Isosceles triangle, angle, area of triangle inscribed in a circle!

**rawrzjaja** · April 26th 2008, 06:18 PM

yeah, i know i'm just wondering where the point R is?

**finch41** · April 26th 2008, 06:21 PM

i have the point r drawn on the bottom right of the triangle and of the circle
p is on the same height as r

q is the top of circle/triangle and contains angle theta

**finch41** · April 26th 2008, 06:21 PM

iceman would you mind to finish the solution

**rawrzjaja** · April 26th 2008, 06:24 PM

i see, i thot we were trying to find the two equal angles xD...? are we? o_o... so basically my P is bottom left, Q is at the top while R is bottom right of the triangle... is that the correct diagram?

**finch41** · April 26th 2008, 06:30 PM

yeah... what it comes down to is showing that when we make a new point S that is on the altitude (and this point s also bisects segment PR) that angles POS and SOR are theta

and therefore you can derive that the area of the triangle is given as

(r+rcostheta)(rsintheta)

although i am not sure myself what to do from here to prove the maximum area given a fixed value theta

**Isomorphism** · April 26th 2008, 08:49 PM

Originally Posted by finch41

yeah... what it comes down to is showing that when we make a new point S that is on the altitude (and this point s also bisects segment PR) that angles POS and SOR are theta

and therefore you can derive that the area of the triangle is given as

(r+rcostheta)(rsintheta)

although i am not sure myself what to do from here to prove the maximum area given a fixed value theta

$r^2 \sin \theta +r^2\sin \theta \cos\theta$

Maximize $\sin \theta + \sin \theta \cos\theta$

Try differentiating and obtaining maxima for above function

**earboth** · April 26th 2008, 08:51 PM

Originally Posted by rawrzjaja

Help please!! T_T

An isosceles triangle is inscribed in a circle of radius R. Determine the value of theta that maximizes the area of the triangle given that theta is the angle contained between the two equal sides.

1. Draw a sketch.

2. The area of the triangle is calculated by: $A=\frac12 \cdot b \cdot h$

3. The angles $\theta_1$ and $\theta_2$ have the same value: $|\theta_1| = |\theta_2|$ (ähemm ... and why?)

4. $\frac12 \cdot b = R \cdot \sin(\theta_2)$

5. $h = R + R \cdot \cos(\theta_2) = R \left(1+\cos(\theta_2)\right)$

6. That means A is a function with respect to $\theta$ :

$A(\theta)=R \cdot \sin(\theta) \cdot R \left(1+\cos(\theta)\right) = R^2 \cdot \sin(\theta) (1+\cos(\theta))~,~0 \leq \theta \leq \pi$

7. Calculate the first derivative. You should get:

$A'(\theta) = R^2\left(2(\cos(\theta))^2+\cos(\theta)-1\right)$

8. Solve for $\boxed{\cos(\theta)}$ :

$A'(\theta) = 0$ Use the substitution $y = \cos(\theta)$ and solve the quadratic equation for y. Afterwards re-substitute and calculate $\theta$
In the (plausible) domain (see #6.) you should get $\theta = \frac{\pi}3$

9. Therefore: The triangle inscribed in a circle with a maximum area is an equilateral triangle.

**rawrzjaja** · April 26th 2008, 10:40 PM

I tried to find the first derivative of A(theta) and i'm getting a different answer (using the product rule) can someone help me >.<
and i have the same question for step 3... not too sure how you got that >.<

**earboth** · April 26th 2008, 11:53 PM

Originally Posted by rawrzjaja

I tried to find the first derivative of A(theta) and i'm getting a different answer (using the product rule) can someone help me >.<
and i have the same question for step 3... not too sure how you got that >.<

1. You'll get further informations to step #3 at

Inscribed angle - Wikipedia, the free encyclopedia

2. From

$A(\theta)= R^2 \cdot \sin(\theta) (1+\cos(\theta))~, ~0 \leq \theta \leq \pi$

you'll get:

$A(\theta)= R^2 \cdot \sin(\theta) +R^2 \cdot \sin(\theta) \cos(\theta)$

Now calculate the first derivative. With the second summand you have to use product rule. During your calculations you'll get the term

$(\cos(\theta))^2-(\sin(\theta))^2 = 2 (\cos(\theta))^2 - 1$

which will finally yield the equation I've posted before.

**finch41** · April 27th 2008, 04:51 AM

so you have:

A(x) = r^2 sinx (1+cosx) where x is the angle

then

A-prime (x) = (r^2 sinx)(prime of (1+cosx)) + (prime of (r^2sinx))(1+cosx)

is that right so far?

so you get

A-prime (x) = (r^2 sinx)(0 + (-sinx)) + (2r sinx + r^2 cosx)(1 + cosx)

A-prime (x) = (r^2 sinx)(-sinx) + (2r sinx + 2r sinxcosx + r^2 cosx + r^2 cos^2 x
A-prime (x) = (-r^2 sin^2 x + 2r sinx + 2r sinx cosx + r^2 cosx + r^2 cos^2 x)

i see to factor out the r^2 and to switch the sin^2 into (1-cos^2 x)

**finch41** · April 27th 2008, 05:00 AM

i have gotten to

A prime (x) = R^2(2cos^2x + cosx + sinx + sinxcosx -1)

**earboth** · April 27th 2008, 08:09 AM

i'm so sorry, but i'm still a little bit confused... not too sure what happened to the R^2 when you are trying to find the first derivative of R^2sintheta + R^2sintheta x costheta

Remember: R is the radius of the circle and therefore it is considered to be constant. And a constant factor doesn't change when calculating the derivative.

$A(\theta)=R^2 \sin(\theta) + R^2 \sin(\theta) \cdot \cos(\theta)$

$A'(\theta)=R^2 \cos(\theta) + R^2\left( \sin(\theta) \cdot (-\sin(\theta))+\cos(\theta)\cdot \cos(\theta)\right)$

$A'(\theta)=R^2 \cos(\theta) + R^2\left( -\left(\sin(\theta)\right)^2+(\cos(\theta))^2\right )$

From $(\sin(\theta))^2+(\cos(\theta))^2=1$ you know that

$-(\sin(\theta))^2=(\cos(\theta))^2-1$

And therefore the first derivative becomes:

$A'(\theta)=R^2 \cos(\theta) + R^2\left( 2(\cos(\theta))^2-1\right)$

Factor out R² and you'll get the result which I've posted before.

**finch41** · April 27th 2008, 10:36 AM

oh yeah the constant!
awesome dude thanks

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