Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.
What are lengths od AP, BP abd CP?
Work so far:
area of ABC = 33SQRT(39) / 4
triangle APB : height from P = (33SQRT(39)/2) / 36
triangle APC : height from P = (33SQRT(39)/2) / 33
triangle BPC : height from P = (33SQRT(39)/2) / 30
From that, how would you calculate lengths of AP, BP and CP?
I keep going around in circles! Thanks.
Alternatively, you could play with co-ordinate geometry.
Placing B at the origin, C is at (10,0).
Then locate the co-ordinates of A.
You've worked out the triangle areas.
You've got the height of P from the line [BC].
Then P lies on the line y=h.
The intersection of the line parallel to [AB] with appropriate perpendicular distance from [AB], with y=h,
or the intersection of the line parallel to [AC] with y=h, correct distance from [AC]
Jul 24th 2010, 10:37 AM
Had a closer look at this; turns out easy enough.
We have triangle ABC with BC = a =10, AC = b = 11 and AB = c = 12 ; point P inside.
Make B(0,0) and A(12,0) ; and C(g,h)
Get the area k.
h = 2k / c
g = SQRT(a^2 - h^2)
x = (c + g) / 3
y = h / 3
NOTE: since 3 inside triangles equal in area, then P is the centroid!