IMO 1969 · Problem 4

A semicircular arc \(\gamma\) is drawn on \(AB\) as diameter. \(C\) is a point on \(\gamma\) other than \(A\) and \(B\), and \(D\) is the foot of the perpendicular from \(C\) to \(AB\). We consider three circles, \(\gamma_1,\gamma_2,\gamma_3\), all tangent to the line \(AB\). Of these, \(\gamma_1\) is inscribed in \(\Delta ABC\), while \(\gamma_2\) and \(\gamma_3\) are both tangent to \(CD\) and to \(\gamma\), one on each side of \(CD\). Prove that \(\gamma_1,\gamma_2\) and \(\gamma_3\) have a second tangent in common.

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

Solution

We shall prove that the three centres are collinear. Reflection of \(AB\) across their line of centres will then give the required tangent.

By scaling and choosing coordinates, take \[ A=(0,0),\qquad B=(1,0), \] with \(\gamma\) above the \(x\)-axis. Put \(a=AC\) and \(b=BC\). Since \(\angle ACB=90^\circ\), the standard altitude relations give \[ a^2+b^2=1,\qquad C=(a^2,ab),\qquad D=(a^2,0). \] In particular, \(0<a,b<1\) and \(a+b>1\).

Let \(I_i\) and \(r_i\) denote the centre and radius of \(\gamma_i\). Comparing the area of \(ABC\) with its inradius times its semiperimeter gives \[ r_1=\frac{ab}{1+a+b}=\frac{a+b-1}{2}. \] Equal tangent lengths from the vertices show that the point where \(\gamma_1\) touches \(AB\) has distance \((1+a-b)/2\) from \(A\). Consequently, \[ I_1=\left(\frac{1+a-b}{2},\frac{a+b-1}{2}\right). \]

Relabel the auxiliary circles, if necessary, so that \(\gamma_2\) lies to the left of \(CD\). Their tangencies to \(AB\) and the segment \(CD\) imply \[ I_2=(a^2-r_2,r_2),\qquad I_3=(a^2+r_3,r_3),\qquad 0<r_2,r_3\le ab. \]

Let \(\Gamma\) be the full circle containing \(\gamma\); its centre is \(O=(1/2,0)\) and its radius is \(1/2\). We first check that its tangencies with the auxiliary circles are internal. External tangency with \(\gamma_2\) would give \[ \left(a^2-r_2-\frac12\right)^2+r_2^2 =\left(\frac12+r_2\right)^2. \] This simplifies to \[ (r_2-a^2)^2=a^2. \] Its only positive solution is \(r_2=a^2+a>ab\), a contradiction. Similarly, external tangency with \(\gamma_3\) would force \(r_3=b^2+b>ab\). Thus both tangencies are internal.

It follows that \[ \left(a^2-r_2-\frac12\right)^2+r_2^2 =\left(\frac12-r_2\right)^2, \] which, using \(a^2+b^2=1\), simplifies to \[ (r_2+b^2)^2=b^2. \] Hence \(r_2=b-b^2\). Interchanging \(A\) and \(B\) gives \(r_3=a-a^2\). Therefore \[ I_2=(1-b,b-b^2),\qquad I_3=(a,a-a^2). \] Their midpoint is \[ \frac{I_2+I_3}{2} = \left(\frac{1+a-b}{2}, \frac{a+b-(a^2+b^2)}{2}\right) =I_1. \] Thus all three centres lie on the line \(\ell=I_2I_3\).

Reflection across \(\ell\) fixes each of the three circles, so it sends their common tangent \(AB\) to another common tangent. This reflected line is distinct from \(AB\): the line \(\ell\) is not \(AB\), since the centres have positive heights, and it is not perpendicular to \(AB\), since \[ x(I_3)-x(I_2)=a+b-1>0. \] These are the only two possibilities for a reflection axis that leaves \(AB\) invariant.

The reflected line is therefore the required second common tangent. The argument also includes \(a=b\), when \(\ell\) is horizontal. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).