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