Diameters and Chords
The applet below illustrates an ancillary problem that is useful for solving a curious problem discussed elsewhere. I preserve most of the notations to make the references easier.
Let O be the center of a circle with diameters BB_{t}, CC_{t} and M_{t}N_{t} and chords BA_{b} and CA_{c}. Assume that BA_{b} intersects M_{t}N_{t} in M and CA_{c} intersects M_{t}N_{t} in N. K_{b} is the second point of intersection of NB_{t} with the circle. K_{c} is the second point of intersection of MC_{t} with the circle. Prove that A_{b} and A_{c} coincide if and only if so do K_{b} and K_{c}. |
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|
Copyright © 1996-2008 Alexander BogomolnySolution
What if applet does not run? |
This is plainly a variant of Pascal's Hexagram Theorem.
Since A's and K's are interchangeable, we only have to prove that if A_{b} = A_{c} = A, say, then, too, K_{b} = K_{c}.
Consider the hexagon ABB_{t}K_{b}C_{t}C. N is the intersection of sides AC and B_{t}K_{b}, O is the intersection of sides BB_{t} and C_{t}C. Let M' be the intersection of AB and K_{b}C_{t}. According to Pascal's Theorem, the three points N, O, and M' are collinear so that M' is the intersection of AB and M_{t}N_{t}. Therefore
As we mentioned at the outset, this result has repercussions to a theorem about a line passing through the circumcenter of a triangle.
Pascal and Brianchon Theorems
- Pascal's Theorem
- Pascal in Ellipse
- Pascal's Theorem, Homogeneous Coordinates
- Projective Proof of Pascal's Theorem
- Pascal Lines: Steiner and Kirkman Theorems
- Brianchon's theorem
- Brianchon in Ellipse
- The Mirror Property of Altitudes via Pascal's Hexagram
- Pappus' Theorem
- Pencils of Cubics
- Three Tangents, Three Chords in Ellipse
- MacLaurin's Construction of Conics
- Pascal in a Cyclic Quadrilateral
- Parallel Chords
- Parallel Chords in Ellipse
- Construction of Conics from Pascal's Theorem
- Pascal: Necessary and Sufficient
- Diameters and Chords
- Chasing Angles in Pascal's Hexagon
- Two Triangles Inscribed in a Conic
- Two Triangles Inscribed in a Conic - with Solution
- Two Pascals Merge into One
- Surprise: Right Angle in Circle
|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|
Copyright © 1996-2008 Alexander Bogomolny62813094 |