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Thread: Circle in Triangle

  1. #1
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    Circle in Triangle

    Hi I have attached the question as a word file. I have worked through it, and need help with the last bit.

    I do not understand how to show the relationship between A and B in ii)

    he first time I did the question, I tried to use the relationship to evaluate the areas asked for next, which ended up being confusing.

    However, this time, I just worked with the kite and sector of the circle to find A. I didn't use B at all.

    Since B was given right before the question to work out the final 2 Areas was asked, I am presuming that it can be used to work out the answers, or else it would have no reason of being added.

    So basically, I dont understand the relationship, and also, can anyone show how they would do the last part using this relationship between area A and B?

    Thanks
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  2. #2
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    Hello Aquafina
    Quote Originally Posted by Aquafina View Post
    ... So basically, I dont understand the relationship, and also, can anyone show how they would do the last part using this relationship between area A and B?

    Thanks
    The function $\displaystyle A(\theta)$ gives the area between any two tangents and the circle expressed as a function of $\displaystyle \theta$, the angle between the tangents.

    So if we denote the angle between the tangents from $\displaystyle R$ to the circle by $\displaystyle \phi$, then the area between these tangents and the circle is $\displaystyle A(\phi)$. But this area is given by $\displaystyle B(\theta)$. Hence $\displaystyle A(\phi) = B(\theta)$.

    But, from $\displaystyle \triangle ORP, \theta = \frac{\pi}{2}-\phi$. Therefore $\displaystyle A(\phi) = B(\frac{\pi}{2}-\phi)$. We can replace $\displaystyle \phi$ by any other parameter; in particular by $\displaystyle \theta$. So $\displaystyle A(\theta) = B(\frac{\pi}{2}-\theta)$.

    For part (iii) I should do as you have done. The calculation is very straightforward for $\displaystyle \theta = \frac{\pi}{3}$. Why complicate it?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello AquafinaThe function $\displaystyle A(\theta)$ gives the area between any two tangents and the circle expressed as a function of $\displaystyle \theta$, the angle between the tangents.

    So if we denote the angle between the tangents from $\displaystyle R$ to the circle by $\displaystyle \phi$, then the area between these tangents and the circle is $\displaystyle A(\phi)$. But this area is given by $\displaystyle B(\theta)$. Hence $\displaystyle A(\phi) = B(\theta)$.

    But, from $\displaystyle \triangle ORP, \theta = \frac{\pi}{2}-\phi$. Therefore $\displaystyle A(\phi) = B(\frac{\pi}{2}-\phi)$. We can replace $\displaystyle \phi$ by any other parameter; in particular by $\displaystyle \theta$. So $\displaystyle A(\theta) = B(\frac{\pi}{2}-\theta)$.

    For part (iii) I should do as you have done. The calculation is very straightforward for $\displaystyle \theta = \frac{\pi}{3}$. Why complicate it?

    Grandad
    Thanks Grandad. I was told I had to mention something like:

    That the reflections would not change the areas and reverses the angles...

    Also, I wanted to try the relationship with B to work out the Area of A just to experiment around. But this didnt work out.

    I did

    A(pi/3) = B(pi/6) = 2*AreaQCR - Area segment
    = root3 - 5pi/12

    rather than the pi/3...

    I get pi/3 if I take the angle ORP to be pi/3, not pi/6
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  4. #4
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    Hello Aquafina
    Quote Originally Posted by Aquafina View Post
    Thanks Grandad. I was told I had to mention something like:

    That the reflections would not change the areas and reverses the angles...

    Also, I wanted to try the relationship with B to work out the Area of A just to experiment around. But this didnt work out.

    I did

    A(pi/3) = B(pi/6) = 2*AreaQCR - Area segment
    = root3 - 5pi/12

    rather than the pi/3...

    I get pi/3 if I take the angle ORP to be pi/3, not pi/6
    I'm not quite sure what you mean here. The straightforward proof that $\displaystyle A(\frac{\pi}{3})= \sqrt3-\frac{\pi}{3}$ is as follows:

    If $\displaystyle C$ is the centre of the circle, then $\displaystyle \angle CPQ = \frac{\pi}{6}$

    $\displaystyle \Rightarrow PQ = \frac{1}{\tan(\tfrac{\pi}{6})} = \sqrt3$

    $\displaystyle \Rightarrow$ area of $\displaystyle \triangle CPQ = \tfrac12\cdot1\cdot\sqrt3 = \frac{\sqrt3}{2}$

    $\displaystyle \Rightarrow$ area of kite $\displaystyle = 2\times\triangle CPQ = \sqrt3$

    The angle at the centre of the circle is $\displaystyle \pi - \theta = \frac{2\pi}{3}$

    $\displaystyle \Rightarrow$ area of sector $\displaystyle = \tfrac12\times1^2\times\frac{2\pi}{3}= \frac{\pi}{3}$

    $\displaystyle \Rightarrow A(\frac{\pi}{3}) =$ area of kite - area of sector $\displaystyle = \sqrt3 - \frac{\pi}{3}$

    Grandad
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