# Proving orthogonal ?

• Jul 19th 2010, 06:23 PM
simplependulum
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 .
• Jul 23rd 2010, 03:45 AM
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 . 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 . (Wondering) ( At the same time , i am planning to start a group concentrated in integrals . (Happy) )
• Aug 8th 2010, 05:51 PM
TheCoffeeMachine
Quote:

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).
Quote:

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 . (Wondering) ( At the same time , i am planning to start a group concentrated in integrals . (Happy))
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). (Happy)
• Aug 8th 2010, 06:27 PM
Also sprach Zarathustra
Be careful!

• Aug 8th 2010, 08:54 PM
simplependulum
Quote:

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 .

Quote:

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 .