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

- Apr 26th 2008, 06:18 PMrawrzjajaHmmmmmmmmmmmmm
yeah, i know i'm just wondering where the point R is?

- Apr 26th 2008, 06:21 PMfinch41
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 - Apr 26th 2008, 06:21 PMfinch41
iceman would you mind to finish the solution

- Apr 26th 2008, 06:24 PMrawrzjajaHmmmmmmmmmmmmm
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?

- Apr 26th 2008, 06:30 PMfinch41
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 - Apr 26th 2008, 08:49 PMIsomorphism
- Apr 26th 2008, 08:51 PMearboth
1. Draw a sketch.

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

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

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

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

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

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

7. Calculate the first derivative. You should get:

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

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

$\displaystyle A'(\theta) = 0$ Use the substitution $\displaystyle y = \cos(\theta)$ and solve the quadratic equation for y. Afterwards re-substitute and calculate $\displaystyle \theta$

In the (plausible) domain (see #6.) you should get $\displaystyle \theta = \frac{\pi}3$

9. Therefore: The triangle inscribed in a circle with a maximum area is an equilateral triangle. - Apr 26th 2008, 10:40 PMrawrzjaja
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 >.< - Apr 26th 2008, 11:53 PMearboth
1. You'll get further informations to step #3 at

Inscribed angle - Wikipedia, the free encyclopedia

2. From

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

you'll get:

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

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

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

which will finally yield the equation I've posted before. - Apr 27th 2008, 04:51 AMfinch41
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) - Apr 27th 2008, 05:00 AMfinch41
i have gotten to

A prime (x) = R^2(2cos^2x + cosx + sinx + sinxcosx -1) - Apr 27th 2008, 08:09 AMearbothQuote:

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

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

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

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

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

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

And therefore the first derivative becomes:

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

Factor out R² and you'll get the result which I've posted before. - Apr 27th 2008, 10:36 AMfinch41
oh yeah the constant!

awesome dude thanks