Expert: Sherman D. - 3/26/2007

Question
QUESTION: Please answer my questions:
The incentre of the triangle with vertices (1,�ã3),(0,0) and (2,0) is?
A pole stands vertically inside a triangular park ABC.If the angle of elevation of the top of the pole from each corner of the park is the same,then in triangle ABC the foot of the pole is the:
1.Centroid  2.Circumcentre 3.incentre  4.orthocentre
Thank you.
ANSWER: On The incentre of the triangle with vertices (1,�ã3),(0,0) and (2,0) is?

do you mean (1,sqrt(3))

since the angle of elevation is the same, then i guess that would make the park an equilateral triangle, and therefore if you were to draw a circle around the triangle connecting each vertice, the place where the pole is standing would be in the perfect center of the triangle.

ANS : Circumcentre

If it had been to each side instead of corner, then it would had been Incentre

For a better look, go to

http://mathworld.wolfram.com/Circumcenter.html
http://mathworld.wolfram.com/Orthocenter.html
http://en.wikipedia.org/wiki/Centroid
http://mathworld.wolfram.com/Incenter.html

---------- FOLLOW-UP ----------

QUESTION: thank you.Yes I do mean sqrt 3.So then how is the question done?

Answer
(1,sqrt(3)),(0,0), and (2,0)

(1,sqrt(3)) and (0,0)
D = sqrt((0 - 1)^2 + (0 - sqrt(3))^2)
D = sqrt((-1)^2 + (-sqrt(3))^2)
D = sqrt(1 + 3)
D = sqrt(4)
D = 2

(0,0) and (2,0)
D = sqrt((2 - 0)^2 + (0 - 0)^2)
D = sqrt((2)^2 + 0^2)
D = sqrt(4)
D = 2

(2,0) and (1,sqrt(3))
D = sqrt((1 - 2)^2 + (sqrt(3) - 0)^2)
D = sqrt((-1)^2 + 3)
D = sqrt(1 + 3)
D = sqrt(4)
D = 2

Since we know this is an equilateral triangle, the distances between the vertices and the incentre are all equal, therefore

Using the distance formula

(0,0) and (x,y)
D = sqrt(x^2 + y^2)

(1,sqrt(3)) and (x,y)
D = sqrt((x - 1)^2 + (y - sqrt(3))^2)

(2,0) and (x,y)
D = sqrt((x - 2)^2 + y^2)

sqrt(x^2 + y^2) = sqrt((x - 2)^2 + y^2)
x^2 + y^2 = (x - 2)^2 + y^2
x^2 = x^2 - 4x + 4
-4x + 4 = 0
-4x = -4
x = 1

sqrt((x - 1)^2 + (y - sqrt(3))^2) = sqrt(x^2 + y^2)
(x - 1)^2 + (y - sqrt(3))^2 = x^2 + y^2
x^2 - 2x + 1 + y^2 - 2sqrt(3)y + 3 = x^2 + y^2
-2x - 2sqrt(3)y + 4 = 0
-2x - 2sqrt(3)y = -4
-2(x + sqrt(3)y) = -4
x + sqrt(3)y = 2
sqrt(3)y = -x + 2
y = (-x + 2)/sqrt(3)
y = (-1 + 2)/(sqrt(3))
y = 1/sqrt(3)
y = (sqrt(3)/3)

Incentre = (1,(sqrt(3)/3))