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Math Help - Difficult Geometry

  1. #1
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    Difficult Geometry

    Let
    ABCD be a quadrilateral having an in-circle. Suppose four circles can be fit inside the quadrilateral such that each circle is externally tangent to two other circles and tangent to two sides of a quadrilateral. Prove that among the four circles at least two are congruent.

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  2. #2
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    Say the 4 smaller circles are Q,R,S,T.
    R is tangent to Q and S

    The quadrilateral is ABCD.
    The incircle is P.

    All the 5 circles are named after their respective centerpoints.

    Incircle P is tangent to AB at point E. So, PE is perpendicular to AB.
    Incircle P is tangent to BC at point F. So, PF is perpendicular to BC.

    Circle R is tangent to side AB at point G, and tangent to BC at point H.
    Circle Q is tangent to side AB at point J....
    Circle S is tangent to side BC at point K....

    -------------------------------
    Consider circles R and Q.

    If R and Q are tangent to each other, then their point of tangency must be along the apothem (inradius) PE. Call that point, X.
    Then line RXQ is a straight line segment.
    And RQ is perpendicular to PE.

    If circlpe Q is tangent to AB, then radius QJ is perpendicular to AB.
    Likewise, if circle R is tangent to AB, then radius RG is perpendicular to AB also.
    Hence, since RG, PE, QJ are all perpendicular to AB, then RG, PE, QJ are parallel.

    Line GEJ is a straight line segment, because points G,E,J are on AB.
    So, GJ is perpendicular to RG, PE, QJ.
    Hence, RQJG is a rectangle.

    In circle R, RG = RX. Both are radii of R.
    In circle Q, QJ = QX. Both are radii of Q.

    In rectangle RQJG, RG = XE = QJ.
    So, RG = QJ
    Therefore, circle R is congruent to circle Q, because they have equal radii.

    -------------------------------
    Repeat that for circles R and S.
    It will be the same. Circles R and S are congruent too.

    Blah, blah, blah, all four smaller circles are congruent.
    And ABCD is a square---cannot be otherwise.
    Last edited by ticbol; September 14th 2007 at 10:03 PM. Reason: J, not H
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  3. #3
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    ticbol wrote

    "Consider circles R and Q.

    If R and Q are tangent to each other, then their point of tangency must be along the apothem (inradius) PE. Call that point, X.
    Then line RXQ is a straight line segment.
    And RQ is perpendicular to PE."

    Why must their point of tangency be along the apothem PE? Not necessarily. Check your diagrams once more, please.
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  4. #4
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    Quote Originally Posted by chessfreak View Post
    ticbol wrote

    "Consider circles R and Q.

    If R and Q are tangent to each other, then their point of tangency must be along the apothem (inradius) PE. Call that point, X.
    Then line RXQ is a straight line segment.
    And RQ is perpendicular to PE."

    Why must their point of tangency be along the apothem PE? Not necessarily. Check your diagrams once more, please.
    Their point of tangency has to be along the apothem PE.
    Circle R needs to be tangent to AB, BC, circle Q and circle S. So R must be confined inside the quadrilateral PEBF. In the same manner, Q is confined inside PEA_. All the 4 smaller circles need to be confined inside their respective quadrilaterals that are each defined by the two adjacent sides of ABCD and the two adjacent apothems.
    Now, where will the circles touch each other?

    I tried a few quadrilateral ABCD's of varying lengths of sides---not squares. Nothing worked except when ABCD is a square. Because, as in the case of circles R and Q, the apothem between them and a radius of each circle muat be parallel, being all aare perpendicular to AB.
    Then RQ must be parallel to AB. That's why the point of tangency of the two circles, X, must be on PE.
    If RQ were not parallel to AB---or GJ---then X need not be on PE. But then there are problems with circles T or S to be able to be tangents to what is said in your question.

    I tried the different diagrams crudely by sketching on paper only as I don't know how to draw in the computer.
    Maybe, if you can draw in your computer, try an ABCD that is not a square and see if it is possible for Q,R,S,T circles to follow what are rtequired of them---each tangent to four surfaces.
    I believe each of Q,R,S,T needs to be confined within adjacent apothems. And their points of tangency are along the apothems only.
    If you could do that, would you please post the diagram here?
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  5. #5
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    Quote Originally Posted by chessfreak View Post
    Let





    ABCD be a quadrilateral having an in-circle. Suppose four circles can be fit inside the quadrilateral such that each circle is externally tangent to two other circles and tangent to two sides of a quadrilateral. Prove that among the four circles at least two are congruent.
    I'm a bad MS Paint artist so I'll try to describe the diagram. Let r denote the inradius of the ABCD. and r_1, r_2, r_3, r_4 the radii of the smaller circles O_1, O_2, O_3, O_4, named clockwise. Let O_1 and O_2 be tangent to segment AB at points X and X' respectively, and denote the point of tangency of the incircle of ABCD as Z. Let AX=s_1 and BX' = s_2. It is readily seen that XX' = 2\sqrt{r_1r_2}. We also have AZ = as_1, \ BZ = bs_2 for some a, b > 1 Thus s_1+2\sqrt{r_1r_2}+s_2 = as_1+bs_2. Defining the points and lengths similarly for the other sides of ABCD, we find that 2(as_1+bs_2+cs_3+ds_4) = 2(s_1+s_2+s_3+s_4) + \Sigma_{cyc}2\sqrt{r_1r_2} or \Sigma_{cyc}2\sqrt{r_1r_2}=2((a-1)s_1+(b-1)s_2+(c-1)s_3+(d-1)s_4) = 4\sqrt{r_1r_2}+4\sqrt{r_3r_4}. So \sqrt{r_2r_3}+\sqrt{r_4r_1} = \sqrt{r_1r_2}+\sqrt{r_3r_4}, and the result follows.
    Last edited by mathisfun1; September 19th 2007 at 04:43 AM.
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