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## correct Blaschke formula (Re: [EMHL] Area of a quadrilateral)

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• Hello Marcello, The formula you are asking for is the following. Let ABCD be (not nesessary convex) quadrilateral with sides AB=k, BC=l, CD=m, DA=n and angles
Message 1 of 2 , Sep 30, 2002
Hello Marcello,

The formula you are asking for is the following. Let ABCD be (not
nesessary convex) quadrilateral with sides AB=k, BC=l, CD=m, DA=n and
angles 2a, 2b, 2c, 2d at vertices A,B,C,D respectively; and let S be the
area of the quadrilateral.

Then:

4*S = (k+l+m+n)^2/(cot(a)+cot(b)+cot(c)+cot(d)) -
(k-l+m-n)^2/(tan(a)+tan(b)+tan(c)+tan(d))

As far as I know, this formula was first discovered by german
mathematition Wilhelm Blaschke. First appeared in 1914 in:

Jahresbericht Deutsch. Mathem. Vereinigung 23, 1914, page 210-234

Plain geometry proof and several computer-based proofs of this formula can
be found in the article:

"Russian killer No. 2, A challenging geometric theorem with human and
machine proofs" by Xiaorong Hou, Hongbo Li, Dongming Wang and Lu Yang
Mathematical Intelligencer, Volume 23, number 1, 2001, pages: 9-15

http://groups.yahoo.com/group/Hyacinthos/files/killers.pdf

Size of file = 1 387 554 bytes, PDF version 1.2 (so you will need Adobe
Acrobat Reader version 3.0 or later).

Sergei Markelov
• Dear Sergei, Since you were the one who sent the Intelligencer article authors the 5 Russian killers, can you post the 5 problems here please? Best regards
Message 2 of 2 , Sep 30, 2002
Dear Sergei,

Since you were the one who sent the Intelligencer article authors the 5
Russian killers,
can you post the 5 problems here please?

Best regards
Sincerely,
Paul

==============
Hyacinthos 2979
From: yiu@...
Date: Thu Jun 7, 2001 11:07 am
Subject: 5 Russian killers

Dear friends,

A recent issue of Mathematical Intelligencer* contains an
article by 4 Chinese authors on Number 2 of 5 Russian killers.
They
wrote that ``these theorems have been used to prepare the
Moscow team
for the all-Russian school mathematics olympiad, and are called
killers to analytic ways of geometric problem-solving''. Here is
Number 2.

Theorem. Let ABCD be an arbitrary quadrilateral with sides AB
= k, BC
= ell, CD = m, DA = n, and internal angles 2a, 2b, 2c, 2d at
vertices
A, B, C, D respectively; and let S be the area of the
Then

4S = (k+ell+m+n)^2/F - (k-ell+m-n)^2/G,

where

F = cot a + cot b + cot c + cot d,

G = tan a + tan b + tan c + tan d.

Does any one know what the other killers are?

Best regards
Sincerely,
Paul

*volume 23 (2001) 9 -- 15.
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