CTK Exchange
Front Page
Movie shortcuts
Personal info
Awards
Reciprocal links
Terms of use
Privacy Policy

Interactive Activities

Cut The Knot!
MSET99 Talk
Games & Puzzles
Arithmetic/Algebra
Geometry
Probability
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
My Logo
Math Poll
Other Math sit's
Guest book
News sit's

Recommend this site

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Products to download and subscription Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: "Mahavira's formula for cyclic quadrilaterals"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange College math Topic #612
Reading Topic #612
melefthe
guest
Apr-11-07, 07:21 PM (EST)
 
"Mahavira's formula for cyclic quadrilaterals"
 
   Does anyone know the proof of Mahavira's formulae for the calculation of the inner diagonals of the cyclic quadrilaterals? Presumably, they must be a continuation of Ptolemy's theorem. They are:

x=Sq.Root<(ad+bc)/(ab+cd)x(ac+bd)> and symmetrically for y.

Thanks.


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

  Subject     Author     Message Date     ID  
Mahavira's formula for cyclic quadrilaterals melefthe Apr-11-07 TOP
  RE: Mahavira's formula for cyclic quadrilaterals Pierre Charland May-11-07 1
     RE: Mahavira's formula for cyclic quadrilaterals melefthe May-19-07 2
         RE: Mahavira's formula for cyclic quadrilaterals alexb May-19-07 3
             RE: Mahavira's formula for cyclic quadrilaterals melefthe May-20-07 4
                 RE: Mahavira's formula for cyclic quadrilaterals alexb May-20-07 5
         RE: Mahavira's formula for cyclic quadrilaterals Pierre Charland May-27-07 6

Conferences | Forums | Topics | Previous Topic | Next Topic
Pierre Charland
Member since Dec-22-05
May-11-07, 06:40 AM (EST)
Click to EMail Pierre%20Charland Click to send private message to Pierre%20Charland Click to view user profileClick to add this user to your buddy list  
1. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #0
 
   I found a partial proof in
Howard Eves, Great Moments in Mathematics Before 1650, 10.5 p.108

Let ABCD be a cyclic quadrilateral of diameter t.
Let BD=m and AC=n.
Let theta = the angle between either diagonal and the perpendicular upon the other.

Then, (using triangle's formula ab=2hR applied to DAB and DCB), we get
mt cos theta = ab+cd
nt cos theta = ad+bc

So m/n = (ab+cd)/(ad+bc)

Also mn = ac+bd (Ptolemy relation)

Multiplying those last 2 equations, we get:
m^2=(ab+cd)(ac+bd)/(ad+bc)

Dividing instead, we get:
n^2=(ac+bd)(ad+bc)/(ab+cd)

===========================
I also found some info there:
https://www.pballew.net/cycquad.html
https://www.vias.org/comp_geometry/geom_quad_cyclic.html
https://www.answers.com/topic/mahavira-mathematician

AlphaChapMtl


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
melefthe
guest
May-19-07, 03:49 PM (EST)
 
2. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #1
 
   I do not understand it unfortunately. What formula are you referring to? What quantity is m.t.cos» ? Could you please explain?


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
alexb
Charter Member
2013 posts
May-19-07, 03:56 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  
3. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #2
 
   Have a look at

https://www.cut-the-knot.org/proofs/ptolemy.shtml#Mahavira

It's the same as Pierre's remark but perhas a little more detailed. What I want you to do is to make a diagram where you mark the angle theta and think of what cos(theta) signifies, OK?


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
melefthe
guest
May-20-07, 10:28 AM (EST)
 
4. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #3
 
   Dear Alex,

Sorry, still nothing! I do not understand anything! Are you sure there is not a flaw in the argument? I have a feeling there is. Could you please send me a diagram and an explanation? You may draw it by hand and fax it through at +30-210-7219225 if this is easier for you.

The formula is proven very easily by using the cosine rule.

m^2=b^2+c^2-2.b.c.cosC
=a^2+d^2-2.a.d.cosA=a^2+d^2-2.a.d.cos(180-C)=a^2+d^2+2.a.d.cosC
Multiply the first with ad and the second with bc, add them up and you will arrive at m^2=(ab+cd).(ac+bd)/(ad+bc)

I am really surprised. I do not understand what cosŤ signifies. There might be an error in the proof.

Have a bash at the probability question I posed. I think I have got an answer, it depends on what one is really asking. I look forward to see the replies.


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
alexb
Charter Member
2013 posts
May-20-07, 04:59 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  
5. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #4
 
  

    ab = hR - can you verify that? It'says that in a triangle the product of two sides equals the product of the diameter circumscribed circle and the altitude to the third side.

  1. In a quadrilateral, you have two pairs of triangles sharing a base (the third side in 1). Bases are the diagonals of the quadrilateral. Consider one pair at a time, say, ab = ht and cd = gt, where h is the altitude to from the ab vertex to the diagonal m, I belive. g is the altitude from the cd vertex to the sam diagonal.

  2. Each of these triangles contains a piece of the diagonal n which plays the hypotenuse, one in a triangle with side h, the other in a triangle with side g. The triangles are similar. You can write

    h = cos(theta)×(one piece) and
    g = cos(theta)×(the other piece).

    On summing up you get

    h + g = cos(theta)·n

    Thus adding ab = ht and cd = gt gives you

    nt·cos(theta) = ab + cd.

Sorry if I confused m and n.


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
Pierre Charland
Member since Dec-22-05
May-27-07, 06:43 PM (EST)
Click to EMail Pierre%20Charland Click to send private message to Pierre%20Charland Click to view user profileClick to add this user to your buddy list  
6. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #2
 
   sorry for my late reply,
the angle theta is the angle between either diagonal and the perpendicular upon the other.
Four of them are marked is the included image.

AlphaChapMtl

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/465a14945fea12de.jpg

  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

Conferences | Forums | Topics | Previous Topic | Next Topic

You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.

Copyright © 1996-2018 Alexander Bogomolny

Search:
Keywords:

Google
Web CTK