, 2007

When a group $X$ consisting of points in the plane is included in band $B$, we say that band $B$ covers $X$. First, we prove the following lemma.

Lemma. For triangle $XYZ$, let $H_X$ be the foot of the perpendicular from $X$ to $YZ$, $H_Y$ be the foot of the perpendicular from $Y$ to $ZX$, and $H_Z$ be the foot of the perpendicular from $Z$ to $XY$. When a band with width $w$ covers the triangle $XYZ$, $\min\{XH_X, YH_Y, ZH_Z\} \le w$.

Proof of Lemma. Let $B$ be a band with width $w$ covering triangle $XYZ$. Then, there exists a line $l$ and a real number $d$ which the following condition holds.
$$B = \{P \mid \text{The distance between } P \text{ and } l \text{ is less than or equal to } \frac{d}{2}\}$$
Let $m_X, m_Y, m_Z$ be the lines perpendicular to $l$ which pass through $X$, $Y$, $Z$. Then, we can say that at least one of the following proposition holds.
* The common point of $m_X$ and line $YZ$ is covered by $B$.
* The common point of $m_Y$ and line $ZX$ is covered by $B$.
* The common point of $m_Z$ and line $XY$ is covered by $B$.
Without loss of generality, assume that $B$ covers the point $W$ which is the common point of $m_Y$ and $XZ$. Since $W$ is a point on line $XZ$, $YH_Y \le YW \le w$. On the other hand, $Y, W \in B$, so we get $YW \le w$. With that, it can be said that $YH_Y \le YW \le w$.

If the convex closure of points $A$, $B$, $C$, $D$ is a triangle, the problem's statement is trivial. So we can assume that $A$, $B$, $C$, $D$ make a convex quadrangle $ABCD$. We are going to show a contradiction, assuming that for any three points among $A$, $B$, $C$, $D$ there exists a band with width $1$ containing them, and that $ABCD$ cannot be covered by a band with width $\sqrt{2}$.

Without loss of generality, assume that the area of the triangle $ABC$ is the largest in all triangles constituted by three points among $A$, $B$, $C$, $D$. (Then note that $D$ locates in shadow area.) Let $H_A$ be the foot of the perpendicular from $A$ to $BC$, $H_B$ be the foot of the perpendicular from $B$ to $CA$, and $H_C$ be the foot of the perpendicular from $C$ to $AB$. By the lemma, $\min\{AH_A, BH_B, CH_C\} \le 1$.

If $CH_C \le 1$ or $AH_A \le 1$, all four points can be covered by a band with width $1$, but this conflicts with our assumption. So we get $BH_B \le 1$.

If $AB \ge \frac{1}{\sqrt{2}}AC$, it follows that $CH_C \le \sqrt{2}BH_B \le \sqrt{2}$, so all four points can be covered with a band with width $\sqrt{2}$. This conflicts with the assumption, so we get $AB < \frac{1}{\sqrt{2}}AC$. Similarly, it follows that $BC < \frac{1}{\sqrt{2}}AC$.
$$\text{Since } AB, BC < \frac{1}{\sqrt{2}}AC,$$
$$\cos \angle ABC = \frac{AB^2 + BC^2 - AC^2}{2AB \cdot AC} < 0.$$
Hence, $\angle ABC > \frac{\pi}{2}$. Thus, at least one of $\angle ABD$, $\angle CBD$ is wider than $\frac{\pi}{4}$. Without loss of generality, we assume that $\angle ABD > \frac{\pi}{4}$. Let $G_A$ be the foot of the perpendicular from $A$ to $BD$, $G_D$ be the foot of the perpendicular from $D$ to $AB$, $G_B$ be the foot of the perpendicular from $B$ to $DA$. If $BG_B \le \sqrt{2}$, all four points can be covered by a band with width $\sqrt{2}$. This conflicts with the assumption, so it follows that $BG_B > \sqrt{2}$. Similarly we get $AH_A > \sqrt{2}$. $AB \ge AH_A$, $BD \ge BG_B$, so it follows that $AB, BD > \sqrt{2}$.

By the lemma, $\min\{AG_A, BG_B, DG_D\} \le 1$. If $AG_A \le 1$, since $AB > \sqrt{2}$, and we get $\sin \angle ABD < \frac{1}{\sqrt{2}}$. Noting that $\angle ABD > \frac{\pi}{4}$, we get $\angle ABD > \frac{3\pi}{4}$. Then, $BG_B < AG_A \le 1$. If $DG_D \le 1$, since $BD > \sqrt{2}$, we get $\sin \angle ABD < \frac{1}{\sqrt{2}}$. So it follows that $\angle ABD > \frac{3\pi}{4}$. Then $BG_B < DG_D \le 1$. With that we get $BG_B \le 1$ in each case. However, when $BG_B \le 1$, all four points can be covered by a band with width $1$. This conflicts with our assumption. With that, all four points $A$, $B$, $C$, $D$ can be covered by a band with width $\sqrt{2}$.