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Bài thi giỏi toán Trung Quốc 2010

[ Wednesday, 07 April, 2010 ]

Bài thi học sinh giỏi toán Trung Quốc 2010 (Hoàng Sa và Trường Sa là của Việt Nam)

Problem 1 of the China Mathematical Olympiad 2010
(Guys! Paracel islands and Spartly islands belong to the Vietnamese – Do not invade our country)

by Vo Duc Dien

Circle Г1 and Г2 intersect at two points A and B. A line through B intersects Г1 and Г2 at points C and D, respectively. Another line through B intersects Г1 and Г2 at points E and F, respectively. Line CF intersects Г1 and Г2 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

Extend AB to meet CD at G. We are going to prove that BG is the bisector of ∠CBF. We have CB x CD = CQ x CF and
FB x EF = FP x CF
Dividing the two above equations, taking into account that CD = EF, we get CB/FB = CQ/FP .
We also have
PG x CG = GB x GA = QG x 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 x IC = IB x IA = IN x IF or IM/IN = IF/IC or the two triangles IMN and IFC are similar, meaning ∠IMN = ∠IFC, but

∠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 Г1 and Г2, respectively. Since ∠CBG = ∠FBG, we have ∠ABD = ∠FBG or AD = AF.
Let K be the midpoint of arc FD, AK is then the diameter of Г2.

∠MCP = ½ ∠BCP = ½ arc (DF – BQ) = arc (KF – NQ) or ∠MCP + ∠NFQ = arc FK. Extend MN to meet Г2 at L, ∠LNF = ∠MCP; therefore, ∠KNL = ∠NFQ = ∠BAN (subtending arc NB = NQ). But ∠ANK = 90° or AN ┴ NK; therefore, NL ┴ AG or MN || O1O2.

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