Planimetrics - radii of incircles

matev91 · #1 $Old$ October 25th, 2009, 01:43 PM

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?

Grandad · #2 $Old$ October 26th, 2009, 11:09 AM

Hello matev91

Welcome to Math Help Forum!

Quote:

Originally Posted by matev91 $View Post$

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, $r$ , is given by the formula

$r=\frac{2\Delta}{a+b+c}$

and the radius of the circumcircle, $R$ , by the formula

$2R = \frac{a}{\sin A}=\frac{b}{\sin C}=\frac{c}{\sin C}$

This second formula - the Sine Rule, of course - gives

$a=2R\sin A, b= 2R\sin B$

and so if $a+b = 2R$ (i.e. $AC + BC = 2CO$ ):

$2R\sin A + 2R\sin B = 2R$

$\Rightarrow \sin A + \sin B = 1$

I denoted the lengths of the line segments as follows:

$SC = x, SB = y, SA = z$ and, as usual, $BC = a, AC = b$

and I also denoted the angle:

$\angle BSC = \theta$

Then, using the Sine Rule on $\triangle SBC$ :

$a = \frac{x\sin\theta}{\sin B}, y = \frac{x\sin(\theta + B)}{\sin B}$

So the area of the triangle is given by:

$\triangle SBC = \tfrac12ay\sin B = \frac{x^2\sin\theta\sin (\theta+B)}{2\sin B}$

and hence the radius, $r_b$ , of its incircle (when simplified) by:

$r_b = \frac{2\triangle SBC}{x+a+y}= \frac{x\sin\theta\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}$

In the same way, the radius, $r_a$ , of the incircle of $\triangle SAC$ is:

$r_a =\frac{x\sin\theta\sin(\theta-A)}{\sin A +\sin\theta+\sin(\theta-A)}$

So the problem can now be re-stated as:

Given that $\sin A + \sin B = 1$ , prove that, for all valid values of $\theta$ , $\frac{\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}=\frac{\sin(\theta-A)}{\sin A +\sin\theta+\sin(\theta-A)}$

It looks promising, but it just ain't so! The LHS can be manipulated as follows:

$\frac{\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}$

$=\frac{2\sin\tfrac12(\theta+B)\cos\tfrac12(\theta+b)}{2\sin\tfrac12(B +\theta)\cos\tfrac12(B-\theta) + 2\sin\tfrac12(\theta+B)\cos\tfrac12(\theta+b)}$

$=\frac{\cos\tfrac12(\theta+B)}{\cos\tfrac12(B-\theta) + \cos\tfrac12(\theta+B)}$

$=\frac{\cos\tfrac12\theta\cos\tfrac12B-\sin\tfrac12\theta\sin\tfrac12B}{2\cos\tfrac12\theta\cos\tfrac12B}$

$=\tfrac12(1-\tan\tfrac12\theta\tan\tfrac12B)$

and, in a similar way, the RHS $= \tfrac12(1-\cot\tfrac12\theta\tan\tfrac12A)$

So, in order for these to be equal, we would need $\tan\tfrac12\theta\tan\tfrac12B = \cot\tfrac12\theta\tan\tfrac12A$

i.e. $\tan^2\tfrac12\theta = \frac{\tan\tfrac12A}{\tan\tfrac12B}$ , for all values of $\theta$ , given $\sin A + \sin B = 1$ . 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

matev91 · #3 $Old$ October 26th, 2009, 06:21 PM

It's hard for me to believe that they are not the same length, but well it may be possible. Your soultion seems to be ok. Thank you for your help

Grandad · #4 $Old$ October 27th, 2009, 04:12 AM

Hello matev91

I am having a second look at this. I am not entirely sure that my reasoning is sound. I assumed that $\angle A$ and $\theta$ could be chosen independently. I'm no longer sure that this is true.

Grandad

matev91 · #5 $Old$ October 27th, 2009, 07:10 AM

Problem solved. I'll post solution later cause I have no time atm. Thanks for your help again

matev91

Grandad · #6 $Old$ October 27th, 2009, 08:03 AM

Hello again matev91

Indeed, I did get it wrong, and I now have the required proof - though it's still somewhat complicated. I wonder whether anyone can find a simpler one?

What I failed to spot was a simple relationship between the angles in the diagram, and it's this:

Produce the radius $CO$ to meet the circle again at $D$ ; join $DB$ . Then:

$\angle CBD = 90^o$ (angle in a semicircle)

$\angle ACD = \angle ABD$ (angles in same segment)

$= 90 - \angle CBA$

Thus, with my original notation, $\theta =A +\angle ACD$ (exterior angle of $\triangle ACS$ )

$\Rightarrow \theta = A + 90 - B$

This gives us:

$\sin\theta = \sin(90-(B-A))=\cos(B-A)$

$\sin(\theta-A)=\sin(90-B) = \cos B$

$\sin(\theta+B)=\sin(90+A)=\cos A$

Now in my original posting I showed that

Quote:

Originally Posted by Grandad $View Post$

So the problem can now be re-stated as:

Quote:

Originally Posted by Grandad $View Post$

Given that $\sin A + \sin B = 1$ , prove that, for all valid values of $\theta$ , $\frac{\sin(\theta+B)}{\sin B + \sin\theta + \sin(\theta+B)}=\frac{\sin(\theta-A)}{\sin A +\sin\theta+\sin(\theta-A)}$

We can now eliminate $\theta$ and $B$ (noting that $\cos B = \sqrt{1-(1-\sin A)^2}=\sqrt{\sin A(2-\sin A)})$ , first from the LHS:

$\frac{\sin(\theta +B)}{\sin B + \sin\theta + \sin(\theta +B)}$

$=\frac{\cos A}{\sin B + \cos(B-A)+\cos A}$

$=\frac{\cos A}{1-\sin A + \cos A \cos B + \sin A \sin B+\cos A}$

$=\frac{\cos A}{1-\sin A + \cos A \sqrt{\sin A(2-\sin A)} + \sin A (1-\sin A)+\cos A}$

$=\frac{\cos A}{1-\sin^2 A + \cos A \sqrt{\sin A(2-\sin A)} +\cos A}$

$=\frac{1}{\cos A + \sqrt{\sin A(2-\sin A)} +1}$

And the RHS:

$\frac{\cos B}{\sin A +\cos(B-A)+\cos B}$

$=\frac{\sqrt{\sin A(2-\sin A)}}{\sin A + \cos A\sqrt{\sin A(2-\sin A)}+(1-\sin A)\sin A +\sqrt{\sin A(2-\sin A)}}$

$=\frac{\sqrt{\sin A(2-\sin A)}}{\sin A(2 - \sin A) + \cos A\sqrt{\sin A(2-\sin A)} +\sqrt{\sin A(2-\sin A)}}$

$=\frac{1}{\sqrt{\sin A(2-\sin A)}+\cos A +1}$

And that completes the proof.

There has to be a simpler solution!

Grandad

PS Ah - I see I'm too late. Well done for finding the solution in spite of me! I feel sure you've got something simpler. Perhaps you'd like to post it here if so, so that everyone can see.

Opalg · #7 $Old$ October 27th, 2009, 03:53 PM

I also spent quite a lot of time on this problem. I came to the conclusion that it is correct, and I have a rather long and clumsy proof of it. However, I have seen the same problem posed in a number of online forums, including here, where it is claimed that it is a problem in the current Polish Math Olympiad, and hints and answers to it should not be given. It seems a shame to delete the posts here, when so much time and effort have been spent on them, but perhaps the Mods ought to do so nonetheless, if the information about the Polish Math Olympiad is correct (I have no idea how one might confirm that).