IMO 1959 · Problem 5

An arbitrary point \(M\) is selected in the interior of the segment \(AB\). The squares \(AMCD\) and \(MBEF\) are constructed on the same side of \(AB\), with the segments \(AM\) and \(MB\) as their respective bases. The circles circumscribed about these squares, with centers \(P\) and \(Q\), intersect at \(M\) and also at another point \(N\). Let \(N'\) denote the point of intersection of the straight lines \(AF\) and \(BC\).

(a) Prove that the points \(N\) and \(N'\) coincide.
(b) Prove that the straight lines \(MN\) pass through a fixed point \(S\) independent of the choice of \(M\).
(c) Find the locus of the midpoints of the segments \(PQ\) as \(M\) varies between \(A\) and \(B\).

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

Answer

(a) \(N=N'\).

(b) Let \(O\) be the midpoint of \(AB\). The fixed point \(S\) lies on the perpendicular to \(AB\) through \(O\), on the side opposite the squares, with \[ OS=\frac{AB}{2}. \]

(c) The locus is an open segment parallel to \(AB\), of length \(AB/2\). Its midpoint lies on the perpendicular through \(O\), on the same side as the squares, at distance \(AB/4\) from \(AB\).

Solution

Put \(AM=a\), \(MB=b\), and \(AB=L=a+b\), so \(a,b>0\). Choose coordinates \[ A=(0,0),\qquad M=(a,0),\qquad B=(L,0), \] with the squares above the \(x\)-axis. Then \[ C=(a,a),\qquad F=(a,b), \] and the circumcenters are \[ P=\left(\frac a2,\frac a2\right),\qquad Q=\left(\frac{L+a}{2},\frac b2\right). \] The two circumcircles therefore have equations \[ \Gamma_1:\quad x^2+y^2-a x-a y=0, \] \[ \Gamma_2:\quad x^2+y^2-(L+a)x-b y+aL=0. \]

(a) The lines \(AF\) and \(BC\) have respective equations \[ bx-ay=0,\qquad ax+by=aL. \] Their determinant is \(a^2+b^2>0\), so they have a unique intersection. Writing \(d=a^2+b^2\), we obtain \[ N'=\left(\frac{La^2}{d},\frac{Lab}{d}\right). \] At this point, \[ x^2+y^2=\frac{L^2a^2}{d}=Lx=a(x+y), \] where the last equality uses \(a+b=L\). Hence \(N'\in\Gamma_1\). Also, since \(ax+by=aL\), \[ x^2+y^2-(L+a)x-by+aL =Lx-(L+a)x-by+aL=0, \] so \(N'\in\Gamma_2\).

The ordinate of \(N'\) is positive, whereas \(M\) has ordinate zero; thus \(N'\ne M\). The circles are distinct because their centers have different abscissae. Consequently their common point other than \(M\) is precisely \(N'\), proving \(N=N'\).

(b) Subtracting the circle equations shows that every common point satisfies \[ Lx+(b-a)y=aL. \] Since \(M\) and \(N\) are distinct, this is the equation of \(MN\).

The point \[ S=\left(\frac L2,-\frac L2\right) \] satisfies this equation for every \(a,b>0\) with \(a+b=L\), because \[ L\frac L2+(b-a)\left(-\frac L2\right) =\frac L2(L-b+a)=aL. \] Thus every line \(MN\) passes through this fixed \(S\).

These calculations include \(a=b\): then \(N=C=F=(L/2,L/2)\), and \(MN\) is the vertical line \(x=L/2\).

(c) If \(R\) is the midpoint of \(PQ\), then \[ R=\left(\frac{L+2a}{4},\frac{a+b}{4}\right) =\left(\frac L4+\frac a2,\frac L4\right). \] As \(M\) varies through the interior of \(AB\), the parameter \(a\) varies through \(0<a<L\). Hence \(R\) ranges over exactly \[ \left\{(x,L/4):\frac L4<x<\frac{3L}{4}\right\}. \] Conversely, every point of this open segment corresponds to the admissible choice \(a=2x-L/2\). This proves the stated locus, with both endpoints excluded.

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