Hello meggy garcia

Welcome to Math Help Forum! Originally Posted by

**meggy garcia** 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

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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:$\displaystyle AB = AC$

$\displaystyle \angle ABC = \angle ACB = \theta$

$\displaystyle AO = OB = r$, the radius of the circumcircle

$\displaystyle M$ is the mid-point of $\displaystyle BC$

$\displaystyle \angle AOB = 2\angle ACB = 2\theta$ (angle at centre)

So if $\displaystyle P$ is the mid-point of $\displaystyle AB$:$\displaystyle AP = PB = r\sin\theta$, from $\displaystyle \triangle APO$

$\displaystyle \Rightarrow AM = 2r\sin^2\theta$ and $\displaystyle BM = 2r\sin\theta\cos\theta$, from $\displaystyle \triangle ABM$

Therefore the area of $\displaystyle \triangle ABC, A,$ is given by:$\displaystyle A = \tfrac12$ base $\displaystyle \times$ height$\displaystyle =BM\times AM $

$\displaystyle = 4r^2\sin^3\theta\cos\theta$

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

$\displaystyle \sin^2\theta(-\sin^2\theta+3\cos^2\theta) = 0$

$\displaystyle \Rightarrow \sin^2\theta = 0$ or $\displaystyle -\sin^2\theta +3\cos^2\theta = 0$

$\displaystyle \sin^2\theta = 0$ gives a minimum value of $\displaystyle A$. The maximum occurs when

$\displaystyle \sin^2\theta = 3\cos^2\theta$

$\displaystyle \Rightarrow \tan^2\theta = 3$

$\displaystyle \Rightarrow \theta = \frac{\pi}{3}$

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

Grandad