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8.5. THE THREE-POINT CONDITION AND QUADRILATERALS 107

Chooses with r < s < t. Since
z1, z2 E 8D n B(zo, s),
we can find for j = 1, 2 an endcut Oj of D which joins Z j to a point zj E D and
which lies in D n B(z 0 , s). Next by (8.4.2) we can find an arc a 3 joining z~ and z~
in D n B ( zo, cs). Then a 1 U a2 U a3 contains a crosscut a of D which joins z 1 , z 2
in B(zo, cs).
The same argument applied to w 1 , w 2 yields a crosscut /3 of D which joins
w1, w2 in D \ B(zo, ct). Then
an /3 c B(z 0 , cs)\ B(z 0 , ct)= 0
since s < t, while the fact that z1, z2 separate w1, w2 in 8D implies that an /3 -1- 0
and we have a contradiction. 0

8.5. The three-point condition and quadrilaterals
We show next that the three-point condition for a Jordan domain D implies
an important relation between conjugate quadrilaterals in D and its exterior D*.

LEMMA 8.5.1. Suppose that DC R^2 is a Jordan domain and that bis a constant
such that for each z 1 , z 2 E 8D \ { oo}
(8.5.2)

where -y 1 , -y 2 are the components of 8D \ {z 1 , z 2 }. If Q and Q* are conjugate
quadrilaterals in D and D* and if mod( Q) = 1, then
(8.5.3) mod( Q*) ~ c,
where c = c(b).

PROOF. Let a 1 and a 2 denote two opposite sides of Q and Q* and let r and
r* denote the family of arcs which join a 1 and a2 in Q and Q*, respectively. Then
mod(Q*) = mod(r*),
and hence it suffices to shows that

mod(r) = mod(Q) = 1


(8.5.4)
Choose z 1 E a 1 and z2 E a2 so that
lz1 - z2I = dist(a1, a2) = r
and let
s = min diam(aj) < oo.
J=l,2

We show first that


(8.5.5)

We may assume that 2b^2 r < s since otherwise (8.5.5) follows trivially.
Let -y 1 , -y 2 denote the components of 8D \ {z 1 , z2}. Then by (8.5.2) and rela-
beling, if necessary, we may assume that
diam('Y1) = min diam('Yj) ~ blz1 - z2I = br.
J=l,2

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