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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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.

Mathematics
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What does max{PA,PB,PC} implies?
It means the one among PA,PB,PC which has the maximum length.
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.

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max{PA,PB,PC} = 1/2(PA+PB+PC) And this whole equation?
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.
im thinking there are no points that fulfill that condition
The locus is the circumcircle of ABC .Can you prove it?
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)
This problem is probably taken from here:http://www.scribd.com/dorh7343/d/23254911-Mathematical-Olympiad-Challenges
I didn't prove it, I know I just did an intuitive proof
|dw:1328360697679:dw| ok if it lies on circumcircle then sqrt(3)x/2 = x ??
Thanks for the link foolformath
how can PB = PA+PC ?
Pompieu's Theorem
@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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