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Thread: Putnam Problem of the Day (3)

  1. #1
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    Putnam Problem of the Day (3)

    This problem was really challenging to me , I've spent much time on this but maybe it is easy to you guys . Enjoy !

    Triangle $\displaystyle ABC$ has the following property : there is an interior point P such that $\displaystyle \angle PAB = 10^o $ , $\displaystyle \angle PBA = 20^o$ , $\displaystyle \angle PCA = 30^o$ and $\displaystyle \angle PAC = 40^o$ . Prove that triangle $\displaystyle ABC$ is isosceles .



    Moderator edit: Approved Challenge question.
    Last edited by simplependulum; Jun 18th 2010 at 08:39 PM. Reason: Approval
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    Quote Originally Posted by simplependulum View Post
    This problem was really challenging to me , I've spent much time on this but maybe it is easy to you guys . Enjoy !

    Triangle $\displaystyle ABC$ has the following property : there is an interior point P such that $\displaystyle \angle PAB = 10^o $ , $\displaystyle \angle PBA = 20^o$ , $\displaystyle \angle PCA = 30^o$ and $\displaystyle \angle PAC = 40^o$ . Prove that triangle $\displaystyle ABC$ is isosceles .



    Moderator edit: Approved Challenge question.
    Maybe I'm close? I don't know how to show without a calculator that

    Spoiler:

    $\displaystyle \sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ}$

    which I got through using law of sines and then guessing which sides are congruent (not very elegant I know).
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    Quote Originally Posted by undefined View Post
    Maybe I'm close? I don't know how to show without a calculator that

    Spoiler:

    $\displaystyle \sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ}$

    which I got through using law of sines and then guessing which sides are congruent (not very elegant I know).
    Morrie's law?
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    Quote Originally Posted by TheCoffeeMachine View Post
    Yes that works perfectly! Thanks.
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    $\displaystyle \sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ} $

    Of course it is correct , by writing $\displaystyle \sin{70^o} = \cos{20^o}~ ~,~ \sin{50^o} = \cos{40^o} $

    $\displaystyle 4\cdot\sin20^{\circ}\cdot\sin50^{\circ}\cdot\sin70 ^{\circ} = (2\sin{20^o}\cos{20^o}) (2\cos{40^o}) $

    $\displaystyle = 2\sin{40^o}\cos{40^o} = \sin{80^o} $

    It is a nice trigonometric solution to the problem , mine is a pure geometrical solution , see if you can find any .
    Last edited by simplependulum; Jun 18th 2010 at 08:40 PM.
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    Quote Originally Posted by simplependulum View Post
    $\displaystyle \sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ} $

    Of course it is correct , by writing $\displaystyle \sin{70^o} = \cos{20^o}~ ~,~ \sin{50^o} = \cos{40^o} $

    $\displaystyle 4\cdot\sin20^{\circ}\cdot\sin50^{\circ}\cdot\sin70 ^{\circ} = (2\sin{20^o}\cos{20^o}) (2\cos{\color{red}50\color{black}^o}) $

    $\displaystyle = 2\sin{40^o}\cos{40^o} = \sin{80^o} $

    It is a nice trigonometric solution to the problem , mine is a pure geometrical solution , see if you can find any .
    Can't believe I've forgotten the double angle formulas already. Only somewhat justifiable by the fact that I haven't used them in about 5 years.

    Not sure I'll have any luck with that pure geometric solution though.

    By the way there was a small typo in your post, in red above.
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    Oh it is a typo , thanks for the correction , it is clear now .
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