compass and straightedge construction of inverse point
Let be a circle in the Euclidean plane with center and let . One can construct the inverse point of using compass and straightedge.
If , then . Thus, it will be assumed that .
The construction of depends on whether is in the interior of or not. The case that is in the interior of will be dealt with first.
-
1.
Draw the ray .
-
2.
Determine such that and .
-
3.
Construct the perpendicular bisector of in order to find one point where it intersects .
-
4.
Draw the ray .
-
5.
Determine such that and .
-
6.
Construct the perpendicular bisector of in order to find the point where it intersects . This is .
Now the case in which is not in the interior of will be dealt with.
-
1.
Connect and with a line segment.
-
2.
Construct the perpendicular bisector of in order to determine the midpoint of .
-
3.
Draw an arc of the circle with center and radius in order to find one point where it intersects . By Thales’ theorem, the angle is a right angle; however, it does not need to be drawn.
-
4.
Drop the perpendicular from to . The point of intersection is .
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 |