I think an "extremal" argument works here. In many problems like this, considering the greatest (or the shortest) distance between two points is a very clever idea. So, denote by $A_1$, $A_2$ two points from $S$ such the distance between them is maximal. Now, if we consider a strip of width $2$ with the borders parallel to $A_1A_2$ such that $A_1A_2$ is exactly the center of the strip (I mean, at distance $1$ from the borders), that strip covers all points from $S$. $ \mathrm{(* * *)}$. Consider now the point, $A_3$, such that the distance from the point $A_3$ to the line $A_1A_2$ is maximal. If we show that this distance $d$ is $\leq 1$, the problem is solved. We know from the hypothesis that the triangle $A_1A_2A_3$ can be covered by a strip of width 1. That means, that at least one altitude of the triangle is less than 1. This is true, because if we construct the perpendicular lines from $A_1$, $A_2$, $A_3$ to the borders of the strip we can notice that at least one meets the opposite side of $\triangle A_1A_2A_3$. Since $A_1A_2$ is the greatest side of the triangle $\Rightarrow$ the distance $d$ is the shortest altitude. So $d \leq 1$. Thus we are done. [geogebra]131ed368d32d50efac785cd0427fbc0ae2af60eb[/geogebra]

Nice problem, wow for the altitude argument, I never thought of that! Here's mine: It can be easily proven that if we want the strip with the least width to cover some triangle, 2 of the vertexes will be one line, the third on other (just use ''Greater angle, greater side''), let's call these 3 strips for every triangle its good strips Now, I couldn't remember that widths of these 3 are the altitudes : OK, let $AB=r$ be the greatest of all distances between any 2 points in S. All of the other points are contained in a figure which is the intersection of the 2 circles with radii $r$, centers $A,B$. 1) $r\le 2$. Then just take the strip whose lines are perpendicular to $AB$ and contain $A$ and $B$ respectively. 2) $r\ge 2$ Let's assume that there is a point $C$ whose distance to $AB$ is greater than 1, WLOG let it be ''above'' $AB$. Draw 2 circles $C_1,C_2$ with centers A and B, radii 1. Obviously the longer of the lines $BC, AC$ will have to intersect both of the circles $C_1,C_2$. Why? Well, WLOG assume that $AC\le BC$. The strip with lines $p(B,C)$ and a parallel line which contains $A$ must have width 1 or less (because it's the shortest of the 3 good strips for $ABC$), so the distance from $A$ to $BC$ is at most 1. So, $BC$ intersects both circles. But this is clearly impossible as C is above the common tangent of the 2 circles $C_1,C_2$ (which is parallel to p(A,B) and at distance 1 from it) This means that all of the points are inside or on the strip which is formed by 2 common tangents of $C_1,C_2$, and its width is 2. Q.E.D. [geogebra]3cd5716d0d5f321b2a91c774bb2e6aa7836b5c13[/geogebra]

sandu2508 wrote: A strip of width $w$ is the set of all points which lie on, or between, two parallel lines distance $w$ apart. Let $S$ be a set of $n$ ($n \ge 3$) points on the plane such that any three different points of $S$ can be covered by a strip of width $1$. Prove that $S$ can be covered by a strip of width $2$. Is it $2$ the minimum?

I feel that this is an old problem that I've seen before, but searching the forum doesn't find it; perhaps I am thinking of http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=166690 or similar.

Take a look here, too: viewtopic.php?p=769818

Someone to solve it by induction?

Here's my solution: Take triangle $ABC$ with maximal area. Draw parallels through $A,B$ and $C$ with respect to the opposite sides. They intersect each other in, let's say, $A', B'$ and $C'$ respectively. We claim that all the points in $S$ are lieing inside triangle $A'B'C'$. Suppose that there is a point $X\in S$ which is in the exterior of this triangle. Then at least one of the areas of triangles $PAB, PBC, PCA$ is greater than the area of $\triangle{ABC}$, contradiction. But from the statement of the problem we have that $\triangle{ABC}$ has an altitude of lenght at most $1$, so $\triangle{A'B'C'}$ has an altitude of lenght at most $2$, from where it follows that it can be covered by a strip of lenght $2$. Thus $S$ can be covered by such a stripe. (because $S\subset Int(\triangle{ABC})$).

Which country proposed this question?!?!

ridgers wrote: Which country proposed this question?!?! Romania i think.

http://mathproblems123.wordpress.com/2010/05/15/bmo-2010-strip-cover-problem/

An easy one,just notice that the minimum width of the strip is for $3$ points is the smallest altitude,so just pick the two points $Ai$ and $Ad$ such that $AiAd$ is maximal.Now,from this it is obvious that all points can be covered with a strip of width $2$,so we are finished.

Since the thread was revived, it is worth mentioning that apparently $2$ might be replaced by $2\cos\frac{\pi}{5}$ as observed here: http://ssmr.ro/gazeta/gmb/2010/6/articol.pdf. No proof of this result is known to me.

$\phantom{I thought it would be both interesting and useful to do it again}$

The author wrote "It seems that the outcome can be improved in the sense that we can replace constant $2$ to $2 \cos \frac{\pi}{5}$. For the time being, the authors of this article do not know a demonstration of this fact and no counter-example."

Posted for storage:Solution: It appears that the solution to this problem is nothing more than constructing parallel lines from a triangle $ABC$ and taking a homothety to show that it works.Actually this construction is motivated by the fact that our strip has width 1 and you want to prove for 2,and assuming that a point $P$ might lie out of the big triangle,Shows that the task is impossible,thus all the points are inside. Also nota that since a triangle is symmetric in A,B,C ,that gives the motivation for taking the parallel from all the vertexs of the triangle

Cool problem. The key claim is that if $\triangle ABC$ can be covered by a strip of width $1$ then its shortest altitude is at most of length $1$. Indeed suppose that $\triangle ABC$ is covered by a strip of minimal length $L$. Clearly at least two vertices of the triangle are on the strip edge, with one on each side. Then if exactly two vertices are on the triangle we can check that we can rotate the triangle about a vertex on the strip edge an infinitesimal amount such that the other two vertices are completely inside the strip, in which case we can make $L$ smaller, contradiction (the details are left to the reader). As such we only have to consider the cases where one strip edge coincides with a side of the triangle and the other is the opposite vertex, in which case the conclusion is obvious as the strip width would be equal to one of the altitudes. Now consider the two points in $S$, $A$ and $B$, such that $AB$ is maximal. Then for any other point $P \in S$, the shortest altitude of $\triangle ABP$ is from $P$. Since this can be at most $1$ unit long, we have $d(P,\overline{AB})\leq 1$ for all $P \in S$. Then the strip of width $2$ "centered" at $\overline{AB}$ covers $S$.

Let $A,B\in S$ be points such that $AB$ is maximal. If $AB\le 1$, we are done because all points in $S$ must lie in/on the circle centered at $A$ with radius $1$. Henceforth, assume $AB>1$ and consider another point $C\in S$. We claim $\delta(C,\overline{AB})\le 1$. Suppose FTSOC the contrary, and let $\ell_1$ and $\ell_2$ form a strip of width one containing $A$, $B$, and $C$. WLOG let $A$ be the the left of $B$. We have three cases: Case 1: The order of our three points is $A$, $C$, $B$ from left to right. [asy][asy] import geometry; size(7cm); pair A,B,C; B=(0,-1); A=(-4,-5); C=(-1,-6); draw(segment((10,1),(-15,1)),blue+white); draw(segment((10,-10),(-15,-10)),blue+white); draw((-4,-1)--B--(0,-6)--(-4,-6)--cycle,red+white); draw(A--B--C--cycle); dot(A^^B^^C); dot((-4,-1)); dot((0,-6)); dot((-4,-6)); label("$C$",C,SW); label("$A$",A,NW); label("$B$",B,SE); label("$W$",(-4,-1),W); label("$Y$",(-4,-6),SW); label("$X$",(0,-6),SE); [/asy][/asy] Then, we can inscribe $\triangle ABC$ in a rectangle with sides parallel to $\ell_1$ such that $B$ is a vertex of the rectangle. Notice $AB>WB$ and $\delta(D,\overline{AB})>1\ge WY$ so $[ABC]>\tfrac{1}{2}[WBXY]$. On the other hand, \begin{align*}[WBXY]-[ABC]&=[WAB]+[BXC]+[AYC]\\&=\frac{WB\cdot WA+AY\cdot YC+CX\cdot BX}{2}\\&=\frac{(YC+CX)(WA)+(AY)(CY)+(CX)(AW+AY)}{2}\\&=\frac{(CX+CY)(AW+AY)+(CX)(WA)}{2}\\&\ge\tfrac{1}{2}[WBXY]\end{align*}a contradiction. Case 2: The order of the vertices is $C$, $A$, $B$. This case is similar to case one except $A$ and $C$ are reversed in the diagram. Then, $AC\ge BC>WB$ so we have a similar area contradiction. Case 3: The order of the vertices is $A$, $B$, $C$. This case is just case two with $A$ and $B$ flipped and the whole diagram reflected through a vertical axis. Hence, it follows from case two. Therefore, we have a contradiction in all scenarios and so $\delta(C,\overline{AB})\le 1$ for all possible $C\in S$. Hence, we can take the strip parallel to $\overline{AB}$ with the both sides of the strip having distance one from $\overline{AB}$, and all points in $S$ will lie on the strip. $\square$

Claim: Three points can be put in a strip of width $1$ if at least one of their altitudes has length less than $1$. Proof. Consider the minimum strip that contains the triangle. This, by shrinking a strip and rotating it until one of the sides coincides with the sides, until the strip's width is the altitude. $\blacksquare$ Now, since there are a finite number of points let $A$ and $B$ be points such that $AB$ is maximal. Then, take the strip of width $2$ with a middle line of $AB$. Then, for any point $C$, it lies within $AB$ of both $A$ and $B$. By the condition, it must lie either in the strip of length $2$, or between the tangents from one of the points of $A$ and $B$ to the circle of radius $1$ at the other. Claim: The tangents from $A$ to the circle of radius $1$ at $B$, the circle of radius $r$ at $A$, and the common external tangent of the circles of radius $1$ at $A$ and $B$ coincide at some $K$. Proof. Take the phantom point $K$ lying on both lines. Tangents and $AB$ from $K$ and $B$ both have feets of length $1$ to each, so $KB = AB$ must hold. $\blacksquare$ Then, as such $C$ lies in thes trip.

Claim. A strip of width $1$ covers a triangle $ABC$ if and only if at least one altitude of $ABC$ has length less than $1$. Proof. This is basically obvious, simply as we can orient the strip for its edge to coincide with the corresponding base in the triangle. $\blacksquare$ Now the finish is quite canonical: consider the longest segment $\overline{AB}$ in $S$. It follows that any triangle containing $\overline{AB}$ as a base has altitude less than $1$, or any point is at most distance $1$ from $\overline{AB}$. This creates a strip of width $2$.

Suppose the maximum edge connecting two vertices in $S$ has length $m$. Then the points in $S$ must lie inside a $2 \times m$ rectangle with the maximal edge as a midline from our maximality assumption and given condition, and thus can be covered by a strip of width 2. $\blacksquare$