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Math Help - Proving orthogonal ?

  1. #1
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    Proving orthogonal ?

    This problem is very special , there is a beautiful theorem behind this , see if you can discover it !


    Problem :

    Outwards along the sides of convex quadrilateral ABCD are constructed equilateral triangles WAB , XBC , YCD , ZDA with centroids S_1 , S_2 , S_3 , S_4 , respectively . Prove that S_1 S_3 is perpendicular to S_2 S_4 if and only if AC = BD .
    ( Mediterranean Mathematical Competition 2000 )

    In fact , equilateral triangles can be changed to squares with centers instead of centroids .
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  2. #2
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    Or this ( the theorem i mentioned ) appears too indirect , by the way , the suggested method is to use complex numbers skilfully and my method is beautiful and symmetric . I've been planning to organize a MHF study group about topics in elementary geometry , especially problems coming from national and international contests . What i lack is not the materials but enthusiasts , here I would like to see how many people being interested in my ideas . ( At the same time , i am planning to start a group concentrated in integrals . )
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    Quote Originally Posted by simplependulum View Post
    Or this ( the theorem i mentioned ) appears too indirect , by the way , the suggested method is to use complex numbers skilfully and my method is beautiful and symmetric .
    Could you post the solution, please? My geometry is not very good, so I couldn't get anywhere (other than drawing a circle and playing with the vectors from the origin for a while).
    I've been planning to organize a MHF study group about topics in elementary geometry , especially problems coming from national and international contests . What i lack is not the materials but enthusiasts , here I would like to see how many people being interested in my ideas . ( At the same time , i am planning to start a group concentrated in integrals . )
    Whatever happened to this idea, by the way? I'm interested, and I'm sure many other members would be. (I'm particularly keen on the Integrals one).
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Be careful!

    The answer in the link!

    Mathematical olympiads 2000-2001 ... - Google
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  5. #5
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Be careful!

    The answer in the link!

    Mathematical olympiads 2000-2001 ... - Google
    Yeah , this is it , i copied the problem from this book .

    However , my method doesn't involve any complex numbers , it is rather elementary .


    Could you post the solution, please? My geometry is not very good, so I couldn't get anywhere (other than drawing a circle and playing with the vectors from the origin for a while).

    The theorem is :

    In convex quardrilateral  ABCD ,  AC is perpendicular to  BD if and only if  AB^2 + CD^2 = BC^2 + DA^2 so if we want to prove  S_1 S_3 being perpendicular to  S_2 S_4 , it is sufficient to show  S_1S_2 ^2 +  S_3 S_4 ^2  = S_2 S_3 ^2 + S_4 S_1 ^2 , vice versa .

    Consider  \Delta S_1 B S_2 ,

     BS_1 = \frac{1}{\sqrt{3}} AB  ,  BS_2 = \frac{1}{\sqrt{3}}  BC and  \angle S_1 B S_2  = 60^o + \angle ABC

    We have  S_1 S_2 ^2 = (BS_1)^2 + (BS_2)^2 - 2(BS_1)(BS_2) \cos(\angle S_1 B S_2 )

     = \frac{1}{3} [ AB^2 + BC^2  -  (AB)(BC) (\cos(\angle ABC) + \sqrt{3} (AB)(BC) \sin(\angle ABC) ) ]

     = \frac{1}{3} [ AB^2 + BC^2  -  \frac{ AB^2 + BC^2 - AC^2 }{2}  + \sqrt{3} (AB)(BC) \sin(\angle ABC) ) ]

     = \frac{1}{3} [  \frac{ AB^2 + BC^2 + AC^2 }{2} + \sqrt{3}  (AB)(BC) \sin(\angle ABC) ) ]

    Similarly ,  S_3 S_4 ^2 =  \frac{1}{3} [  \frac{ DA^2 + CD^2 + AC^2 }{2} + \sqrt{3} (DA)(CD) \sin(\angle ADC) ) ]


    Therefore ,  S_1S_2 ^2 +  S_3 S_4 ^2

     = \frac{1}{3} [ \frac{ AB^2 + BC^2 + CD^2 + DA^2 }{2}  + AC^2 + \sqrt{3} ( (AB)(BC) \sin(\angle ABC) + (DA)(CD) \sin(\angle ADC) ) ]

     = \frac{1}{3} [  \frac{ AB^2 + BC^2 + CD^2 + DA^2 }{2}  + AC^2 + 2 \sqrt{3} S ] where  S denotes the area of  ABCD .

    It is not hard to show that :

     S_2 S_3 ^2 + S_4 S_1 ^2  =  \frac{1}{3} [  \frac{ AB^2 + BC^2 + CD^2 + DA^2 }{2}  + BD^2 + 2 \sqrt{3} S ]

    Finally ,  (S_1S_2 ^2 +  S_3 S_4 ^2) - ( S_2 S_3 ^2 + S_4 S_1 ^2)  = \frac{1}{3} ( AC^2 - BD^2 )  so by this theorem we get the answer .
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