Circle $ \Gamma_{1},$ with radius $ r,$ is internally tangent to circle $ \Gamma_{2}$ at $ S.$ Chord $ AB$ of $ \Gamma_{2}$ is tangent to $ \Gamma_{1}$ at $ C.$ Let $ M$ be the midpoint of arc $ AB$ (not containing $ S$), and let $ N$ be the foot of the perpendicular from $ M$ to line $ AB.$ Prove that $ AC\cdot CB=2r\cdot MN.$
Problem
Source: China Girls Mathematical Olympiad 2009, Problem 6
Tags: geometry, similar triangles, China
18.08.2009 18:26
Some hints SC is the bisector of <ASB N is midpoint of AB Use the tangent of half-angle
18.08.2009 19:40
let R be big radius, r be small. MN=x CN=sqrt(2(R-r)x+x^2) NB=sqrt(2Rx+x^2) then ACxCB=2rx
18.08.2009 20:10
personally, i think this is too easy for CGMO.
19.08.2009 17:37
according to the head of the question committee this year, who also happens to be China's IMO team leader this year. According to him, the first 2 questions of each day (namely problems 1, 2, 5, and 6) are "warm-up" problems while the last two are meant to differentiate the good ones from the really good ones:)
20.08.2009 05:05
You can also solve this by noting that C, S, and M are collinear. Since they are collinear, then by power of a point, we have to prove that SC*CM=2r*MN, rearrange this to get SC/2r=MN/CM. By similar triangles in a way, we can get the result.
11.07.2017 16:15
My solution: Let $O_1,O_2$ be the centers of this circle, respectively,and $O_1O_2\cap\Gamma_{1}=Q.$ We know $O_2,O_1,S$ collinear.Let $M'=SC\cap \Gamma_{2}$ and $N=M'O_2\cap AB.$ We know $\angle QCS=90,$ and $$\angle O_2M'S=\angle O_2SM'=^{\text{from tangent line theorem}} \angle QCN=90-\angle NCM'\to \angle NM'C+\angle NCM'=90,$$then $M'\equiv M.$ Or $S,C,M$ are collinear.Alao from $PoP$ $AC\cdot CB=SC\cdot CM,$ and $$\triangle NMC\sim\triangle CSQ\to QS\cdot MN=MC\cdot SC.$$As desired.
14.09.2024 11:25
Certainly,it's quit easy for such a huge competetition,especaily as a p6!