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| More Concyclic Points (Posted on 2014-01-26) |
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Prove the following:
If Γ 1, Γ 2, Γ 3, and Γ 4 are circles or
straight lines ( see Note below )
such that
Γ 1∩Γ 2 = {A,K},
Γ 2∩Γ 3 = {B,L},
Γ 3∩Γ 4 = {C,M}, and
Γ 4∩Γ 1 = {D,N}
then
the points A, B, C, and D are concyclic if and only if
the points K, L, M, and N are concyclic.
Note: At most only one of the points in {A,B,C,D} is the intersection
of two straight lines. In that case the corresponding
point in
{K,L,M,N} is the point at infinity ( ∞ ).
For extra credit, use this theorem to prove the theorem in
On the same line.
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Submitted by Bractals
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Rating: 5.0000 (1 votes)
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Solution:
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Let A, B, C, and D be distinct points in the extended complex plane
C∪{∞} and define the cross-ratio:
(A-C)(B-D)
[A,B;C,D] = ------------
(A-D)(B-C)
where if one of the points is the symbol ∞ , the corresponding
two differences are removed from the formula.
If the four points are finite, then there are six different values which
the cross ratio may take, depending on the order in which the points are
chosen. Let λ = [A,B;C,D]. Possible values of the cross-ratio are then
λ, 1/λ, 1-λ, 1/(1-λ), (λ-1)/λ, and λ/(λ-1).
If one of the six values is real ( ∈ℜ ), then all six are real and likewise
if one is not real ( ∉ℜ ), then all six are not real.
THEOREM: The four points are concyclic or collinear if and only if
the cross-ratio [A,B;C,D] is real.
( Without proof. It is referenced all over the web,
but I cannot find a nice proof. )
PROOF OF OUR THEOREM:
From
Γ1∩Γ2 = {A,K},
Γ2∩Γ3 = {B,L},
Γ3∩Γ4 = {C,M}, and
Γ4∩Γ1 = {D,N}
we have
{A,D,K,N} ⊂ Γ1,
{B,A,L,K} ⊂ Γ2,
{C,B,M,L} ⊂ Γ3, and
{D,C,N,M} ⊂ Γ4,
Therefore,
[A,L;B,K], [B,M;C,L], [C,N;D,M], and [D,K;A,N] are real.
But,
[A,L;B,K][C,N;D,M]
[A,C;B,D][K,M;L,N] = --------------------
[B,M;C,L][D,K;A,N]
∴ [A,C;B,D][K,M;L,N] is real.
∴ [A,C;B,D] is real if and only if [K,M;L,N] is real.
∴ A, B, C, and D are concyclic or collinear if and only if
K, L, M, and N are concyclic or collinear.
QED
EXTRA CREDIT: ON THE SAME LINE.
Since ∠PYA = ∠PZA = 90°, points P, A, Y, and Z lie on a circle;
let Γ1 be that circle. Let lines AB and BC be Γ2 and Γ3 respectively.
Since ∠PXC = ∠PYC = 90°, points P, C, X, and Y lie on a circle;
let Γ4 be that circle.
Γ1∩Γ2 = {A,Z},
Γ2∩Γ3 = {B,∞},
Γ3∩Γ4 = {C,X}, and
Γ4∩Γ1 = {P,Y}
Therefore,
P lies on the circumcircle of ΔABC
⇔ A, B, C, and P are concyclic
⇔ X, Y, Z, and ∞ are concyclic
⇔ X, Y, and Z are collinear.
QED
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