IMO 2008 · Problem 6

Let \(ABCD\) be a convex quadrilateral with \(|BA| \neq |BC|\). Denote the incircles of triangles \(ABC\) and \(ADC\) by \(\omega_1\) and \(\omega_2\) respectively. Suppose that there exists a circle \(\omega\) tangent to the ray \(BA\) beyond \(A\) and to the ray \(BC\) beyond \(C\), which is also tangent to the lines \(AD\) and \(CD\). Prove that the common external tangents of \(\omega_1\) and \(\omega_2\) intersect on \(\omega\).

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

Solution

Let \(O\) and \(R\) be the centre and radius of \(\omega\). Denote its points of tangency with \(BA,BC,AD,CD\) by \(E,F,G,K\), respectively.

We first establish the positions of these points. Since \(E,F\) lie on the rays \(BA,BC\), the centre \(O\) lies on the internal bisector of \(\angle ABC\), and the disk bounded by \(\omega\) lies inside that angle. Consequently, \(G\) and \(D\) lie on the same side of \(AB\), so \(G\) lies on the ray \(AD\). Similarly, \(K\) lies on the ray \(CD\).

The centre \(O\) lies between the two tangent rays \(AE,AG\). Since \(AE\) is opposite to \(AB\), this puts \(O\) on the opposite side of \(AD\) from \(B\). Likewise, \(O\) lies on the opposite side of \(CD\) from \(B\). Thus \(O\) lies in the angle vertically opposite to \(\angle ADC\), on its bisector. Its perpendicular projections onto \(AD,CD\) therefore lie on the rays from \(D\) away from \(A,C\). Hence the orders are \[ A,D,G \qquad\text{and}\qquad C,D,K. \] Equal tangent lengths now give \[ BE=BA+AD+DG,\qquad BF=BC+CD+DK. \] Since \(BE=BF\) and \(DG=DK\), we obtain \[ BA+AD=BC+CD. \tag{1} \]

Choose coordinates \[ A=(0,0),\qquad C=(b,0), \] with \(B\) above the horizontal axis and \(D\) below it.

Let the tangency point of \(\omega_1\) with \(AC\) divide \(AC\) into lengths \(p,q\), measured from \(A,C\), respectively. Then \[ p,q>0,\qquad p+q=b,\qquad BA-BC=p-q. \] By (1), \[ AD-CD=q-p. \] The equal-tangent-length relations in triangle \(ADC\) consequently show that its incircle touches \(AC\) at distance \(q\) from \(A\) and \(p\) from \(C\).

Writing the two inradii as \(r,s>0\), their centres are therefore \[ I=(p,r),\qquad J=(q,-s). \tag{2} \] Also, \(p\ne q\), because \(BA\ne BC\).

Write \(O=(h,k)\), and put \[ u=k+R,\qquad v=k-R. \] Thus \(T=(h,u)\) is the uppermost point of \(\omega\). We will prove that both common external tangents pass through \(T\).

The equations expressing the signed distances of \(O\) from the four sides are \[ \begin{aligned} 2prh&=p^2u-r^2v, & 2qr(b-h)&=q^2u-r^2v,\\ 2qsh&=-q^2u+s^2v, & 2ps(b-h)&=-p^2u+s^2v. \end{aligned} \tag{3} \] Here is a derivation, including the signs. The line \(AB\) has equation \[ 2prx-(p^2-r^2)y=0. \] Indeed, it passes through \(A\), and its distance from \(I=(p,r)\) is \(r\); it is therefore the tangent from \(A\) to \(\omega_1\) other than \(AC\). Its normal has length \(p^2+r^2\), and its expression is positive on the side containing \(C\). Since \(O\) lies on that side, \[ 2prh-(p^2-r^2)k=R(p^2+r^2), \] which is the first equation of (3).

The other three side equations are \[ \begin{aligned} BC:\quad&2qr(b-x)-(q^2-r^2)y=0,\\ AD:\quad&2qsx+(q^2-s^2)y=0,\\ CD:\quad&2ps(b-x)+(p^2-s^2)y=0. \end{aligned} \] Their expressions are positive toward the corresponding triangles. The signed distances of \(O\) are \(R,-R,-R\), respectively, by the positions established above. These give the remaining equations of (3). These equations also apply when any side is vertical.

Divide the first two equations of (3) by \(p,q\), respectively, and add. Using \(p+q=b\), we obtain \[ 2r=u-\frac{r^2}{pq}v. \] Similarly, the last two equations give \[ 2s=-u+\frac{s^2}{pq}v. \] Adding these identities yields \[ 2(r+s)=\frac{(s-r)(s+r)}{pq}v. \] In particular, \(r\ne s\), and \[ v=\frac{2pq}{s-r},\qquad u=\frac{2rs}{s-r}. \] Substitution into the first equation of (3) gives \[ h=\frac{ps-qr}{s-r}. \] Together with (2), these formulas say precisely that \[ T=(h,u)=\frac{rJ-sI}{r-s}. \tag{4} \]

Finally, \[ IJ^2=(p-q)^2+(r+s)^2>(r+s)^2, \] so the two incircles are strictly disjoint and have two distinct common external tangents.

Let \(\ell\) be either such tangent. Orient signed distance from \(\ell\) so that both centres have positive distances. Those distances are \(r,s\), respectively. Since signed distance to a fixed line is an affine function, (4) gives \[ d_\ell(T) =\frac{r\,d_\ell(J)-s\,d_\ell(I)}{r-s} =\frac{rs-sr}{r-s} =0. \] Thus both common external tangents pass through \(T\). Since \(T=(h,k+R)\) lies on \(\omega\), their intersection lies on \(\omega\), as required. \(\square\)

The solution was checked against the official IMO shortlist solutions.