IMO Shortlist 2010 - Problem G7
Source:
July 17, 2011
geometrygeometric transformationhomothetyincenterIMO Shortlist
Problem Statement
Three circular arcs and connect the points and These arcs lie in the same half-plane defined by line in such a way that arc lies between the arcs and Point lies on the segment Let , and be three rays starting at lying in the same half-plane, being between and For denote by the point of intersection of and (see the Figure below). Denote by the curved quadrilateral, whose sides are the segments and arcs and We say that this quadrilateral is if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals are circumscribed, then the curved quadrilateral is circumscribed, too.Proposed by Géza Kós, Hungary[asy]
pathpen=black;
size(400);
pair A=(0,0), B=(4,0), C=(10,0);
draw(L(A,C,0.3));
MP("A",A); MP("B",B); MP("C",C);
pair X=(5,-7);
path G1=D(arc(X,C,A));
pair Y=(5,7), Z=(9,6);
draw(Z--B--Y);
struct T {pair C;real r;};
T f(pair X, pair B, pair Y, pair Z)
{
pair S=unit(Y-B)+unit(Z-B);
real s=abs(sin(angle((Y-B)/(Z-B))/2));
real t=10, r=abs(X-A);
pair Q;
for(int k=0;k<30;++k)
{
Q=B+t*S;
t-=(abs(X-Q)-r)/abs(S)-s*t;
}T T=new T;
T.C=Q; T.r=s*t*abs(S);
return T;
}
void g(pair Q, real r)
{
real t=0;
for(int k=0;k<30;++k)
{
X=(5,t);
t+=(abs(X-Q)+r-abs(X-A));
}
}
pair Z1=(1.07,6);
draw(B--Z1);
T T=f(X,B,Y,Z1);
draw(CR(T.C,T.r));
T T=f(X,B,Y,Z);
draw(CR(T.C,T.r));
g(T.C,T.r);
path G2=D(arc(X,C,A));
T T=f(X,B,Y,Z1);
draw(CR(T.C,T.r));
T=f(X,B,Y,Z);
draw(CR(T.C,T.r));
g(T.C,T.r);
path G3=D(arc(X,C,A));
pen p=black+fontsize(8);
MC("\gamma_1",G1,0.85,p);
MC("\gamma_2",G2,0.85,NNW,p);
MC("\gamma_3",G3,0.85,WNW,p);
MC("h_1",B--Z1,0.95,E,p);
MC("h_2",B--Y,0.95,E,p);
MC("h_3",B--Z,0.95,E,p);
path[] G={G1,G2,G3};
path[] H={B--Z1,B--Y,B--Z};
pair[][] al={{S+SSW,S+SSW,3*S},{SE,NE,NW},{2*SSE,2*SSE,2*E}};
for(int i=0;i<3;++i)
for(int j=0;j<3;++j)
MP("V_{"+string(i+1)+string(j+1)+"}",IP(H,G[j]),al[j],fontsize(8));[/asy]