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Thread: Circle inscribed of triangle

  1. #1
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    Circle inscribed of triangle

    The circle O is an inscribed circle of \Delta ABC and points P, Q and R are the points of tangency of sides BC, CA and AB respectively.
    AB = AC = 13, BC = 10.

    The scalar product of two vectors \vec{AB}.\vec{AO}, \vec{AB}.\vec{BC}?

    Any tips?


    Look at the picture.
    Circle inscribed of triangle-hi.jpg
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    Re: Circle inscribed of triangle

    Quote Originally Posted by provasanteriores View Post
    The circle O is an inscribed circle of \Delta ABC and points P, Q and R are the points of tangency of sides BC, CA and AB respectively.
    AB = AC = 13, BC = 10.
    The scalar product of two vectors \vec{AB}.\vec{AO}, \vec{AB}.\vec{BC}?
    Any tips? Look at the picture.
    Click image for larger version. 

Name:	hi.jpg 
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ID:	36999
    From the given, it follows that $|BP|=|PC|=5~$ or $\Delta BAC$ is isosceles.
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    Re: Circle inscribed of triangle



    $\vec{u} \cdot \vec{v} = |u| \, |v| \, \cos{\theta}$, where $\theta$ is the angle between $\vec{u} \text{ and } \vec{v}$

    note triangles BPA and CPA are 5-12-13 right triangles ... circle radius is rather simple to determine.
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    Re: Circle inscribed of triangle

    Quote Originally Posted by skeeter View Post


    $\vec{u} \cdot \vec{v} = |u| \, |v| \, \cos{\theta}$, where $\theta$ is the angle between $\vec{u} \text{ and } \vec{v}$

    note triangles BPA and CPA are 5-12-13 right triangles ... circle radius is rather simple to determine.
    thanks, I got it now

    I found 104 e 50. The answers are 104 e -50 (why minus?)

    \vec{AB}= -\vec{BA}?
    Last edited by provasanteriores; Feb 12th 2017 at 12:45 PM.
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    Re: Circle inscribed of triangle

    Quote Originally Posted by Plato View Post
    From the given, it follows that $|BP|=|PC|=5~$ or $\Delta BAC$ is isosceles.
    I saw, thanks
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    Re: Circle inscribed of triangle

    $\vec{AB} \cdot \vec{BC} = |AB| \, |BC| \, \cos(\pi - \angle{ABC}) = 13 \cdot 10 \cdot \left(-\dfrac{5}{13}\right) = -50$

    that's why
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    Re: Circle inscribed of triangle

    Quote Originally Posted by skeeter View Post
    $\vec{AB} \cdot \vec{BC} = |AB| \, |BC| \, \cos(\pi - \angle{ABC}) = 13 \cdot 10 \cdot \left(-\dfrac{5}{13}\right) = -50$

    that's why

    why pi?


    \vec{AB} \cdot \vec{BC} = |AB| \, |BC| \, \cos(\angle{ABC}) = 13 \cdot 10 \cdot \left(\dfrac{5}{13}\right) = 50
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    Re: Circle inscribed of triangle

    [QUOTE=provasanteriores;915999]why pi?/QUOTE]

    see diagram ...
    Attached Thumbnails Attached Thumbnails Circle inscribed of triangle-vector3.png  
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