In ΔABC the altitude, angle bisector, and median from C divide ∠C into four equal angles.
What is the ratio of sides a and b (shorter:longer)?
Call the places where the altitude, angle bisector and median intersect AB by D, E and F respectively.
angles BCD = DCE = ECF = FCA = x
h=the altitude CD
so we have the following lengths:
BD = h*tan x
ED = h*tan x
FD = h*tan 2x
AD = h*tan 3x
and since F bisects AB
h*tan(3x)  h*tan(2x) = h*tan(x) + h*tan(2x)
sin(3x)/cos(3x)  sin2x)/cos(2x) = sin(x)/cos(x) + sin(2x)/cos(2x)
cos(x)cos(2x)sin(3x)  cos(x)sin(2x)cos(3x) = sin(x)cos(2x)cos(3x) + cos(x)sin(2x)cos(3x)
cos(x)[cos(2x)sin(3x)sin(2x)cos(3x)] = cos(3x)[sin(x)cos(2x)+cos(x)sin(2x)]
cos(x)sin(x) = cos(3x)sin(3x)
.5sin(2x) = .5sin(6x)
let y = 2x
sin(y) = sin(3y)
sin(y) = 3sin(y)  4sin(y)^3
2sin(y)  4sin(y)^3 = 0
12sin(y)^2 = 0
sin(y) = +/ sqrt(2)/2
y = 45 degrees = 2x
x = 22.5 degrees
So angle BCA = 4x = 90 degrees (a right triangle!)
angle BAC = 903x = 22.5 degrees
The ratio of a to b = BC/AC = tan(22.5) = tan(45/2) = (1cos(45))/sin(45) = (1  1/sqrt(2))/(1/sqrt(2)) = sqrt(2)  1
or about .4142

Posted by Jer
on 20101202 15:34:48 