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Thread: 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 $\displaystyle ABCD$ are constructed equilateral triangles $\displaystyle WAB , XBC , YCD , ZDA$ with centroids $\displaystyle S_1 , S_2 , S_3 , S_4$ , respectively . Prove that $\displaystyle S_1 S_3$ is perpendicular to $\displaystyle S_2 S_4$ if and only if $\displaystyle AC = BD$ .
    ( Mediterranean Mathematical Competition 2000 )

    In fact , equilateral triangles can be changed to squares with centers instead of centroids .
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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
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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 $\displaystyle ABCD $ , $\displaystyle AC $ is perpendicular to $\displaystyle BD $ if and only if $\displaystyle AB^2 + CD^2 = BC^2 + DA^2 $ so if we want to prove $\displaystyle S_1 S_3 $ being perpendicular to $\displaystyle S_2 S_4 $ , it is sufficient to show $\displaystyle S_1S_2 ^2 + S_3 S_4 ^2 = S_2 S_3 ^2 + S_4 S_1 ^2 $ , vice versa .

    Consider $\displaystyle \Delta S_1 B S_2 $ ,

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

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

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

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

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

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


    Therefore , $\displaystyle S_1S_2 ^2 + S_3 S_4 ^2 $

    $\displaystyle = \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) ) ] $

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

    It is not hard to show that :

    $\displaystyle 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 , $\displaystyle (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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