NMO Selection Tests for the Balkan and International Mathematical Olympiads - Second Selection Test

The problem at hand is a special case of the following situation: Let $U$ be a point on the diagonal $AD$ and let $V = UP \cap BC$ and $W = UQ \cap EF$. Then $UV \le \max(AC, BD)$ and, similarly, $UW \le \max(AE, DF)$, so $VW \le UV + UW \le \max(AC, BD) + \max(AE, DF)$. In particular, if $U = AD \cap PQ$, then $V = X$ and $W = Y$, and the conclusion follows.

We now prove that $UV \le \max(AC, BD)$ in three different ways. The first proof is elementary; the other two both require some acquaintance with analysis and are intimately related.

1st Proof. Let $\alpha = \angle UPA = \angle VPC$ and let $\beta = \angle DPU = \angle BPV$. In what follows, $[KLM]$ stands for the area of a generic triangle $KLM$. Write
$$\begin{aligned} \frac{1}{2} \cdot PD \cdot PA \cdot \sin(\beta + \alpha) &= [PDA] = [PDU] + [PUA] \\ &= \frac{1}{2} \cdot PD \cdot PU \cdot \sin \beta + \frac{1}{2} \cdot PU \cdot PA \cdot \sin \alpha, \end{aligned}$$
to get $PU = \frac{PA \cdot PD \cdot \sin(\alpha+\beta)}{PA \cdot \sin \alpha + PD \cdot \sin \beta} \le \frac{PA \cdot PD \cdot (\sin \alpha + \sin \beta)}{PA \cdot \sin \alpha + PD \cdot \sin \beta} \le \frac{PA \cdot \sin \beta + PD \cdot \sin \alpha}{\sin \alpha + \sin \beta}$, where the last inequality holds since equivalent to $(PA - PD)^2 \cdot \sin \alpha \cdot \sin \beta \ge 0$. Similarly,
$$PV \le \frac{PB \cdot \sin \alpha + PC \cdot \sin \beta}{\sin \alpha + \sin \beta},$$
so
$$\begin{aligned} UV &= PU + PV \le \frac{PA \cdot \sin \beta + PD \cdot \sin \alpha}{\sin \alpha + \sin \beta} + \frac{PB \cdot \sin \alpha + PC \cdot \sin \beta}{\sin \alpha + \sin \beta} \\ &= \frac{AC \cdot \sin \beta + BD \cdot \sin \alpha}{\sin \alpha + \sin \beta} \le \max(AC, BD), \end{aligned}$$
as desired. This completes the proof.

2nd Proof. Let $\mathcal{L}$ be the pencil of lines through $P$ crossing the sides $BC$ and $DA$ of the convex quadrangle $ABCD$. As before, we prove that the length of each segment $\ell \cap ABCD$, $\ell$ in $\mathcal{L}$, does not exceed $\max(AC, BD)$.
Clearly, no intersection segment has a length greater than the diameter of $ABCD$, so $\sup_{\ell \in \mathcal{L}} |\ell \cap ABCD|$ is finite; here and hereafter, $|s|$ denotes the length of the line segment $s$.
Suppose, if possible, that $\sup_{\ell \in \mathcal{L}} |\ell \cap ABCD| > \max(AC, BD)$ and consider a line $\ell_0$ in $\mathcal{L}$ such that
$$|\ell_0 \cap ABCD| > \frac{1}{2} \sup_{\ell \in \mathcal{L}} |\ell \cap ABCD| + \frac{1}{2} \max(AC, BD) > \frac{1}{2}(AC + BD).$$
Recall that the length of the internal bisectrix of an angle of a (possibly degenerate) triangle does not exceed the arithmetic mean of the lengths of the sides forming that angle, to infer that $\ell_0$ does not internally bisect $\angle BPC = \angle DPA$; say, $\angle(\ell_0, AC) < \angle(\ell_0, BD)$.
Finally, reflect the line $AC$ in $\ell_0$ to obtain a line $\ell_1$ in $\mathcal{L}$ such that
$$\begin{aligned} \frac{1}{2}|\ell_1 \cap ABCD| + \frac{1}{2} \max(AC, BD) &\ge \frac{1}{2}(|\ell_1 \cap ABCD| + AC) \ge |\ell_0 \cap ABCD| \\ &> \frac{1}{2} \sup_{\ell \in \mathcal{L}} |\ell \cap ABCD| + \frac{1}{2} \max(AC, BD), \end{aligned}$$
and thereby reach a contradiction; the inequality in the middle comes from the above mentioned fact about the length of an internal bisectrix in a triangle. This ends the proof.

3rd Proof. Notice that $UV$ depends continuously on $U$. Since the closed segment $AD$ is compact, $UV$ achieves a maximum at some position $U_0V_0$. We will show that $U_0V_0 \le \max(AC, BD)$, so $UV \le \max(AC, BD)$ for all positions.
Suppose, if possible, that $U_0V_0 > \max(AC, BD) \ge \frac{1}{2}(AC + BD)$.
As before, it then follows that $U_0V_0$ does not internally bisect $\angle BPC = \angle DPA$;
say, $\angle(U_0V_0, AC) < \angle(U_0V_0, BD)$.
Reflexion of $AC$ in $U_0V_0$ then provides a segment $U_1V_1$ through $P$, where $U_1$ lies on $AD$ and $V_1$ lies on $BC$, such that
$$\begin{aligned} U_0V_0 &\le \frac{1}{2}(U_1V_1 + AC) \le \frac{1}{2} \cdot U_1V_1 + \frac{1}{2} \max(AC, BD) \le \\ &\le \frac{1}{2} \cdot U_0V_0 + \frac{1}{2} \max(AC, BD) < \frac{1}{2} \cdot U_0V_0 + \frac{1}{2} \cdot U_0V_0 = U_0V_0, \end{aligned}$$
which is a contradiction. This ends the proof.