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IsTim
 4 years ago
Triangle ABC is an equilateral triangle, with O as its centroid. Show that OA+OB+OC=0.
IsTim
 4 years ago
Triangle ABC is an equilateral triangle, with O as its centroid. Show that OA+OB+OC=0.

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Hero
 4 years ago
Best ResponseYou've already chosen the best response.1The distance is zero? How can that be?

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.2That's what I think it looks like.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1Yeah, but distance is always positive so how can you add three positive distances and get zero?

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, you should post them as vectors.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1I'll try to get an answer to this for you. No promises though

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.2I don't seem to follow the process.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1I'm sorry to hear that. I thought it would have been obvious

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.2Is there a specific order you did this from?

Hero
 4 years ago
Best ResponseYou've already chosen the best response.11. ABC is an equilateral Triangle 2. Segments OA, OB and OC are equidistant from the center 3. AB = BC = AC 4. AB, BC, and AC are labeled side s. 5. OA, OB and OC bisect angles A, B, and C. 6. OA bisects BC. 7. Create point D. 8. BD = s/2 ; CD = s/2 9. x compnent OA = 0 because it's x component has no distance x compoent OB = s/2 x component OC = s/2 10. x component 0 + s/2  s/2 = 0 11. y component OB = s/2 tan(30) = sqrt(3)/6 y component OB = y component OC = sqrt(3)/6 12. y component OA = OA = OB 13. OB = s/2 cos(30) = s*sqrt(3)/3 14. y component OA + OB + OC = sqrt(3)/3  sqrt(3)/6  sqrt(3)/6 = 0 QED

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1I'll take questions concerning specific steps you have questions on

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.2I'm just interpreting it right now. I guess I'm good for now. Thanks.
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