1. ## 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 .

2. 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 . )

3. Originally Posted by simplependulum
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).

4. Be careful!

5. Originally Posted by Also sprach Zarathustra
Be careful!

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 .