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anonymous

  • 4 years ago

Challenge: Let ABC be an equilateral triangle and P an arbitrary point inside ABC such that : max{PA,PB,PC} = 1/2(PA+PB+PC) Find the locus of P.

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  1. anonymous
    • 4 years ago
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    What does max{PA,PB,PC} implies?

  2. anonymous
    • 4 years ago
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    It means the one among PA,PB,PC which has the maximum length.

  3. anonymous
    • 4 years ago
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    max{PA,PB,PC} = the maximum distance between P to either of A, B or C And if P lies on the circumcircle , then one of PA, PB or PC is sum of the other 2.

  4. anonymous
    • 4 years ago
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    max{PA,PB,PC} = 1/2(PA+PB+PC) And this whole equation?

  5. anonymous
    • 4 years ago
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    this means that , the max distance from P to any pt ( A, B or C) is equal to half of the sum of all three distances.

  6. dumbcow
    • 4 years ago
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    im thinking there are no points that fulfill that condition

  7. anonymous
    • 4 years ago
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    The locus is the circumcircle of ABC .Can you prove it?

  8. anonymous
    • 4 years ago
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    max{PA,PB,PC} = a which makes it lie on the circumcircle. Now as thinker said if one of them lie on circumcircle then the sum of other two is equal to the third. max{PA,PB,PC} = 1/2 (PA+PB+PC) a = 1/2 (a+a)

  9. anonymous
    • 4 years ago
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    This problem is probably taken from here: http://www.scribd.com/dorh7343/d/23254911-Mathematical-Olympiad-Challenges

  10. anonymous
    • 4 years ago
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    I didn't prove it, I know I just did an intuitive proof

  11. dumbcow
    • 4 years ago
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    |dw:1328360697679:dw| ok if it lies on circumcircle then sqrt(3)x/2 = x ??

  12. anonymous
    • 4 years ago
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    Thanks for the link foolformath

  13. dumbcow
    • 4 years ago
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    how can PB = PA+PC ?

  14. anonymous
    • 4 years ago
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    Pompieu's Theorem

  15. dumbcow
    • 4 years ago
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    @Aron sorry but could you elaborate, im not familiar with that Theorem i dont see how any point on the circumcircle meets the requirement of one length being the sum of the other 2 lengths

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