Straight Edge Only Construction of Polar
Given point P and a circle with radius R centered at point O. One needs only a straight edge to construct the polar of P.
Draw two secants through P. Assume that one intersects the circle at points A and C, the other at points B and D.
Let M be the intersection of AD and BC. Let N be the intersection of AB and CD.
NM is the desired polar of P.
To justify the construction, we proceed in two steps. First observe that the four lines CDN, ABN, BC, and AD form a complete quadrilateral, for which P serves as the intersection of a pair of diagonals, AC and BD. The latter are cut by the third diagonal MN at points conjugate to P with respect to A, C and B, D, respectfully. Therefore, MN is the polar of P with respect to the circle. Let ROQP be the line through P and the center O; and assume MN meets OP at S. Then S is conjugate of P with respect to R, Q. We have only to show that S is the inverse of P in the given circle. But this is a straightforward property of the inversion.
Since A and B might have been chosen symmetrically with respect to OP, MN is necessarily perpendicular to OP. MN is thus the polar of P with respect to the given circle.
(While toying with the applet, it's instructive to note that moving either A or B does not affect the line MN.)
- R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1996
- D. Wells, The Penguin Dictionary of Curious and Intersting Geometry, Penguin Books, 1991
Poles and Polars
- Poles and Polars
- Brianchon's Theorem
- Complete Quadrilateral
- Harmonic Ratio
- Harmonic Ratio in Complex Domain
- Joachimsthal's Notations
- La Hire's Theorem
- La Hire's Theorem, a Variant
- La Hire's Theorem in Ellipse
- Nobbs' Points, Gergonne Line
- Polar Circle
- Pole and Polar with Respect to a Triangle
- Poles, Polars and Quadrilaterals
- Straight Edge Only Construction of Polar
- Tangents and Diagonals in Cyclic Quadrilateral
- Secant, Tangents and Orthogonality
- Poles, Polars and Orthogonal Circles
- Seven Problems in Equilateral Triangle, Solution to Problem 1