Show circle with points coloured red and blue must have monochromatic red equilateral triangle

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Colour each point on a circle of radius $frac12$ red or blue, such that the region of blue points has length $1$. Prove that we can inscribe an equilateral triangle in the circle such that all three vertices are red.



I think the Pigeonhole Principle will be involved, but don't quite see how to apply it. The length condition also seems a bit hard to work with, so any hints or suggestions would be much appreciated.










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    What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable?
    – John Hughes
    Dec 9 at 21:41










  • Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length.
    – Prasiortle
    Dec 9 at 21:53














up vote
4
down vote

favorite
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Colour each point on a circle of radius $frac12$ red or blue, such that the region of blue points has length $1$. Prove that we can inscribe an equilateral triangle in the circle such that all three vertices are red.



I think the Pigeonhole Principle will be involved, but don't quite see how to apply it. The length condition also seems a bit hard to work with, so any hints or suggestions would be much appreciated.










share|cite|improve this question



















  • 1




    What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable?
    – John Hughes
    Dec 9 at 21:41










  • Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length.
    – Prasiortle
    Dec 9 at 21:53












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





Colour each point on a circle of radius $frac12$ red or blue, such that the region of blue points has length $1$. Prove that we can inscribe an equilateral triangle in the circle such that all three vertices are red.



I think the Pigeonhole Principle will be involved, but don't quite see how to apply it. The length condition also seems a bit hard to work with, so any hints or suggestions would be much appreciated.










share|cite|improve this question















Colour each point on a circle of radius $frac12$ red or blue, such that the region of blue points has length $1$. Prove that we can inscribe an equilateral triangle in the circle such that all three vertices are red.



I think the Pigeonhole Principle will be involved, but don't quite see how to apply it. The length condition also seems a bit hard to work with, so any hints or suggestions would be much appreciated.







combinatorics discrete-mathematics






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edited Dec 9 at 22:16









Jean Marie

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asked Dec 9 at 21:31









Prasiortle

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1075







  • 1




    What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable?
    – John Hughes
    Dec 9 at 21:41










  • Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length.
    – Prasiortle
    Dec 9 at 21:53












  • 1




    What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable?
    – John Hughes
    Dec 9 at 21:41










  • Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length.
    – Prasiortle
    Dec 9 at 21:53







1




1




What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable?
– John Hughes
Dec 9 at 21:41




What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable?
– John Hughes
Dec 9 at 21:41












Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length.
– Prasiortle
Dec 9 at 21:53




Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length.
– Prasiortle
Dec 9 at 21:53










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Make all the red points that are a distance exactly $frac 2pi3$ away from a blue point blue. The measure of the blue points is now no more than $3$, but the circumference of the circle is $pi$. There is at least $pi-3$ of the circle still colored red and any of the red points is on an all red equilateral triangle.






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    Make all the red points that are a distance exactly $frac 2pi3$ away from a blue point blue. The measure of the blue points is now no more than $3$, but the circumference of the circle is $pi$. There is at least $pi-3$ of the circle still colored red and any of the red points is on an all red equilateral triangle.






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      up vote
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      Make all the red points that are a distance exactly $frac 2pi3$ away from a blue point blue. The measure of the blue points is now no more than $3$, but the circumference of the circle is $pi$. There is at least $pi-3$ of the circle still colored red and any of the red points is on an all red equilateral triangle.






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        up vote
        10
        down vote










        up vote
        10
        down vote









        Make all the red points that are a distance exactly $frac 2pi3$ away from a blue point blue. The measure of the blue points is now no more than $3$, but the circumference of the circle is $pi$. There is at least $pi-3$ of the circle still colored red and any of the red points is on an all red equilateral triangle.






        share|cite|improve this answer














        Make all the red points that are a distance exactly $frac 2pi3$ away from a blue point blue. The measure of the blue points is now no more than $3$, but the circumference of the circle is $pi$. There is at least $pi-3$ of the circle still colored red and any of the red points is on an all red equilateral triangle.







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        share|cite|improve this answer








        edited Dec 10 at 5:22









        Acccumulation

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        6,6512616










        answered Dec 9 at 21:41









        Ross Millikan

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        290k23196369



























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