IMO 1962 · Problem 5

On the circle \(K\) there are given three distinct points \(A, B, C\). Construct (using only straightedge and compasses) a fourth point \(D\) on \(K\) such that a circle can be inscribed in the quadrilateral thus obtained.

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

Solution

Let \(M\) be the midpoint of the arc \(AC\) not containing \(B\). It is constructible by intersecting the perpendicular bisector of \(AC\) with \(K\) and selecting the intersection on that arc. Put \[ m=AM=CM. \]

If \(AB=BC\), choose \(D=M\).

If \(AB\ne BC\), construct the length \[ \ell=\frac{m\,|AB-BC|}{AC}. \] This is a straightedge-and-compass construction by similar triangles: mark \(T\) on \(AC\) with \(AT=|AB-BC|\), and draw through \(T\) the parallel to \(CM\), meeting \(AM\) at \(N\). Then \(AN=\ell\).

Draw the circle with centre \(M\) and radius \(\ell\). Choose its intersection with \(K\) as follows:

  • If \(AB>BC\), choose \(D\) on the subarc \(MC\) of the arc \(AC\) not containing \(B\).
  • If \(AB<BC\), choose \(D\) on the subarc \(AM\) of that arc.

We prove that this construction always works.

Since three distinct points on a circle are noncollinear, the strict triangle inequality gives \[ |AB-BC|<AC. \] Consequently, in the unequal case, \[ 0<\ell<m. \] Each of the subarcs \(AM\) and \(MC\) is shorter than a semicircle. Along either subarc, the chord length from \(M\) increases strictly from \(0\) to \(m\). Thus the auxiliary circle meets each subarc in exactly one interior point. In every case, \(D\) is distinct from \(A,B,C\), and \(ABCD\) is a convex quadrilateral in circular order.

We next show that \[ AB+CD=BC+DA. \tag{1} \]

If \(AB=BC\), this follows immediately from \(D=M\) and \(AM=CM\).

If \(AB>BC\), the points \(A,M,D,C\) occur in that circular order. Ptolemy’s theorem gives \[ AD\cdot MC=AM\cdot DC+MD\cdot AC. \] Since \(AM=MC=m\) and \(MD=\ell\), we obtain \[ AD-CD=\frac{\ell\,AC}{m}=AB-BC, \] which is (1).

If \(AB<BC\), apply Ptolemy’s theorem to the points \(A,D,M,C\): \[ AM\cdot DC=AD\cdot MC+DM\cdot AC. \] Hence \[ CD-AD=\frac{\ell\,AC}{m}=BC-AB, \] again proving (1).

To finish, we prove the needed converse of Pitot’s theorem: a convex quadrilateral whose opposite side lengths have equal sums has an inscribed circle.

Let \(PQRS\) be such a quadrilateral, so that \[ PQ+RS=QR+SP. \] Start with a sufficiently small circle inside the quadrilateral tangent to \(PQ\) and \(QR\). Enlarge it homothetically about \(Q\) until it first touches \(PS\) or \(RS\). Such a first contact exists because the quadrilateral is bounded.

If it touches both, it is the required circle. Otherwise, reversing the labels \(P,R\) if necessary, suppose it touches \(PS\) but not \(RS\).

Move the line \(RS\) inward, parallel to itself, until it becomes tangent to the circle. It cuts \(QR\) and \(PS\) at interior points \(R'\) and \(S'\): the circle already touches those two segments, so the moving line reaches the circle before passing either \(Q\) or \(P\).

The circle is inscribed in \(PQR'S'\). Equality of tangent lengths from each vertex therefore gives \[ PQ+R'S'=QR'+S'P. \] Subtracting this from the assumed equality yields \[ RS=RR'+R'S'+S'S. \] But the broken path \(R,R',S',S\) is not straight, so the strict triangle inequality gives \[ RS<RR'+R'S'+S'S, \] a contradiction. Thus the circle must touch all four original sides.

Applying this result to (1), the constructed quadrilateral \(ABCD\) has an inscribed circle, as required.

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