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Math Help - Area of the triangle through the circumcenter

  1. #1
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    Area of the triangle through the circumcenter

    Tryin' to help my niece on her homework: The area of triangle ABC is 133. The angle bisectors meet at point I. AB:BC:CA=4:7:8. What is the area of triangle BCI?

    We have that angle bisectors meet at the circumcenter, which is the center of the circumscribed circle. We have a bunch of things about medians, perpendicular bisectors, and altitudes. It seems logical that BCI is about 1/3 of ABC, but you know the math teacher isn't going to accept that for an answer.

    Tx
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  2. #2
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    Re: Area of the triangle through the circumcenter

    Quote Originally Posted by mathDad View Post
    Tryin' to help my niece on her homework: The area of triangle ABC is 133. The angle bisectors meet at point I. AB:BC:CA=4:7:8. What is the area of triangle BCI?

    We have that angle bisectors meet at the circumcenter, which is the center of the circumscribed circle. <-- unfortunately no
    We have a bunch of things about medians, perpendicular bisectors, and altitudes. It seems logical that BCI is about 1/3 of ABC, but you know the math teacher isn't going to accept that for an answer.

    Tx
    1. The point I is the centre of the inscribed circle (therefore the name I). Let r denote the radius of the incircle.

    2. Then the area of the complete triangle is calculated by:

    A_{\Delta} = \frac12 a r+\frac12 b r+\frac12 c r = \frac12 r (a+b+c)

    3. You don't know the actual lengthes of the sides but

    c = 4k
    a = 7k
    b = 8k

    That means the area of the triangle is

    A_{\Delta} = \frac12 r k (4+7+8)

    4. The area of the triangle BCI is

    A_{BCI} = \frac12 r k \cdot 7

    5: The proportion \frac{A_{BCI}}{A_{\Delta}} = \frac{\frac12 r k \cdot 7}{\frac12 r k (4+7+8)} = \frac7{19}

    6. Now you can determine the area of \Delta(BCI). I've got 49.
    Attached Thumbnails Attached Thumbnails Area of the triangle through the circumcenter-rhoaus3eckfl.png  
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  3. #3
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    Re: Area of the triangle through the circumcenter

    Thanks, earboth. Good solution.

    Did they really think a 10th grader could do this problem? Wow.
    Last edited by mathDad; November 22nd 2011 at 08:16 PM.
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