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

- Apr 26th 2008, 07:18 PMrawrzjajaHmmmmmmmmmmmmm
yeah, i know i'm just wondering where the point R is?

- Apr 26th 2008, 07:21 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, 07:21 PMfinch41
iceman would you mind to finish the solution

- Apr 26th 2008, 07:24 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, 07:30 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, 09:49 PMIsomorphism
- Apr 26th 2008, 09:51 PMearboth
1. Draw a sketch.

2. The area of the triangle is calculated by:

3. The angles and have the same value: (ähemm ... and why?)

4.

5.

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

7. Calculate the first derivative. You should get:

8. Solve for :

Use the substitution and solve the quadratic equation for y. Afterwards re-substitute and calculate

In the (plausible) domain (see #6.) you should get

9. Therefore: The triangle inscribed in a circle with a maximum area is an equilateral triangle. - Apr 26th 2008, 11:40 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 27th 2008, 12:53 AMearboth
1. You'll get further informations to step #3 at

Inscribed angle - Wikipedia, the free encyclopedia

2. From

you'll get:

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

which will finally yield the equation I've posted before. - Apr 27th 2008, 05:51 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, 06:00 AMfinch41
i have gotten to

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

From you know that

And therefore the first derivative becomes:

Factor out R² and you'll get the result which I've posted before. - Apr 27th 2008, 11:36 AMfinch41
oh yeah the constant!

awesome dude thanks