IMO 1992 · Problem 4

In the plane let \(C\) be a circle, \(L\) a line tangent to the circle \(C\), and \(M\) a point on \(L\). Find the locus of all points \(P\) with the following property: there exists two points \(Q, R\) on \(L\) such that \(M\) is the midpoint of \(QR\) and \(C\) is the inscribed circle of triangle \(PQR\).

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

Answer

Let \(O\) be the centre of \(C\), let \(T\) be its point of tangency with \(L\), and let \(N\) be the point diametrically opposite \(T\).

The locus is the open ray starting at \(N\), parallel to and in the same direction as \(\overrightarrow{MO}\): \[ \boxed{\left\{P:\overrightarrow{NP}=t\,\overrightarrow{MO},\quad t>0\right\}.} \] The endpoint \(N\) is excluded.

Solution

Choose coordinates so that \[ L:\ y=0,\qquad T=(0,0),\qquad O=(0,r),\qquad M=(m,0), \] where \(r>0\) is the radius of \(C\). Thus \(N=(0,2r)\).

First suppose that \(PQR\) is a triangle satisfying the conditions. Since \(C\) is its incircle, \(P=(x,h)\) has \(h>0\), and \(T\) lies strictly between \(Q\) and \(R\). Interchanging \(Q\) and \(R\) if necessary, write \[ Q=(m-d,0),\qquad R=(m+d,0), \] where \(d>|m|\). Put \[ a=PQ,\qquad b=PR. \]

Let \(U,V\) be the points where \(C\) touches \(PQ,PR\), respectively. Equal tangent lengths give \[ QU=QT=d-m,\qquad RV=RT=d+m,\qquad PU=PV. \] Consequently, \[ a-b=-2m. \] On the other hand, the distance formula gives \[ a^2-b^2 =(x-m+d)^2+h^2-(x-m-d)^2-h^2 =4d(x-m). \] Therefore \[ 4d(x-m)=-2m(a+b), \] and hence \[ x=m-\frac{m(a+b)}{2d}. \tag{1} \]

The area of \(PQR\) is \(dh\). Joining \(O\) to the three vertices divides the triangle into three triangles whose altitudes to its sides are all \(r\). Thus \[ dh=\frac r2(2d+a+b), \] so \[ h=r\left(1+\frac{a+b}{2d}\right). \tag{2} \] The strict triangle inequality \(a+b>QR=2d\) implies \[ h>2r. \] Combining (1) and (2), we obtain \[ x=-\frac mr(h-2r). \] Setting \(t=(h-2r)/r>0\), this becomes \[ P=(-mt,\,2r+rt). \] Since \(\overrightarrow{MO}=(-m,r)\), we have \[ \overrightarrow{NP}=t\,\overrightarrow{MO}. \] This proves that every admissible point lies on the stated open ray.

Conversely, take any point \[ P=(-mt,\,2r+rt),\qquad t>0. \] Its height is strictly greater than \(2r\), so \(P\) lies above the entire circle. Draw the two tangents from \(P\) to \(C\). Neither is parallel to \(L\), because the only tangents to \(C\) parallel to \(L\) are \(y=0\) and \(y=2r\), neither of which contains \(P\). Let these tangents meet \(L\) at \(Q\) and \(R\).

The two tangency points lie strictly between heights \(0\) and \(2r\): a tangency point at either extreme would have a horizontal tangent. Thus the rays from \(P\) through the tangency points meet \(L\) beyond those points. The disk bounded by \(C\) lies in the angle between these rays and above \(L\), hence inside triangle \(PQR\). The points \(Q,R\) are distinct, since two distinct lines through \(P\) cannot have another common point. Therefore \(PQR\) is nondegenerate and \(C\) is its incircle.

Let the midpoint of \(QR\) be \(M'=(m',0)\). Applying the relation already proved to this triangle yields \[ x=-\frac{m'}r(h-2r). \] But our chosen \(P\) satisfies \[ x=-\frac mr(h-2r). \] Since \(h-2r>0\), it follows that \(m'=m\). Thus \(M'=M\), as required.

The endpoint \(N\) is impossible because every admissible triangle has \(h>2r\). The argument also includes \(M=T\); in that case the locus is the open ray perpendicular to \(L\), starting at \(N\) and directed away from \(L\).

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