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_tN_t and chords BA_b and CA_c. Assume that BA_b intersects M_tN_t in M and CA_c intersects M_tN_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.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

Solution

Solution

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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_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

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