Diagram p14 |

__Problem 14__: Every space point is colored either red or blue. Show that among the squares with side 1 in this space there is at least one with three red vertices or at least one with four blue vertices.

__Solution__: (With reference to diagram p14)

Proof: With
reference to diagram 1 on this page – draw parallel unit circles around the 2
red vertices; the circles must be perpendicular to the line joining the 2 red
vertices. No point on these 2 circles can be coloured red since otherwise that
point, along with the 2 red center points and the counterpart point on the
other circle would form a unit square having 3 red vertices.

Lemma 2: If
no 2 red vertices lie at a distance of 1 unit apart, then there must exist a
unit square with 4 blue vertices.

Proof:

(a) If there is no red vertex in the space, then all points must be blue and an all-blue unit square exists obviously.

(b) If a red vertex does exist, draw a unit sphere around it. All points on the sphere must be blue and it is easy to construct a unit square having its 4 vertices on the spherical surface and hence being an all-blue unit square.

(a) If there is no red vertex in the space, then all points must be blue and an all-blue unit square exists obviously.

(b) If a red vertex does exist, draw a unit sphere around it. All points on the sphere must be blue and it is easy to construct a unit square having its 4 vertices on the spherical surface and hence being an all-blue unit square.

Combining lemmas 1 and 2, it follows that
there must be a unit square having either 4 blue vertices or (at least) 3 red
vertices.

## No comments:

## Post a Comment