>Dear Hyacinthians,

>

>May be some of you will find the following interesting:

>

>In a triangle ABC with a-side and b-bisector m_b

predefined,

>consider the relation between B-angle an C-bisector t_c:

>t_c = F(B). Prove that F is monotonic.

>

>I believe I have a simple argument in case m_b >= a.

>Unfortunately, it breaks up when the condition is

reversed...

>

>Regards,

>Barukh.

If m and t are the b-bisector and C-bisector respectively

then

4m^2 = 2(a^2+c^2)-b^2 hence c^2 = (4m^2-2a^2+b^2)/2

t^2 = a*b*(a+b+c)*(a+b-c)/(a+b)^2=

= a*b*(4a^2-4m^2+4ab+b^2)/(a+b)^2 = f(b)

where a, m are constants. This function has positive

derivative and hence is increasing.

If O is the reflection of C about B then BA = 2m and the

locus of A is the circle with center O and radius 2m.

If OC meets this circle at A1, A2 (A1 closer to C) then

it is known that the distance b = AC is increasing as A

moves on the circle from A1 to A2 and also the angle B

is increasing and conversely. Hence

B1 > B2 => b1 > b2 => f(b1) > f(b2) => t1 > t2 =>

=> F(B1) > F(B2).

Hence the function F(B) is increasing.

Best regards

Nikos Dergiades