THEOREM: If ABCD is a quadrilateral on a circle, then
|AB| |BC| |CA| + |AC| |CD| |DA| = |BC| |CD| |DB| + |AB| |BD| |DA|
Proof. (This is a Mathematica version of a Maple proof found by
Shalosh B. Ekhad,
the computer collaborator of Doron Zeilberg
at Rutgers university).
H[A_,B_]:=(A[[1]]-B[[1]])^2+(A[[2]]-B[[2]])^2;
P[t_]:={t+1/t,(t-1/t)/I}/2;
T[A_,B_,C_,D_]:=((A+B-C-D)^2-4*(A*B+C*D))^2-64*A*B*C*D;
U[A_,B_,C_]:=H[A,B]*H[B,C]*H[C,A];
K={P[t[1]],P[t[2]],P[t[3]],P[t[4]]};
S=T[U[P[t[2]],P[t[3]],P[t[4]]],U[P[t[1]],P[t[3]],P[t[4]]],
U[P[t[1]],P[t[2]],P[t[4]]],U[P[t[1]],P[t[2]],P[t[3]]]]==0;
Simplify[S]
Side remark:
Ekhads Maple Proof
uses a 360 Character Maple code, which proves it in 0.05 seconds. The above Mathematica code has 336
characters and needed 0.01 seconds to complete on a Pentium D 840.
Remark added Mai 4, 2007: The swiss mathematician
Roland Stärk
informed us of an elegant proof of Sadovs theorem: a triangle A,B,C with sides a,b,c has area
A = a b sin(gamma)/2 = abc/(4r)
if r is the radius of the circle through A,B,C.
[This formula uses a basic property of the cross product and
the sin-theorem sin(gamma)/c = 2/r, applied to the right angle triangle A,(A+B)/2,(A+B+C)/3.]
Sadovs formula just expresses the area of the quadrilateral in two different ways.
Stärk wrote us that conversely, the formula
|AB| |BC| |CA| + |AC| |CD| |DA| = |BC| |CD| |DB| + |AB| |BD| |DA|
implies that the quadrilateral triangle is on a circle but that this is harder to prove.
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