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Math Help - Planimetrics - radii of incircles

  1. #1
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    Planimetrics - radii of incircles

    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?
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  2. #2
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    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\the  ta\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
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  3. #3
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    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
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  4. #4
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    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
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    Problem solved. I'll post solution later cause I have no time atm. Thanks for your help again

    matev91
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  6. #6
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    Solution!

    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.
    Last edited by Grandad; October 27th 2009 at 07:05 AM. Reason: Add PS
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  7. #7
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    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).
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