Isosceles triangle inscribed in a circle

**meggy garcia** · January 7th 2010, 05:24 AM

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)

**Amer** · January 7th 2010, 06:58 AM

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

**Grandad** · January 7th 2010, 07:13 AM

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
(

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

**meggy garcia** · January 7th 2010, 09:02 AM

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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