Math 348 (2006): Assignment #3: Solutions
1. Let
δ
be a circle passing through
O
. Let
A
be the point on
δ
diametrically opposite from
O
. Let
A
be the inverse of
A
. Now consider any point
P
on
δ
other than
O
and
A
. Since
∠
OPA
is an angle inscribed in a semicircle, we know that it is a right angle. Let
P
be
the inverse of
P
. Then
OP
·
OP
=
k
2
=
OA
·
OA
, and hence we have
OP
OA
=
OA
OP
from which it follows by
SAS
for similarity that
OPA
∼
OA P
.
Since the corre
sponding angles are congruent, this means
∠
OA P
= 90
◦
. Since
P
was arbitrary, we have
shown that the inverse of all the points (other than
O
) on the circle
δ
is the line
through
A
which is perpendicular to line
←→
OA
.
2. Suppose
β
is a circle passing through
O
and
C
. Then we know that the inverse
β
of
β
with respect to inversion in
γ
will be a line not through
O
, and since
β
passes through
C
, the line
β
must pass through
C
, the inverse of
C
. Since
C
is the centre of the circle
δ
, and
β
is a line through
C
,
β
is perpendicular to
δ
. Since inversion preserves angles,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '06
 Karigiannis
 Math, Geometry, Inverse, Euclidean geometry

Click to edit the document details