IMO 2006 Shortlisted Problems

We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $ABCD$. Then: $$\begin{array}{rl}& \text{Circles }(BCT)\text{ and }(DAT)\text{ are tangent at }T\\ & \text{if and only if}\angle ADT+\angle BCT=\angle ATB.\end{array}\text{\hspace{1em}(1)}$$ Indeed, if the two circles touch each other then their common tangent at $T$ intersects the segment $AB$ at a point $Z$, and so $\angle ADT=\angle ATZ$, $\angle BCT=\angle BTZ$, by the tangent-chord theorem. Thus $\angle ADT+\angle BCT=\angle ATZ+\angle BTZ=\angle ATB$. And conversely, if $\angle ADT+\angle BCT=\angle ATB$ then one can draw from $T$ a ray $TZ$ with $Z$ on $AB$ so that $\angle ADT=\angle ATZ$, $\angle BCT=\angle BTZ$. The first of these equalities implies that $TZ$ is tangent to the circle $(DAT)$; by the second equality, $TZ$ is tangent to the circle $(BCT)$, so the two circles are tangent at $T$.



So the equivalence (1) is settled. It will be used later on. Now pass to the actual solution. Its key idea is to introduce the circumcircles of triangles $ABP$ and $CDP$ and to consider their second intersection $Q$ (assume for the moment that they indeed meet at two distinct points $P$ and $Q$).

Since the point $A$ lies outside the circle $(BCP)$, we have $\angle BCP+\angle BAP<180^{\circ }$. Therefore the point $C$ lies outside the circle $(ABP)$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same $\mathrm{arc}CD$ of the circle $(BCP)$.



By symmetry, $P$ and $Q$ lie on the same arc $AB$ of the circle $(ABP)$. Thus the point $Q$ lies either inside the angle $BPC$ or inside the angle $APD$. Without loss of generality assume that $Q$ lies inside the angle $BPC$. Then $$\begin{array}{c}\angle AQD=\angle PQA+\angle PQD=\angle PBA+\angle PCD\le 90^{\circ }.\end{array}\text{\hspace{1em}(2)}$$ by the condition of the problem.

In the cyclic quadrilaterals $APQB$ and $DPQC$, the angles at vertices $A$ and $D$ are acute. So their angles at $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $BPC$ but in fact inside the triangle $BPC$, hence also inside the quadrilateral $ABCD$.

Now an argument similar to that used in deriving (2) shows that $$\begin{array}{c}\angle BQC=\angle PAB+\angle PDC\le 90^{\circ }.\end{array}\text{\hspace{1em}(3)}$$ Moreover, since $\angle PCQ=\angle PDQ$, we get $$\angle ADQ+\angle BCQ=\angle ADP+\angle PDQ+\angle BCP-\angle PCQ=\angle ADP+\angle BCP.$$ The last sum is equal to $\angle APB$, according to the observation (1) applied to $T=P$. And because $\angle APB=\angle AQB$, we obtain $$\angle ADQ+\angle BCQ=\angle AQB.$$ Applying now (1) to $T=Q$ we conclude that the circles $(BCQ)$ and $(DAQ)$ are externally tangent at $Q$. (We have assumed $P\ne Q$; but if $P=Q$ then the last conclusion holds trivially.)

Finally consider the halfdiscs with diameters $BC$ and $DA$ constructed inwardly to the quadrilateral $ABCD$. They have centres at $M$ and $N$, the midpoints of $BC$ and $DA$ respectively. In view of (2) and (3), these two halfdiscs lie entirely inside the circles $(BQC)$ and $(AQD)$; and since these circles are tangent, the two halfdiscs cannot overlap. Hence $MN\ge \frac{1}{2}BC+\frac{1}{2}DA$.

On the other hand, since $\overrightarrow{MN}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{CD})$, we have $MN\le \frac{1}{2}(AB+CD)$. Thus indeed $AB+CD\ge BC+DA$, as claimed.