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
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
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.
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
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.
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)
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.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.