Japan 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 $XYZ$, $\min \{X{H}_X,Y{H}_Y,Z{H}_Z\}\le w$.

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 $XZ$, $Y{H}_Y\le YW\le w$. On the other hand, $Y,W\in B$, so we get $YW\le w$. With that, it can be said that $Y{H}_Y\le YW\le w$.

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, $\min \{A{H}_A,B{H}_B,C{H}_C\}\le 1$.



If $C{H}_C\le 1$ or $A{H}_A\le 1$, all four points can be covered by a band with width $1$, but this conflicts with our assumption. So we get $B{H}_B\le 1$.

If $AB\ge \frac{1}{\sqrt{2}}AC$, it follows that $C{H}_C\le \sqrt{2}B{H}_B\le \sqrt{2}$, 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 $BC<\frac{1}{\sqrt{2}}AC$. $$\text{Since }AB,BC<\frac{1}{\sqrt{2}}AC,$$ $$\cos \angle ABC=\frac{A{B}^2+B{C}^2-A{C}^2}{2AB\cdot AC}<0.$$ 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 $B{G}_B\le \sqrt{2}$, all four points can be covered by a band with width $\sqrt{2}$. This conflicts with the assumption, so it follows that $B{G}_B>\sqrt{2}$. Similarly we get $A{H}_A>\sqrt{2}$. $AB\ge A{H}_A$, $BD\ge B{G}_B$, so it follows that $AB,BD>\sqrt{2}$.

By the lemma, $\min \{A{G}_A,B{G}_B,D{G}_D\}\le 1$. If $A{G}_A\le 1$, 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, $B{G}_B<A{G}_A\le 1$. If $D{G}_D\le 1$, since $BD>\sqrt{2}$, we get $\sin \angle ABD<\frac{1}{\sqrt{2}}$. So it follows that $\angle ABD>\frac{3\pi }{4}$. Then $B{G}_B<D{G}_D\le 1$. With that we get $B{G}_B\le 1$ in each case. However, when $B{G}_B\le 1$, 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 $\sqrt{2}$.