Two
medians of a triangle have lengths of 12 and 15. How long is the third median when the area of the triangle is a maximum? (Try to solve this without calculus.)
Let ABC be the triangle, X' the midpoint of the side
opposite vertex X, and AA' and BB' the given medians.
If the coordinates of the centroid are (0,0), then
the coordinates of the vertices and midpoints are
A = (2a,0) A' = (a,0)
B = (2b*cos(x),2b*sin(x)) B' = (b*cos(x),b*sin(x))
C = (2b*cos(x)2a,2b*sin(x)) C' = (b*cos(x)+a,b*sin(x))
Therefore,
Area(ABC) = Area(ABA') + Area(ACA')
= 2*Area(ABA')
= 6ab*sin(x)
Clearly, the maximum area occurs when x = 90 degrees.
Therefore,
C = (2a,2b) C' = (a,b)
and
CC' = 3*sqrt(a^2+b^2)
For our problem,
CC' = 3*sqrt(5^2+4^2) = 3*sqrt(41)

Posted by Bractals
on 20080203 12:13:42 