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Subject: "Tell Pythagoras that V. Poncelet says 'Hi.'"     Previous Topic | Next Topic
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Conferences The CTK Exchange This and that Topic #694
Reading Topic #694
Maj. Pestich
guest
Mar-23-06, 02:11 PM (EST)
 
"Tell Pythagoras that V. Poncelet says 'Hi.'"
 
   Hello civil(ian) friends!

Let's use this time the Poncelet's theorem as a starting point.

I did not know that this theorem was associated with V.Poncelet,

but read about it in here

https://agutie.homestead.com/


There is an interesting connection between L. Euler and V. Poncelet,

and it is Russia: LE worked in St.Petersburg in Russian Academy

of Sciences, VP was a prisoner of Napoleonic war of 1812 in

Voronezh for a couple of years. It did him a lot of good, because

there he developed what has been peddled as a geometry of Poncelet,

and smuggled it accross the border after his status changed from POW

to a frenchman free to go home.


The Poncelet theorem states that in a right (angled) triangle

the radius of ex-circle touching hypotenuse c is equal to the sum

of inradius and the radii of the other two ex-circles

Rc = r + Ra + Rb

Multiplying both sides by r we have area of triangle ABC

r * Rc = r^2 + r * Ra + r * Rb

(BTW, the line connecting the touch points of ex-circles A and B

with sides AC and BC divides area of ABC in two equal parts -

nice fact in itself).

Draw lines through the incenter I parallel to sides a and b of

the triangle ABC and complete this rectangle in point D - the

antipode of C on the circumcircle of ABC. It is easy to see that

this rectangle is equal in area to triangle ABC (the area of

overlap is equal to the area of the two small 'missing' triangles

on the sides. Draw a circle through I and vertices A and B and

by intersecting chords theorem

r * Rc = (p - a) * (p - b) where p = ( a + b + c ) / 2


Doing the multiplication (p - a) * (p - b) =

(c + b - a)/2 * (c + a - b)/2 = 1/4 * (c^2 - a^2 - b^2 + 2*a*b)

which has to be equal ( as area of ABC) to a*b / 2 and for this to

happen c^2 - a^2 - b^2 has to be equal to zero.


Now I have to ask myself a question:

How come a simple nesessary condition in this theorem or a very

particular case in Euler's identity became what has been known for

so long as Pythagora's theorem ? It'seems the answer is that he was

an expert in the ancient con art of being a great salesman,

somewhat comparable to his letter-sake

modern days salesman-artist P (icasso). I'm sure that without

these marketing abilities his theorem would not have lasted

past the day when a routine scheduled circumcision brought the

welcome change from B.C. to A.D. Really.



Salute,

Maj. Pestich


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alexbadmin
Charter Member
1811 posts
Mar-27-06, 02:29 PM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
1. "RE: Tell Pythagoras that V. Poncelet says 'Hi.'"
In response to message #0
 
   Dear Major:

Could you please make a drawing for this derivation. I am getting stuck somehow and, at this point, it takes to much of an effort to paddle through.

Many thanks,
Alex


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Maj. Pestich
guest
Mar-27-06, 07:52 PM (EST)
 
2. "RE: Tell Pythagoras that V. Poncelet says 'Hi.'"
In response to message #1
 
   Hello Alex,


The drawing is very simple:

We have a triangle ABC with C=90. Complete it to a rectangle ACBD.

Inscribe an incircle into ABC. Draw parallels through incenter

to the rectangle sides. You get a smaller rectangle that shares

vertex D with ACBD, (it overlaps ABC over a small triangle),

the small rectangle has area that equals to ABC,

the sides of the small rectangle are the segments of hypotenuse made

by tangency point of incircle (they also equal to radii of ex-circles).

The core of the proof is in the expression (p-a)*(p-b)=a*b/2

where p = (a+b+c)/2 In doing multiplication for the equality

to hold the c^2 = a^2 + b^2 must be true.


This proof is very close to #33 and #42.


I hope this time you got the configuration correctly

(probably you constructed point D as reflection of C in AB,

instead of antipode, or reflection in circumcenter)

so there should be no need in the file, but anyway

how do I send a drawing in a Sketchpad file ?


Salute,

Maj. Pestich


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alexbadmin
Charter Member
1811 posts
Mar-30-06, 11:11 AM (EST)
Click to EMail alexb Click to send private message to alexb Click to view user profileClick to add this user to your buddy list  
4. "RE: Tell Pythagoras that V. Poncelet says 'Hi.'"
In response to message #2
 
   > but anyway
> how do I send a drawing in a Sketchpad file ?

You can attache a file to your message. Just below the edit control into which you type your messages there us an Attachment box with the words "Click here to choose your file" beneath it. The way to do that is to "Click here" and choose the file.

If possible make if a graphics file, not Sketchpad's.

Thank you.
Alex


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Maj. Pestich
guest
Mar-27-06, 07:52 PM (EST)
 
3. "RE: Tell Pythagoras that V. Poncelet says 'Hi.'"
In response to message #1
 
  
I think now that mine is practically the same as

the proof # 42, so just forget about it.

It is hard to come up with something different

from what's in the list already, if the proof is

from 90 deg. angle to c*c = a*a + b*b.

It is more challenging to prove it in the direction

from cc = aa + bb to the 90 deg. angle.

For example, in a circle draw a diameter AB and a

chord CD parallel to this diameter. Take a fixed point X

on diameter (not the center) and connect it to

the ends of the chord ( line segments XC, XD).

Then XC^2 + XD^2 is invariant (obviously equal

to AX^2 + BX^2. Can make a proof of P.theorem

from this configuration, but need other than P. theorem

tools to prove orthogonality.


Also, in triangle ABC with altitudes AD from A and BE

from B we have AB^2 = AC * AE + BC * BD.

Seems to be a good raw material for P. proof.


I still do not know how to prove P.T. from

parabola (to show that x-coordimates make 90 deg

triangle when y-coordinates are Yc = Ya + Yb).

A whole lot of identities can be turned into

P. theorem proof. It's like distilling alcohol,

one can do it practically from anything,

including kitchen tabouret, some even prefer

'tabouretovka' taste to 'plain vanila' wheat grains.

It was much easier for Euclid, he had the proper

sequence of the flow from one theorem to the other,

he knew exactly what follows from what. We were

taught from different

textbooks, with different flows, that is why there are

some proofs that use power of a point and inversion

to prove P. theorem, although most inversion stuff

has P. theorem. as it's foundation.


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