IMO 1989 · Problem 4

Let \(ABCD\) be a convex quadrilateral such that the sides \(AB, AD, BC\) satisfy \(AB = AD + BC\). There exists a point \(P\) inside the quadrilateral at a distance \(h\) from the line \(CD\) such that \(AP = h + AD\) and \(BP = h + BC\). Show that: \[ \frac{1}{\sqrt{h}}\geq \frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}. \]

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

Solution

Put \(a=AD\), \(b=BC\), and \(s=a+b=AB\). Since \(P\) is interior, \(a,b,h>0\).

First, we show that the closed disk \(K\) with centre \(P\) and radius \(h\) lies inside the quadrilateral. Let \(K_A,K_B\) be the closed disks with centres \(A,B\) and radii \(a,b\). For every \(X\in K_A\), \[ PX\geq PA-AX\geq(a+h)-a=h, \] and similarly \(PX\geq h\) for every \(X\in K_B\). The segments \(AD\) and \(BC\) belong to these respective disks, while \(AB\) is covered by their union because \(AB=a+b\). Thus none of these three sides meets the open disk with centre \(P\) and radius \(h\). Neither does \(CD\), by the definition of \(h\).

If a point of that open disk were outside the quadrilateral, the segment joining it to the interior point \(P\) would meet the boundary at distance less than \(h\) from \(P\), a contradiction. Taking closures proves the assertion.

Choose coordinates \[ A=(0,0),\qquad B=(s,0),\qquad P=(x,y),\qquad y>0. \] The two prescribed distances give \[ x^2+y^2=(a+h)^2,\qquad (s-x)^2+y^2=(b+h)^2. \] Subtracting, and then substituting back, yields \[ x=a+\frac{(a-b)h}{s}, \qquad y=\frac{2\sqrt{ab}}{s}\sqrt{h(s+h)}. \tag{1} \]

Consider the line \[ \ell:\quad (a-b)u+2\sqrt{ab}\,v=as. \] Its normal vector has length \(s\), since \[ (a-b)^2+4ab=s^2. \] The distances from \(A\) and \(B\) to \(\ell\) are respectively \(a\) and \(b\), and both centres lie in the half-plane \[ (a-b)u+2\sqrt{ab}\,v\leq as. \] Consequently, both disks \(K_A,K_B\) lie in this half-plane. All four vertices of the quadrilateral belong to their union, so the whole quadrilateral—and hence \(K\)—lies in this half-plane.

The distance from \(P\) to \(\ell\) is therefore at least \(h\), giving \[ (a-b)x+2\sqrt{ab}\,y+sh\leq as. \] Substitution from (1) and simplification give \[ (a^2+b^2)h+2ab\sqrt{h(s+h)}\leq abs. \] Dividing by \(abh>0\), we obtain \[ \frac{a^2+b^2}{ab} +2\sqrt{\frac{s+h}{h}} \leq \frac{s}{h}. \]

Set \(t=\sqrt{(s+h)/h}>1\). Since \(s/h=t^2-1\), the last inequality becomes \[ (t-1)^2\geq \frac{a^2+b^2}{ab}+2 =\frac{s^2}{ab}. \] Thus \(t\geq 1+s/\sqrt{ab}\), and consequently \[ 1+\frac{s}{h} =t^2 \geq \left(1+\frac{s}{\sqrt{ab}}\right)^2. \] Subtracting \(1\) and dividing by \(s>0\) gives \[ \frac1h\geq \frac{s}{ab}+\frac{2}{\sqrt{ab}} =\left(\frac1{\sqrt a}+\frac1{\sqrt b}\right)^2. \] Taking positive square roots proves \[ \boxed{\frac1{\sqrt h}\geq \frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}}. \]

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