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Thread: Finding maximum area of triangle

  1. #1
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    Finding maximum area of triangle

    Hi, I have the following problem.

    "An isosceles triangle ABC, where AB = AC and BāC = 2(alpha) is inscribed in a circle of radius r. Show that the area of the triangle ABC is a maximum if alpha = pi/6."

    To be honest I have no idea where to start. Also, I'm not sure if I'm in the right subforum, so move me if need be.
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  2. #2
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    Re: Finding maximum area of triangle

    Well, I know at least that I'm looking for dA/d(alpha), which is equal to dA/dr * dr/d(alpha)
    dA/dr would be the derivative of the area respect to r, but I'm not sure what the formula for the area of this triangle is.. heheh..
    dr/d(alpha) I -think- I can get with the cosine rule, from where I get the relationship arccos((a^2 -2r^2)/(-2r^2)) = alpha.
    Derivating that I get dr/da =$\displaystyle -2*a/(r*sqrt(2*r-a)*sqrt(2*r+a))$

    So yup, any help would be appreciated.
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  3. #3
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    Re: Finding maximum area of triangle

    Wikipedia has the following formula for area: $\displaystyle S=2r^2\sin(2\alpha)\sin^2\beta$ where $\displaystyle \beta$ is the measure of the other two angles.
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  4. #4
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    Re: Finding maximum area of triangle

    The following link might be useful
    Isosceles Triangle Equations Formulas Calculator - Area Geometry

    Has many useful formulas for various triangles. One of which is the radius of a circumscribed circle
    $\displaystyle r=\frac{a}{2 \sin(A)}$

    Switching it around you get
    $\displaystyle \sin(A)=\frac{a}{2 r}$

    You can plug those into the area formula and go from there.
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  5. #5
    Member sbhatnagar's Avatar
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    Re: Finding maximum area of triangle

    Recall that in any triangle, $\displaystyle \sin{A}=\frac{2\Delta}{bc}$. "$\displaystyle \Delta$" is the area of the trianlgle.
    If k is one of the equal sides, $\displaystyle \sin{2\alpha}=\frac{2\Delta}{k^2}$.........[1]



    Also, $\displaystyle \cos{(\pi-2\alpha)}=\frac{2r^2-k^2}{2r^2}$.........[Cosine Formula in $\displaystyle \Delta AOC$]

    $\displaystyle -\cos{(2\alpha)}=\frac{2r^2-k^2}{2r^2}$

    $\displaystyle k^2=2r^2[1+\cos{2\alpha}]$.........[2]

    Substitute [2] into [1]:
    $\displaystyle \sin{2\alpha}=\frac{2\Delta}{2r^2[1+\cos{2\alpha}]}$

    This is the relation between $\displaystyle \alpha$ and area of the triangle.
    $\displaystyle \Delta = r^2[\sin{2\alpha}+ \sin{2\alpha}\cos{2\alpha}]$

    $\displaystyle \frac{d\Delta}{d\alpha} = r^2[4\cos^2{\alpha}(2\cos{2\alpha}-1)]$

    I hope you can easily proceed from here...
    Last edited by sbhatnagar; Nov 16th 2011 at 03:23 AM.
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