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Math Help - Unusual proof.

  1. #1
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    Unusual proof.

    I was thinking of an unusual problem in geometry. Given triangle ABC. Define a "rank" as a line-segment drawn from any vertex to any point on the opposite side. To prove: That any two ranks intersect.

    Why, do I need this? When math books on geometry prove for example that medians pass though a common point they assume that they intersect first. But if it were not always true then by contropositive their proof is destroyed. Thus, to prove that the medians pass through a common point we must first prove that the medians intersect somewhere.

    I believe I have a proof to this from analytic geometry. Construct a triangle on (0,0),(a,b),(c,0) where these points are not collinear. Then you could perhaps use elementary algebra to prove that such a point exists. But if anyone has a geometric proof I would greatly appreciate it.
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    Do you mean to specify ranks FROM DIFFERENT VERTICES?
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  3. #3
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    Quote Originally Posted by ctelady
    Do you mean to specify ranks FROM DIFFERENT VERTICES?
    Ranks from the same vertex intersect trivially, don't they?

    RonL
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    Quote Originally Posted by ThePerfectHacker
    I was thinking of an unusual problem in geometry. Given triangle ABC. Define a "rank" as a line-segment drawn from any vertex to any point on the opposite side. To prove: That any two ranks intersect.

    Why, do I need this? When math books on geometry prove for example that medians pass though a common point they assume that they intersect first. But if it were not always true then by contropositive their proof is destroyed. Thus, to prove that the medians pass through a common point we must first prove that the medians intersect somewhere.

    I believe I have a proof to this from analytic geometry. Construct a triangle on (0,0),(a,b),(c,0) where these points are not collinear. Then you could perhaps use elementary algebra to prove that such a point exists. But if anyone has a geometric proof I would greatly appreciate it.
    Consider a triangle ABC, and a "rank" a_q from vertex A which meets BC at Q
    between B and C, and another "rank" b_r from vertex B which meets AC at R
    between A and C.

    As \angle CAB = \angle CAQ + \angle QAB, \angle CAQ< \angle CAB.
    Similarly \angle RBA< \angle CBA. So \angle CAQ + \angle RBA<180^{\circ}
    and so the "ranks" are not parallel, and so meet.

    RonL
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    Quote Originally Posted by CaptainBlack
    Consider a triangle ABC, and a "rank" a_q from vertex A which meets BC at Q
    between B and C, and another "rank" b_r from vertex B which meets AC at R
    between A and C.

    As \angle CAB = \angle CAQ + \angle QAB, \angle CAQ< \angle CAB.
    Similarly \angle RBA< \angle CBA. So \angle CAQ + \angle RBA<180^{\circ}
    and so the "ranks" are not parallel, and so meet.

    RonL
    CaptainBlack, that is not a bad demonstration however I made the same mistake! I said a line-segment. Thus, the fact that these "ranks" are not parallel does not prove anything, because they have finite length.
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    Quote Originally Posted by ThePerfectHacker
    CaptainBlack, that is not a bad demonstration however I made the same mistake! I said a line-segment. Thus, the fact that these "ranks" are not parallel does not prove anything, because they have finite length.
    For most purposes you only need that the lines intersect, not that they
    intersect inside the triangle. In fact the example you quote only requires
    that there be a point of intersection not that it be inside the triangle.

    However I'm sure that we can show that the point of intersection is inside
    the triangle. (Once we decide what it means for a point to be inside a
    triangle, and develop some theorems based on the definition we plump
    for)

    RonL
    Last edited by CaptainBlack; February 2nd 2006 at 10:10 PM.
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    Quote Originally Posted by CaptainBlack
    For most purposes you only need that the lines intersect, not that they
    intersect inside the triangle. In fact the example you quote only requires
    that there be a point of intersection not that it be inside the triangle.

    However I'm sure that we can show that the point of intersection is inside
    the triangle. (Once we decide what it means for a point to be inside a
    triangle, and develop some theorems based on the definition we plump
    for)

    RonL
    I agree with you that showing that extended "ranks" interesect is the important fact, not the "ranks" themselves.

    I already in the first post said we may show it algebraically buy constructing it in the x-y plane, but this is not fun. And messy
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    Quote Originally Posted by ThePerfectHacker
    I agree with you that showing that extended "ranks" interesect is the important fact, not the "ranks" themselves.

    I already in the first post said we may show it algebraically buy constructing it in the x-y plane, but this is not fun. And messy
    It might be more tricky than you think. Consider what it means for a point
    to be inside a triangle (or better still any simple closed curve).

    You may well be making an assumption about how you know a point is
    inside the triangle.

    RonL
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    Quote Originally Posted by CaptainBlack
    It might be more tricky than you think. Consider what it means for a point
    to be inside a triangle (or better still any simple closed curve).

    You may well be making an assumption about how you know a point is
    inside the triangle.

    RonL
    I believe that to show that they intersect inside, you need to show that a unique solution exists on a certain interval then the problem of showing that they exist inside is resolved.
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  10. #10
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    Quote Originally Posted by Rich B.
    Gentlemen:

    Def(triangle interior): Let "I" denote "interior of triangle(ABC)".
    then, I = {P,Q in Plane(ABC) : sgmnt(PQ) intrsct trngl(ABC) = {} for all P,Q in I.
    This is OK for the triangle, but it would be better that the interior be
    defined for more general simple closed curves than the triangle.

    RonL
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  11. #11
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    Okay?

    Def(interior of a triangle): Let point P lie in [Plane(ABC) - triangle(ABC)]. P lies in the Int(tri(ABC)) iff there exist points Q and R s.t. {Q,R} s.set tri(ABC) and P lies in sgmt(QR). [Restated: The interior of tri(ABC) is the set of all points, P, such that P lies between two points on the triangle (recall definition of betweenness). Of course, P lies on infinitely many segments, the endpoints of which lie on the triangle but, one is sufficient.]

    Case 1: Two ranks "emanating" from the same triangle vertex intersect, trivially, "on" the triangle.

    Case 2: By definition, the endpoints of a triangle's rank are elements of that triangle. Hence, according to the definition of "triangle interior", if sgmt(AR) is a rank of tri(ABC), then, sgmt(AR) - {AR} s.set Int(tri(ABC)). By PPT (previously proved theorem), the intersection of two distinct ranks is non-empty (Ron L.). Hence, provided those ranks do not share a common endpoint (case 1), it follows that such intersection lies within the triangle's interior, thus completing the proof.

    Regards,

    Rich B.
    Last edited by Rich B.; February 11th 2006 at 09:48 AM.
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