## Re: [EMHL] Monotonicity

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• ... predefined, ... reversed... ... 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
Message 1 of 1 , Sep 2, 2002
>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