compass and straightedge construction of inverse point
Let c be a circle in the Euclidean plane with center O and let P≠O. One can construct the inverse point
P′ of P using compass and straightedge.
If P∈c, then P=P′. Thus, it will be assumed that P∉c.
The construction of P′ depends on whether P is in the interior of c or not. The case that P is in the interior of c will be dealt with first.
-
1.
Draw the ray →OP.
-
2.
Determine Q∈→OP such that Q≠O and ¯OP≅¯PQ.
-
3.
Construct the perpendicular bisector
of ¯OQ in order to find one point T where it intersects c.
-
4.
Draw the ray →OT.
-
5.
Determine U∈→OP such that U≠O and ¯OT≅¯TU.
-
6.
Construct the perpendicular bisector of ¯OU in order to find the point where it intersects →OP. This is P′.
Now the case in which P is not in the interior of c will be dealt with.
-
1.
Connect O and P with a line segment
.
-
2.
Construct the perpendicular bisector of ¯OP in order to determine the midpoint
M of ¯OP.
-
3.
Draw an arc of the circle with center M and radius ¯OM in order to find one point T where it intersects C. By Thales’ theorem, the angle ∠OTP is a right angle
; however, it does not need to be drawn.
-
4.
Drop the perpendicular
from T to ¯OP. The point of intersection is P′.
A justification for these constructions is supplied in the entry inversion of plane.
If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.
Title | compass and straightedge construction of inverse point |
---|---|
Canonical name | CompassAndStraightedgeConstructionOfInversePoint |
Date of creation | 2013-03-22 17:13:17 |
Last modified on | 2013-03-22 17:13:17 |
Owner | Wkbj79 (1863) |
Last modified by | Wkbj79 (1863) |
Numerical id | 12 |
Author | Wkbj79 (1863) |
Entry type | Algorithm |
Classification | msc 51K99 |
Classification | msc 53A30 |
Classification | msc 51M15 |