IMO 1972 · Problem 2

Prove that if \(n \geq 4\), every quadrilateral that can be inscribed in a circle can be dissected into \(n\) quadrilaterals each of which is inscribable in a circle.

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

Solution

We shall prove that every cyclic quadrilateral can be dissected into four cyclic quadrilaterals, one of which is a rectangle. Dividing that rectangle into \(n-3\) smaller rectangles then gives exactly \(n\) cyclic quadrilaterals.

Let \(ABCD\) be the given quadrilateral, with its vertices in cyclic order. Its opposite angles are supplementary.

Suppose first that one angle, say \(\angle A\), is a right angle. Then \(BD\) is a diameter of the circumcircle, so its center \(O\) lies inside the quadrilateral. Join \(O\) to the midpoints of all four sides. Each resulting quadrilateral has two opposite right angles, because the line from a circle’s center to a chord’s midpoint is perpendicular to the chord. Thus all four pieces are cyclic. The piece containing \(A\) has three right angles and therefore is a rectangle.

Now suppose that no angle is right. Exactly two angles are acute, and they are adjacent. Relabel the vertices so that \(\angle A,\angle B\) are acute and \(\angle C,\angle D\) are obtuse. Place \(AB\) horizontally, with \(A\) to the left of \(B\) and the quadrilateral above \(AB\).

From \(D\), draw the ray perpendicular to \(AD\) that enters the quadrilateral; it points downwards and to the right. From \(C\), draw the ray perpendicular to \(BC\) that enters the quadrilateral; it points downwards and to the left. These rays enter the quadrilateral because the angles at \(D\) and \(C\) are obtuse.

We claim that a horizontal line below both \(C\) and \(D\) can meet these rays at interior points \(P,Q\), respectively, with \(P\) to the left of \(Q\).

To justify this, interchange the labels \(A,B\) and \(D,C\), if necessary, so that \(C\) is no higher than \(D\). The ray from \(D\) lies strictly inside \(\angle ADC\), so it meets \(AC\) at a point \(S\) strictly between \(A\) and \(C\). In particular, \(S\) is lower than \(C\). At the height of \(C\), this ray meets the horizontal section of triangle \(ADC\) at a point \(R\) strictly to the left of \(C\); if \(C,D\) have the same height, then \(R=D\). Consequently, a horizontal line sufficiently slightly below \(C\) meets the two rays at interior points \(P,Q\) close to \(R,C\), respectively, and still has \(P\) to the left of \(Q\). This proves the claim.

Let \(E,F\) be the perpendicular projections of \(P,Q\) onto \(AB\). Since \(\angle A,\angle B\) are acute, \(D\) is to the right of \(A\), and \(C\) is to the left of \(B\). The directions of the two rays therefore give the strict horizontal order \[ A,\ E,\ F,\ B. \] The segments \(DP,PE,PQ,QF,QC\) dissect the quadrilateral into \[ AEPD,\qquad EFQP,\qquad FBCQ,\qquad PQCD. \] All four pieces are nondegenerate: \(P,Q\) are interior points at the same positive height, strictly below \(C,D\), and \(E,F\) are distinct interior points of \(AB\).

The piece \(EFQP\) is a rectangle. Moreover, \[ \angle ADP=\angle AEP=90^\circ, \qquad \angle BCQ=\angle BFQ=90^\circ, \] so \(AEPD\) and \(FBCQ\) are cyclic.

Finally, put \(\beta=\angle ABC\). Since the original quadrilateral is cyclic, \[ \angle CDP=\angle CDA-90^\circ =(180^\circ-\beta)-90^\circ =90^\circ-\beta. \] Also, \(CQ\perp BC\) and \(QP\parallel BA\), with the directions specified above, give \[ \angle CQP=90^\circ+\beta. \] These opposite angles sum to \(180^\circ\), so \(PQCD\) is cyclic as well.

Thus the required four-piece dissection always exists, and subdividing its rectangle as described proves the result for every integer \(n\geq4\).

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