Let ABC be a rightangled triangle with hypotenuse c = AB.
Let Wa and Wb be the lengths of the angle bisectors from A & B, respectively.
Prove that Wa + Wb ≤ 2c*sqrt( 2−√2).
Outline
proof:
The bisector of angle A has length W(A) = c*cos(A)/cos(A/2)
=
c(2cos(A/2) – sec(A/2))
Thus W’(A) = c*sin(A/2)(1 + sec^{2}(A/2)/2)
For 0 < A < pi/2, both sin(A/2) and sec(A/2) are increasing,
so W’(A) is a decreasing function of A. i.e. W is concave.
Length of bisector of angle B is
W(B) = W(pi/2 – A),
whose graph is a reflection of W(A) in the line
A = pi/4.
It follows that W(B) is also concave wrt A and that their
sum will therefore be concave, with a maximum point where
they cross at A = pi/4.
Thus: W(A) +
W(B) <= 2*W(pi/4)
<=
2c*cos(pi/4)/cos(pi/8)
<=
2c*cos(pi/4)/sqrt((1 + cos(pi/4))/2)
<= 2c/sqrt(1
+ 1/sqrt(2))
<=
2c*sqrt(2 – sqrt(2))
Now, is there a simpler approach using geometry?

Posted by Harry
on 20170928 11:57:05 