Finding maximum area of triangle

• Nov 15th 2011, 09:58 AM
Capricious
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.
• Nov 15th 2011, 10:13 AM
Capricious
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.
• Nov 15th 2011, 01:53 PM
emakarov
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.
• Nov 15th 2011, 02:02 PM
takatok
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.
• Nov 16th 2011, 01:20 AM
sbhatnagar
Re: Finding maximum area of triangle
Quote:

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]

Quote:

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

Quote:

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...