem" in, for example, Richard Courant
    & Herbert Robbins, "What is Mathematics?" (Look up the index for the
    page numbers.)

    Ken Pledger.
    Ken Pledger, Mar 15, 2005
  4. Gunnar G

    Philippe 92 Guest


    Bob Pease wrote :
    Expressely as it seems to be false...
    D is OK, but E and F should be intersections of *GH* with AB, AC

    Given any point D on BC.
    G symetric of D wrt AB, H symetric of D wrt AC
    Choose any two points E on AC and F on AB
    FG = FD, EH = ED hence DF + FE + ED = GF + FE + EH
    is minimum when G, F, E, H in line, shortest line between G and H.

    Now the problem is to find D for which GH is minimum.
    AG = AD = AH
    G,D,H are on a circle centered in A going through D
    G lie on the symetric BS of BC wrt AB
    H on the symetric CR of BC wrt AC, BS and CR are fixed lines.
    G and H are then be constructed as intersections of circle centered in
    and radius AD with lines BS and CR.
    GH will be minimum when this circle is the smallest possible,
    that is tangent to BC (in D, and to BS in G and to CR in H).
    Hence AD is the altitude of ABC.
    Symetrically (exchanging A,B,C), CF and BE are also the altitudes
    and DEF is the so called "orthic" triangle of ABC.

    This triangle is inside ABC iff ABC is an acute triangle.
    The problem is also known as Fagano's problem

    I suspect an analytic solution to be much harder...

    Best regards.
    Philippe 92, Mar 15, 2005
  5. Gunnar G

    Bob Pease Guest


    I got a nice proof from

    Quoting some more...i

    "This shows that the orthic triangle solves Fagnano's problem.

    The solution is unique because any triangle with the minimum perimeter
    should possess the mirror property. As we know, the mirror property is a
    characteristic feature of the orthic triangle."

    The problem is that the "Mirror Property" is used to prove minimality
    because of some kind of Metric Euclidean Triangle axiom..( OK, so far...)

    Although it SEEMS obvious, and I believe it, I don't follow the logic of the


    Bob Pease, Mar 15, 2005
  6. Gunnar G

    Philippe 92 Guest

    Bob Pease wrote :
    correction : Fagnano, my fault...
    I don't either, that's why the H.A.Schwarz's solution seems to be a messed
    up logic statement :
    The searched triangle has "Mirror Property" [MP]
    The orthic triangle has the [MP]
    hence the searched triangle is the orthic triangle. [wrong logic]

    The cheating trick is they say without a proof (or the proof is in
    another page... "as we know") :
    "the [MP] is *characteristic* of the orthic triangle"
    That is no other triangle than the orthic triangle has [MP].
    Or in other words :
    any triangle with [MP] is the orthic triangle
    the searched triangle has [MP]
    hence the searched triangle is the orthic triangle [true]

    For a one page full proof of Fagnano, the best is L.Fejér's proof
    (a little bit lower in Fagnano.shtml page)
    .... but discard the [MP] :
    Choose any D and search min triangle *for that D* :
    the min triangle has perimeter P'P" (cut the knot notation), or GH (OP notation)
    just because the shortest P'QRP" = GEFH line is a straight line.
    Then find P = D for which P'P" = GH is minimum : AP = AD minimum.
    Hence AD is the altitude.
    As the problem is symetric in A,B,C, the same proof choosing first E or F results into :
    BE and CF are also altitudes. QED.

    That was roughly the proof I gave in my previous post.

    Philippe 92, Mar 15, 2005
  7. Gunnar G

    Bob Pease Guest

    Thanks for sharing my concern here .
    I guess I need to see an IFF proof of thie orthic triangle business
    Or it may be beyond the scope of this ng to persue this

    Bob Pease, Mar 15, 2005
  8. Gunnar G

    Philippe 92 Guest

    Bob Pease wrote :
    Everything is in the cut-the-knot pages...
    with some personal work left to the reader !

    First we have to define exactly what is the "Mirror Property" [MP].

    Given a triangle ABC and three points DEF on the sides.
    (A on BC, D on AC, F on AB)
    Triangle DEF is said to have the [MP] iff :
    a light ray DE reflects on the side AC as ray EF
    EF reflects on AB as FD and FD reflects on BC as DE.
    That is the light ray turns for ever along the DEF triangle sides !

    I. The orthic triangle has the [MP].
    But I had to rework it entirely to figure out exactly what angles
    are concerned (with the above notations) :
    AEDB concyclic (AEB and ADB = 90deg) => angles ADE = ABE
    BFEC concyclic (BFC and BEC = 90deg) => angles FBE = FCE
    AFDC concyclic (AFC and ADC = 90deg) => angles FCA = FDA
    that is ADE = FDA (reflected = incident)
    Same proof for other vertices. QED.

    II. The orthic triangle is the only triangle with the [MP].
    Here also some work is left to the reader...
    Consider a [MP] triangle.
    Angles FDB = EDC = u (complement of incident/reflected angles)
    Angles DEC = FEA = v
    Angles AFE = DFB = w
    In triangle AFE : A + v + w = 180
    similarly B + u + w = 180 and C + u + v = 180
    Angle EDF = 180 - 2u
    similarly FED = 180 - 2v and EFD = 180 - 2w
    In triangle EDF : EDF + DEF + EFD = 180 => 2(u + v + w) = 2*180
    => u = A and similarly B = v and C = w
    (the A' = alpha ... missing proof in cut-the-knot)
    => all [MP] triangles have their sides parallel to fixed directions
    And now the figure in the ABisector.shtml#mirror page clearly
    shows there can be only one [MP] triangle.

    III. Back to Fagnano... pure recreational way.
    Consider a rubber band DEF constrained by DEF lying on the ABC sides.
    Let DEF freely move on the sides.
    At equilibrum, DEF is the Fagnano triangle (minimum perimeter).
    Consider the forces at point D.
    The resultant of the tensions in DE and DF is perpendicular to
    the side BC (equilibrum).
    But the two tensions being equal (tension of rubber band), the resultant
    lies on the angle bissector of EDF.
    => the Fagnano triangle has the [MP].

    As I. and II. say that the only [MP] triangle is the orthic triangle
    => Fagnano triangle is the orthic triangle ! (yet another proof)

    And yet another "eXtended Mirror Property" [XMP]:
    A set of points DEFGH...XYZ lying on the sides of a triangle ABC
    has the [XMP] iff :
    DE ray reflects to EF, EF to FG ... XY to YZ and YZ to ZD.
    (the light ray turns more than once before going back to first point, then
    repeats forever)

    Find sets of [XMP] points, if any, with more than 3 points.
    Or proove the only solution is 3 points and the orthic triangle above.

    Philippe 92, Mar 15, 2005
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