IMO 1985 · Problem 1

A circle has center on the side \(AB\) of the cyclic quadrilateral \(ABCD\). The other three sides are tangent to the circle. Prove that \(AD+BC=AB\).

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

Solution

Let \(O\) and \(r>0\) be the center and radius of the circle, and let \(P,Q,T\) be its points of tangency with \(AD,BC,CD\), respectively. Write \[ \alpha=\angle DAB,\qquad \beta=\angle ABC. \]

The point \(O\) lies strictly between \(A\) and \(B\): if it equaled either endpoint, one of the tangent sides would pass through the circle’s center. Since \(OP\perp AD\) and \(OQ\perp BC\), perpendicular projection gives \[ AP=AO\cos\alpha,\qquad r=AO\sin\alpha, \] \[ BQ=BO\cos\beta,\qquad r=BO\sin\beta. \] These formulas remain valid if \(P=A\) or \(Q=B\).

Because the quadrilateral is convex, \(O\) lies inside the angles at \(C\) and \(D\). The right triangles \(OCQ\) and \(OCT\) have a common hypotenuse and equal legs \(OQ=OT=r\), so they are congruent. Thus \(OC\) bisects \(\angle BCD\). Likewise, \(OD\) bisects \(\angle CDA\).

Opposite angles of a cyclic quadrilateral are supplementary. Consequently, \[ \angle OCQ=\frac{\pi-\alpha}{2}, \qquad \angle ODP=\frac{\pi-\beta}{2}. \] The corresponding right triangles therefore give \[ CQ=r\cot\frac{\pi-\alpha}{2}=r\tan\frac{\alpha}{2}, \qquad DP=r\tan\frac{\beta}{2}. \] Neither triangle is degenerate: tangency at \(C\) or \(D\) would make two distinct sides tangent to the circle at the same point, which is impossible.

Using \(\sin\theta\tan(\theta/2)=1-\cos\theta\), we obtain \[ AP+CQ =AO\cos\alpha+AO\sin\alpha\tan\frac{\alpha}{2} =AO. \] Similarly, \[ BQ+DP=BO. \] Since the tangency points lie on their respective sides, \[ AD+BC =(AP+DP)+(BQ+CQ) =AO+BO =AB. \] This includes tangency at \(A\) or \(B\), where the corresponding projection length is zero. \(\square\)

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