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Math Help - Isosceles triangle, angle, area of triangle inscribed in a circle!

  1. #16
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    Hmmmmmmmmmmmmm

    yeah, i know i'm just wondering where the point R is?
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  2. #17
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    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
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  3. #18
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    iceman would you mind to finish the solution
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  4. #19
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    Hmmmmmmmmmmmmm

    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?
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  5. #20
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    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
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  6. #21
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    Quote Originally Posted by finch41 View Post
    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
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  7. #22
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    Quote Originally Posted by rawrzjaja View Post
    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.
    Attached Thumbnails Attached Thumbnails Isosceles triangle, angle, area of triangle inscribed in a circle!-dreieck_in_kreis.gif  
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  8. #23
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    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 >.<
    Last edited by rawrzjaja; April 26th 2008 at 11:00 PM.
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  9. #24
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    Quote Originally Posted by rawrzjaja View Post
    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.
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  10. #25
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    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)
    Last edited by finch41; April 27th 2008 at 05:15 AM.
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  11. #26
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    i have gotten to

    A prime (x) = R^2(2cos^2x + cosx + sinx + sinxcosx -1)
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  12. #27
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    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.
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  13. #28
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    oh yeah the constant!
    awesome dude thanks
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