Problem 1 of the China Mathematical Olympiad 2010

Circle Γ1 and Γ2 intersect at two points A and B. A line through B intersects Γ1 and Γ2 at points C and D, respectively. Another line through B intersects Γ1 and Γ2 at points E and F, respectively. Line CF intersects Γ1 and Γ2 at points P and Q, respectively. Let M and N be the midpoints of arcs PB and QB, respectively. Prove that if CD = EF, then C, F, M and N are concyclic.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Mimi Nguyen of IBM)

Extend AB to meet CF at G. We are going to prove that BG is the bisector of ∠CBF. We have CB×CD = CQ×CF and FB× EF = FP×CF. Now dividing the two equations and taking into account CD = EF, we get CB/FB = CQ/FP . We also have

PG × CG = GB × GA = QG × FG or QG/PG = CG/FG = (QG + CG)/(PG + FG) = CQ/PF.

It follows that CG/FG = CB/FB or ∠CBG = ∠FBG and BG is the bisector of ∠CBF.

So now the three bisectors CM, FN and BG concur. Let them meet at I on BG. We now have IM × IC = IB × IA = IN × IF or IM/IN = IF/IC, meaning that the two triangles IMN and IFC are similar which implies ∠IMN = ∠IFC. On the other hand,

∠IMN + ∠CMN = 180° or ∠CMN + ∠NFC = 180° and C, F, M and N are concyclic.

Extension of the problem

Let’s prove that MN || O1O2 where O1 and O2 are centers of Γ1 and Γ2, respectively. Since ∠CBG = ∠FBG, we have ∠ABD = ∠FBG, i.e., AD = AF. Let K be the midpoint of arc FD, AK is then the diameter of Γ2.

∠MCP = ½ ∠BCP = ½ arc (DF – BQ) = arc (KF – NQ) or ∠MCP + ∠NFQ = arc FK. Extend MN to meet Γ2 at L, ∠LNF = ∠MCP; therefore, ∠KNL = ∠NFQ = ∠BAN (subtending arc NB = NQ). But ∠ANK = 90° or AN ┴ NK; therefore, NL ┴ AG or MN || O1O2.