Tuesday, 5 May 2015

Bi-coloured Infinite Plane

There are several problems on this. In this post we will look for some simple solutions to a few elementary ones.
The problems have a common premises. Let there is an infinite plane, all the points of which are coloured with two colours, say red and blue. It is understood that none of the colours are finite in number, in which case the problems becomes trivial. It is clear that we always can assume as axiom that we can always find a line segment with two extremities of the same colour. The problems are to prove that it is always possible find the followings on this plane :

1. Three collinear points, the inner one being the midpoint of the two outer points, of the same colour.
2. An equilateral triangle with the vertices of the same colour.
3. An isosceles right triangle with the vertices of the same colour.
4. A circle is given on the above bi-coloured plane. It is always possible to inscribe an isosceles triangle with vertices of one colour.

More may be added later. Let us look for some simple solutions now.

SOLUTIONS
1. Let $A$ and $B$ be two points of same colour, say blue. Consider two more points $C$ and $D$ on the same line $AB$, such that the points are in the order $CABD$, with $CA=AB=BD$.
Now if any one of $C$ and $D$ is blue the we are done.
Now let both $C$ and $D$ are red. Let the midpoint of $AB$ is $M$. Note that whatever may be the color of $M$, we have the required set of points, either in $C, M, D$ or $A, M, B$.

2. Let $A, M, B$ are three collinear points of the same colour, say blue, where $AM=MB$ (see 1). Complete the equilateral triangle $ABC$. Let $E, F$ be the midpoints of $CA$ and $CB$ respectively, giving $EF||AB$.
Now if any one of $E$ and $F$ is blue the we are done.
Now let both $E$ and $F$ are red. Note that whatever may be the color of $C$, we have the required triangle, either $CEF$ or $CAB$.

3. Take 2 points $A$ and $B$ of the same color, say blue. Complete the square on $AB$. Let $C$ and $D$ be the other two vertices. If any one of $C$ and $D$ is blue we are done.
Now say both $C$ and $D$ are red.Then consider the color of the centre of the square $O$. Depending on its color, either of the triangles $AOB$or $COD$, which are right-triangles, will be of the same color vertices.
And we are done. :) :)
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