Circle in Triangle

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October 27th 2009, 11:08 AM
Aquafina

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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
October 28th 2009, 07:20 AM
Grandad

Hello Aquafina

Quote:

Originally Posted by Aquafina

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

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

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

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

Grandad
October 29th 2009, 12:47 AM
Aquafina

Quote:

Originally Posted by Grandad

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

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

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

For part (iii) I should do as you have done. The calculation is very straightforward for $\theta = \frac{\pi}{3}$ . 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
October 29th 2009, 11:20 AM
Grandad

Hello Aquafina

Quote:

Originally Posted by Aquafina

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 $A(\frac{\pi}{3})= \sqrt3-\frac{\pi}{3}$ is as follows:

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

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

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

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

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

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

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

Grandad