IMO 1988 · Problem 5

\(ABC\) is a triangle right-angled at \(A\), and \(D\) is the foot of the altitude from \(A\). The straight line joining the incenters of the triangles \(ABD\), \(ACD\) intersects the sides \(AB\), \(AC\) at the points \(K\), \(L\) respectively. \(S\) and \(T\) denote the areas of the triangles \(ABC\) and \(AKL\) respectively. Show that \(S \geq 2T\).

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

Solution

Write \[ AB=b,\qquad AC=c,\qquad BC=a=\sqrt{b^2+c^2},\qquad AD=h. \] Computing the area of \(ABC\) in two ways gives \[ h=\frac{bc}{a}. \] Choose coordinates \(A=(0,0)\), \(B=(b,0)\), \(C=(0,c)\). We shall prove that the line joining the two incenters has equation \[ x+y=h. \]

Let \(I\) be the incenter of \(ABD\), with inradius \(r\). Denote the incircle’s contact points with \(AD,AB,BD\) by \(E,F,G\), respectively. Since \(\angle ADB=90^\circ\) and the radii \(IE,IG\) are perpendicular to the corresponding sides, \(DEIG\) is a square. Thus \(DE=r\).

The tangent segments from \(A\) to this incircle are equal, so \[ AF=AE=AD-DE=h-r. \] As \(IF\perp AB\) and \(IF=r\), we obtain \[ I=(h-r,r). \] Likewise, if \(J\) is the incenter of \(ACD\), with inradius \(s\), the same argument gives \[ J=(s,h-s). \]

Both centers therefore lie on \(x+y=h\). They are distinct, since they lie in the interiors of the two triangles on opposite sides of \(AD\). Consequently, this is precisely the line \(IJ\), and \[ K=(h,0),\qquad L=(0,h). \] Moreover, \(a>b,c\), so \(0<h<\min(b,c)\); hence these intersections lie strictly inside the stated sides.

It follows that \[ T=\frac{h^2}{2} =\frac{b^2c^2}{2(b^2+c^2)}, \qquad S=\frac{bc}{2}. \] Therefore \[ S-2T =\frac{bc}{2}-\frac{b^2c^2}{b^2+c^2} =\frac{bc(b-c)^2}{2(b^2+c^2)} \geq 0. \] Thus \(S\geq 2T\), with equality exactly when \(AB=AC\).

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