Let A_{1}, A_{2}, A_{3}, and A_{4} be distinct points on a circle Γ. Let m
be any line in the plane of Γ such that m does not contain any
of the A's and intersects lines A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, and A_{4}A_{1}.
Let P_{i} = m ∩ line A_{i}A_{i+1} for i = 1, 2, 3, and 4 ( with A_{5} = A_{1} ).
Let B_{1} be a point on Γ distinct from A_{1} and line m. Let line B_{i}P_{i}
intersect Γ again at point B_{i+1} for i = 1, 2, and 3.
Prove that line B_{4}P_{4}
intersects Γ again at point B_{1}.
No solution yet.
The problem would have been more easily read if the
last four lines were replaced with
"Let line B_iP_i intersect Gamma again at point B_(i+1)
for i = 1, 2, 3, and 4. Prove that B_5 = B_1."
Note: If you do an inversion with respect to the
circle Gamma, then all the lines ( not through the
center ) are inverted into circles through the center
of Gamma.

Posted by Bractals
on 20150702 16:42:45 