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$y=kx, \;\; k>0.\;$ $A>B>0$ are 2 different points on $x$ ass.

I want to prove that $ \forall x_1,x_2:$ if $A\leq x_1<x_2\leq B$ then incircle for $\bigtriangleup AC_1B$ is smaller than incircle for $\bigtriangleup AC_2B .$ Where $C_1=(x_1,kx_1),C_2=(x_2,kx_2)$

Since radius for incircle is: $r=\dfrac{2S}{P}$. One ends up comparing $\dfrac{x_1}{a_1+b_1+c_1}?<?\dfrac{x_2}{a_2+b_2+c_2}.$ It is not difficult to express $a_1,b_1,c_1$, but things got unmanagable when I tried to compare them. I would like to prove this with elementary methods (If that is possible). P.S. I am not entirely sure that this ''theorem'' is true. And sorry for no picture.

Karlis Olte
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    I think things can get easier if you consider a line like $y=kx+m \ k,m>0$ and then setting $B=0$. Also you can work on the derivative of $r(x)$ – hhsaffar Dec 14 '13 at 00:11

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it is not true.and I suppose the A>B is typo. I use $0< A\leq x_1<x_2\leq B$ as the condition:$A(a,0),B(b,0),b>a>0$

let $x$ on$(a,b),C(x,kx) \to r=\dfrac{kx(b-a)}{2(\sqrt{(x-a)^2+(kx)^2}+\sqrt{(x-b)^2+(kx)^2}+b-a)}$ ,

if $a=2,b=3,k=0.5, x_1=2.95,x_2=2.99 \implies r_1=0.174337,r_2=0.174319 \implies r_1>r_2$,

it is because $r(x)$ has min point between $(a,b)$ for some $a,b$

chenbai
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