Hello matev91
Welcome to Math Help Forum! Originally Posted by
matev91 Triangle ABC (where ACB>90) is inscribed into circle with centre O. Line CO
intersects segment AB at point S. Prove that if AC+BC=2CO then radii of incircles of triangles ASC and BSC have the same length.
Could anyone help me with that please?
I have spent some time looking at this question, and have to report that it's not true - the radii of the incircles are not equal in length.
With the usual notation, the radius of the incircle of a triangle, $\displaystyle r$, is given by the formula$\displaystyle r=\frac{2\Delta}{a+b+c}$
and the radius of the circumcircle, $\displaystyle R$, by the formula$\displaystyle 2R = \frac{a}{\sin A}=\frac{b}{\sin C}=\frac{c}{\sin C}$
This second formula - the Sine Rule, of course - gives$\displaystyle a=2R\sin A, b= 2R\sin B$
and so if $\displaystyle a+b = 2R$ (i.e. $\displaystyle AC + BC = 2CO$):$\displaystyle 2R\sin A + 2R\sin B = 2R$
$\displaystyle \Rightarrow \sin A + \sin B = 1$
I denoted the lengths of the line segments as follows:$\displaystyle SC = x, SB = y, SA = z$ and, as usual, $\displaystyle BC = a, AC = b$
and I also denoted the angle:$\displaystyle \angle BSC = \theta$
Then, using the Sine Rule on $\displaystyle \triangle SBC$:$\displaystyle a = \frac{x\sin\theta}{\sin B}, y = \frac{x\sin(\theta + B)}{\sin B}$
So the area of the triangle is given by:$\displaystyle \triangle SBC = \tfrac12ay\sin B = \frac{x^2\sin\theta\sin (\theta+B)}{2\sin B}$
and hence the radius, $\displaystyle r_b$, of its incircle (when simplified) by:$\displaystyle r_b = \frac{2\triangle SBC}{x+a+y}= \frac{x\sin\theta\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}$
In the same way, the radius, $\displaystyle r_a$, of the incircle of $\displaystyle \triangle SAC$ is:$\displaystyle r_a =\frac{x\sin\theta\sin(\theta-A)}{\sin A +\sin\theta+\sin(\theta-A)}$
So the problem can now be re-stated as:Given that $\displaystyle \sin A + \sin B = 1$, prove that, for all valid values of $\displaystyle \theta$, $\displaystyle \frac{\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}=\frac{\sin(\theta-A)}{\sin A +\sin\theta+\sin(\theta-A)}$
It looks promising, but it just ain't so! The LHS can be manipulated as follows:
$\displaystyle \frac{\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}$$\displaystyle =\frac{2\sin\tfrac12(\theta+B)\cos\tfrac12(\theta+ b)}{2\sin\tfrac12(B +\theta)\cos\tfrac12(B-\theta) + 2\sin\tfrac12(\theta+B)\cos\tfrac12(\theta+b)}$
$\displaystyle =\frac{\cos\tfrac12(\theta+B)}{\cos\tfrac12(B-\theta) + \cos\tfrac12(\theta+B)}$
$\displaystyle =\frac{\cos\tfrac12\theta\cos\tfrac12B-\sin\tfrac12\theta\sin\tfrac12B}{2\cos\tfrac12\the ta\cos\tfrac12B}$
$\displaystyle =\tfrac12(1-\tan\tfrac12\theta\tan\tfrac12B)$
and, in a similar way, the RHS $\displaystyle = \tfrac12(1-\cot\tfrac12\theta\tan\tfrac12A)$
So, in order for these to be equal, we would need $\displaystyle \tan\tfrac12\theta\tan\tfrac12B = \cot\tfrac12\theta\tan\tfrac12A$
i.e. $\displaystyle \tan^2\tfrac12\theta = \frac{\tan\tfrac12A}{\tan\tfrac12B}$, for all values of $\displaystyle \theta$, given $\displaystyle \sin A + \sin B = 1$. But this simply is not so.
I have put some calculations together into an Excel spreadsheet to confirm this - that we can fulfill all the conditions and yet produce different radii. I attach the Excel file.
Grandad