IMO 1999 · Problem 5

Two circles \(G_1\) and \(G_2\) are contained inside the circle \(G\), and are tangent to \(G\) at the distinct points \(M\) and \(N\), respectively. \(G_1\) passes through the center of \(G_2\). The line passing through the two points of intersection of \(G_1\) and \(G_2\) meets \(G\) at \(A\) and \(B\). The lines \(MA\) and \(MB\) meet \(G_1\) at \(C\) and \(D\), respectively.

Prove that \(CD\) is tangent to \(G_2\).

Official text of the problem. Official problem paper (PDF)

Solution

Let \(O,O_1,O_2\) be the centers of \(G,G_1,G_2\), and let their radii be \(R,r,s\), respectively. Internal tangency gives \[ 0<r,s<R,\qquad OO_1=R-r,\qquad OO_2=R-s. \] Since \(O_2\) lies on \(G_1\), we also have \(O_1O_2=r\).

First, the construction is nondegenerate. The point \(M\) lies strictly outside \(G_2\), since \(G_2\) meets the boundary of \(G\) only at \(N\ne M\). Hence \[ s<O_2M\le O_2O_1+O_1M=2r. \] Consequently, \[ |r-s|<O_1O_2=r<r+s, \] so \(G_1\) and \(G_2\) intersect at two distinct points. Both points lie strictly inside \(G\); therefore their line \(\ell\) meets \(G\) at two distinct points \(A,B\).

Take \(O\) as the origin, and denote the position vectors of \(O_1,O_2,M\) by \(\mathbf p,\mathbf q,\mathbf m\). The internal tangency at \(M\) gives \[ \mathbf p=\left(1-\frac rR\right)\mathbf m. \] Put \(\mathbf v=\mathbf q-\mathbf p\), so that \(|\mathbf v|=r\).

At either common point of \(G_1\) and \(G_2\), with position vector \(\mathbf x\), we have \[ |\mathbf x-\mathbf p|^2-r^2 = |\mathbf x-\mathbf q|^2-s^2. \] Expanding gives \[ 2\mathbf v\cdot\mathbf x =|\mathbf q|^2-|\mathbf p|^2+r^2-s^2 =(R-s)^2-(R-r)^2+r^2-s^2 =2R(r-s). \] Thus the line \(\ell=AB\) has equation \[ \mathbf v\cdot\mathbf x=R(r-s). \tag{1} \] Moreover, \(M\notin\ell\): otherwise the equality of powers above, together with \(M\in G_1\), would imply \(M\in G_2\), a contradiction.

Now consider the homothety with center \(M\) and ratio \(r/R\). In vector form it is \[ H(\mathbf x) =\mathbf m+\frac rR(\mathbf x-\mathbf m) =\mathbf p+\frac rR\mathbf x. \] It sends the circle \(G\) onto \(G_1\) and preserves every line through \(M\). Since \(A,B\ne M\), it therefore sends \(A\) to \(C\) and \(B\) to \(D\). In particular, \(C,D\) are distinct.

Applying this homothety to (1), we find that \(CD\) has equation \[ \mathbf v\cdot(\mathbf y-\mathbf p)=r(r-s). \] The perpendicular distance from \(O_2\) to this line is consequently \[ \frac{\left|\mathbf v\cdot(\mathbf q-\mathbf p)-r(r-s)\right|} {|\mathbf v|} = \frac{|r^2-r(r-s)|}{r} =s. \] This equals the radius of \(G_2\), so \(CD\) is tangent to \(G_2\). \(\square\)

The solution was checked against Evan Chen's published IMO solution notes.