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Math Help - Isosceles triangle inscribed in a circle

  1. #1
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    Unhappy Isosceles triangle inscribed in a circle

    Determine the dimensions of the isosceles triangle inscribed in a circle of radius "r" that will give the triangle a maximum area. express the dimensions in terms of the circle's radius r.
    figure:


    determine the isosceles triangle's dimensions a, b,c
    (sorry guys, hope someone can help me out here, i'm gonna fail if I can't solve this one and I really don't know how to..unlucky for me, I don't really have any friend at this clas..please help)
    Last edited by mr fantastic; January 8th 2010 at 02:42 AM. Reason: Changed post title
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  2. #2
    MHF Contributor Amer's Avatar
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    we need to find the area with respect of s and r then derive it to find the maximum values

    ok

    Attachment 14705

    I divide the triangle to three equal triangles the hight of the small triangle is
    using Pythagorean theorem

    h = \sqrt{r^2 - \frac{s^2}{4}}

    the area of every small triangle is

    \frac{h \cdot s }{2}

    so the area of the large triangle is

    A = \frac{3s\cdot \sqrt{r^2 - \frac{s^2}{4}}}{2}

    A' = \left(\frac{3}{2}\right) \left( \sqrt{r^2 - \frac{s^2}{4}} - s\cdot \frac{s}{4} \cdot \frac{1}{\sqrt{r^2 - \frac{s^2}{4}}}\right)

    A' = \left(\frac{3}{2}\right) \left( \sqrt{r^2 - \frac{s^2}{4}} -   \frac{s^2}{4\sqrt{r^2 - \frac{s^2}{4}}}\right)

    A' = \frac{4r^2 - 2s^2 }{4\sqrt{r^2 - \frac{s^2}{4}}}

    critical points

    4r^2 - s^2 = 0 \Rightarrow s = \mp r \sqrt{2} but we have length so ignore negatives

    r^2 - \frac{s^2}{4} \Rightarrow s = 2r

    we have two positive critical points study the sign of the Area around these two
    the function is not define for s>=2r and s<=-2r

    for s values -r sqrt(2) < s< r sqrt(2) the first derivative is positive so the function is increasing

    for s> r sqrt(2) take for example 1.5r

     4r^2 - 2(1.5r)^2 = 4r^2 - 2(2.25 r^2) = 4r^2 - 4.5 r^2 <0

    for in this interval the first derivative is negative the function is decreasing

    so you have s= r sqrt(2) as a maximum point

    is there something you can't understand
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  3. #3
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    Hello meggy garcia

    Welcome to Math Help Forum!
    Quote Originally Posted by meggy garcia View Post
    Determine the dimensions of the isosceles triangle inscribed in a circle of radius "r" that will give the triangle a maximum area. express the dimensions in terms of the circle's radius r.
    figure:


    determine the isosceles triangle's dimensions a, b,c
    (sorry guys, hope someone can help me out here, i'm gonna fail if I can't solve this one and I really don't know how to..unlucky for me, I don't really have any friend at this clas..please help)
    In the attached diagram:
    AB = AC

    \angle ABC = \angle ACB = \theta

    AO = OB = r, the radius of the circumcircle

    M is the mid-point of BC

    \angle AOB = 2\angle ACB = 2\theta (angle at centre)
    So if P is the mid-point of AB:
    AP = PB = r\sin\theta, from \triangle APO

    \Rightarrow AM = 2r\sin^2\theta and BM = 2r\sin\theta\cos\theta, from \triangle ABM
    Therefore the area of \triangle ABC, A, is given by:
    A = \tfrac12 base \times height
    =BM\times AM

    = 4r^2\sin^3\theta\cos\theta
    So we need to find the value of \theta that makes A a maximum. The simplest method uses calculus:
    \frac{dA}{d\theta}= 4r^2(-\sin^4\theta +3\sin^2\theta\cos^2\theta)
    = 0 when
    \sin^2\theta(-\sin^2\theta+3\cos^2\theta) = 0

    \Rightarrow \sin^2\theta = 0 or -\sin^2\theta +3\cos^2\theta = 0
    \sin^2\theta = 0 gives a minimum value of A. The maximum occurs when
    \sin^2\theta = 3\cos^2\theta

    \Rightarrow \tan^2\theta = 3

    \Rightarrow \theta = \frac{\pi}{3}
    So (perhaps unsurprisingly) the area is maximum when the triangle is equilateral, with sides 2r\sin\left(\frac{\pi}{3}\right)=r\sqrt3

    Grandad
    Attached Thumbnails Attached Thumbnails Isosceles triangle inscribed in a circle-untitled.jpg  
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  4. #4
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    thank you...

    hi amer and grand dad..thanks a lot for the help..wow..you guys are really great,. this site is really helpful for someone like me,.

    you guys just saved me from not graduating next month..to be honest i don't really understand the equation because this is the only thing i have to do for my trigo class, coz of exemptions and the fact that my school isnt that good..well, thanks again..you guys completed my day..
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