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,(t1/t)/I}/2;
T[A_,B_,C_,D_]:=((A+BCD)^24*(A*B+C*D))^264*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 sintheorem 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.
