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Three Tangents, Three Secants: What is this about?
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The applet purports to suggest the following problem by Peter Y. Woo (#1557, Math Magazine, 1998):
| Let PQ be a diameter of a circle, with A and B two distinct points on the circle on the same side of PQ. Let C be the intersection of the tangents to the circle at A and B. Let the tangent to the circle at Q meet PA, PB, and PC at A', B', and C', respectively. Prove that C' is the midpoint of A'B'. |
| Buy this applet What if applet does not run? |
Two solutions have been published Math Magazine, Vol. 72, No. 4, October 1999. The one by Richard E. Pfiefer is outright beautiful in its simplicity.
The key is to make an inversion with center P and radius PQ. Point Q remains invariant; the given circle inverts into its tangent at Q. The circle with center at C and radius AC, which is orthogonal to the given circle, inverts into a circle orthogonal to that tangent. Under the inversion, A' and B' are the images of A and B, respectively. The only circle through A', B' orthogonal to the tangent has its center at the midpoint of A'B'. Therefore
Inversion - Introduction
- Angle Preservation Property
- Apollonian Circles Theorem
- Archimedes' Twin Circles and a Brother
- Bisectal Circle
- Chain of Inscribed Circles
- Circle Inscribed in a Circular Segment
- Circle Inversion: Reflection in a Circle
- Circle Inversion Tool
- Feuerbach's Theorem: a Proof
- Four Touching Circles
- Hart's Inversor
- Inversion in the Incircle
- Inversion with a Negative Power
- Miquel's Theorem for Circles
- Peaucellier Linkage
- Polar Circle
- Poles and Polars
- Radical Axis of Circles Inscribed in a Circular Segment
- Steiner's porism
- Stereographic Projection and Inversion
- Tangent Circles and an Isosceles Triangle
- Tangent Circles and an Isosceles Triangle II
- Three Tangents, Three Secants
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