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# RGB plane

## Question

Each point of the plane is painted either red, green or blue.

Prove, that there exists segment of length 1 with both ends of the same color.

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Here's the simplest, probably most elegant solution...that I feel dumb for not seeing earlier:

Choose a point of one color, say red, call it the origin. Draw a unit circle around it of points that cannot be red. Each set of two points on this circle that are a distance 1 from each other must be different colors, one blue and one green.

Now draw a circle of radius sqrt(3) with the red dot as the origin. Each point on this circle forms an equilateral triangle with two points on the unit circle, so all the points on this circle must be red. But there are infinite number of points on this circle that are distance 1 from each other, hence there exists two points with the same color that are distance 1 from each other. ##### Share on other sites

• 0 10pts.JPG ##### Share on other sites

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If a 1 unit length circle with red center point is made of green and blue points,

there are 2 points with same color on the circle that is 1 unit away.
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If a 1 unit length circle with red center point is made of green and blue points,

there are 2 points with same color on the circle that is 1 unit away.

This is not true.
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Choose a point, color it one color, say blue. Choose a point a distance of one from the first point, it must be a different color, say green. Now find the points that are one unit away from both B and G, they must be the third color, red.

Each point has a unit circle around which represents the points that cannot be the same color as the point. Consider the ones for the two red points (colored in light turquoise in the diagram).

Pick two points on one of the unit circles that are a distance one from each other, call them 1 and 2. They are one unit away from the center of the circle as well as an outside point, call it 3. Since there are an infinite number of points of types 1 and 2, there are a corresponding infinite number of points of type 3, and they draw out a circle of radius sqrt(3) around R.

This circle (in light purple) intersects with the other unit circle. Hence there exists an equilateral unit triangle (four actually) for which all three points lie on the unit circles of points that cannot be red, designated 1', 2', and 3'. Since there are three points and only two colors left, two of these points must be the same color, hence two points of the same color a distance of one from each other must exist. ##### Share on other sites

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If a 1 unit length circle with red center point is made of green and blue points,

there are 2 points with same color on the circle that is 1 unit away.

This is not true.

If you pick a point on the green-blue circle..there is a chance that there are 2 or 1 that is 1 length away from it or 25% that there is none..with that probability..applying to all the points in the circle, you get almost 100% for the whole circle to have at least one. Suppose at first there is no same color,then next,then next...so on, until you are back to the first point - the probability approaches zero. Therefore the statement above is not 100% True but not False..Unless statistics is out of the question it proves it

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If a 1 unit length circle with red center point is made of green and blue points,

there are 2 points with same color on the circle that is 1 unit away.

This is not true.

If you pick a point on the green-blue circle..there is a chance that there are 2 or 1 that is 1 length away from it or 25% that there is none..with that probability..applying to all the points in the circle, you get almost 100% for the whole circle to have at least one. Suppose at first there is no same color,then next,then next...so on, until you are back to the first point - the probability approaches zero. Therefore the statement above is not 100% True but not False..Unless statistics is out of the question it proves it

Pick a random real number from interval [0,100]. What is the probability that picked number is not 7? Clearly 100%. Does it follow, that picking 7 is impossible or that number 7 does not exists? Clearly not.

Divide unit circle into 6 equal arcs. Paint those arc blue and green alternately (so that six end points of the arcs are painted alternately too).

You have an "impossible" blue-green circle.

In fact you can take just one arc (1/6 of circle) and paint it any way you want. Then for each point P of the arc take a regular hexagon H inscribed in the unit circle such that P is a vertex of H, and paint the other 5 vertices of H alternately (regarding color of P). Again you have an impossible blue-green circle.

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• 0

If a 1 unit length circle with red center point is made of green and blue points,

there are 2 points with same color on the circle that is 1 unit away.

This is not true.

If you pick a point on the green-blue circle..there is a chance that there are 2 or 1 that is 1 length away from it or 25% that there is none..with that probability..applying to all the points in the circle, you get almost 100% for the whole circle to have at least one. Suppose at first there is no same color,then next,then next...so on, until you are back to the first point - the probability approaches zero. Therefore the statement above is not 100% True but not False..Unless statistics is out of the question it proves it

Pick a random real number from interval [0,100]. What is the probability that picked number is not 7? Clearly 100%. Does it follow, that picking 7 is impossible or that number 7 does not exists? Clearly not.

Divide unit circle into 6 equal arcs. Paint those arc blue and green alternately (so that six end points of the arcs are painted alternately too).

You have an "impossible" blue-green circle.

In fact you can take just one arc (1/6 of circle) and paint it any way you want. Then for each point P of the arc take a regular hexagon H inscribed in the unit circle such that P is a vertex of H, and paint the other 5 vertices of H alternately (regarding color of P). Again you have an impossible blue-green circle.

"Divide unit circle into 6 equal arcs. Paint those arc blue and green alternately (so that six end points of the arcs are painted alternately too)."

Then "all red centered circles" are painted that way. Making 0 deg-green, 60 deg-blue. But the points are alternated so 30 deg is green (even).

Therefore all ends of 60 deg arcs is two colored while all ends of 30 deg arcs is one colored.

If two red centered circles , where one has twice the radius of the other intersects at P2 and P1 ...

P2

r1 60deg 30 deg r2

P1

are P1 and P2 same colored or not ? or 1 is colored cyan?

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