Let $B\not= A$ be a point on the tangent to circle $S_1$ through the point $A$ on the circle. A point $C$ outside the circle is chosen so that segment $AC$ intersects the circle in two distinct points. Let $S_2$ be the circle tangent to $AC$ at $C$ and to $S_1$ at some point $D$, where $D$ and $B$ are on the opposite sides of the line $AC$. Let $O$ be the circumcentre of triangle $BCD$. Show that $O$ lies on the circumcircle of triangle $ABC$.
Problem
Source: Italy TST 2003
Tags: geometry, circumcircle, geometry unsolved
jayme
10.11.2010 14:16
Dear WakeUp, can you send if you have some time a figure? Sincerely jean-Louis
oneplusone
14.11.2010 12:41
Does not seem true. $B$ is a variable point on the tangent at $A$. By moving $B$, $\angle BAC$ remains constant but $\angle BDC$ changes and so does $\angle BOC$. So $ABCO$ is not necessarily cyclic.
Sunrise01
03.09.2016 18:25
Let point T be a intersection of half line BO1 and CO1. Then ⦟TBA = ⦟TCA = 90 〫and point A,B,C,T are concyclic points. Now, to show point T,B,O,C are concyclic points will solve this problem. Let ⦟O1BD = ⦟O1DB = α and ⦟DCA = β. Then ⦟TO1O2 = 2α, ⦟CO2O1 = 2β, ⦟O2DC = 90-β Therefore, ⦟O1TO2 = 2β-2α…(1), ⦟BOC = 2(180-⦟BDC) = 360-2⦟BDC = 360-2(180-⦟O1DB-⦟O2DC) = 2(⦟O1DB+⦟O2DC) = 2(α+90-β) = 180+2α-2β…(2) ∴ ⦟O1TO2+⦟BOC = 180 (∵(1),(2)) Therefore, T,B,O,C are concyclic points then the problem is solved. QED
Attachments:
