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.

Solution

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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_tK_bC_tC. N is the intersection of sides AC and B_tK_b, O is the intersection of sides BB_t and C_tC. Let M' be the intersection of AB and K_bC_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_tN_t. Therefore M = M'. The fact that, by the construction of K_c, M is also the intersection of M_tN_t with K_cC_t, implies that K_cC_t and K_bC_t are one and the same line so that, too, K_b and K_c are one and the same point.

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

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