Maths question! (1 Viewer)

jet · Jan 10, 2010

I did it. Got an equation in sin(x) and x. I cbb with approximating at that point in time, so I left it there

I will post my solution later, let me just work it out again

EDIT: Because this problem is extremely difficult, though not unrelated to the HSC, I'm moving it to the 4-unit Maths forums.

jet · Jan 10, 2010

$\text{For the circle at point B, } x^2 + y^2 = \frac{1}{\pi} \Rightarrow y = \pm\sqrt{\frac{1}{\pi} - x^2} \\ \text{For the circle at point A,} (x - \frac{1}{\sqrt{\pi}})^2 + y^2 = L^2 \Rightarrow y = \pm \sqrt{L^2 - \left (x - \frac{1}{\sqrt{\pi}} \right )^2} \\ \text{Solving the two equations, } x^2 - \left (x - \frac{1}{\sqrt{\pi}} \right )^2 = \frac{1}{\pi} - L^2 \\ x^2 - x^2 + \frac{2}{\sqrt{\pi}}x - \frac{1}{\pi} = \frac{1}{\pi} - L^2 \\ \frac{2}{\sqrt{\pi}}x = \frac{2}{\pi}-L^2 \\ x = \frac{1}{\sqrt{\pi}} - \frac{\sqrt{\pi}L^2}{2} \Rightarrow x = \frac{2 - \pi L^2}{2\sqrt{\pi}} = a \text{ for now} \\ \text{Therefore the x-coordinate where I split the areas is at } E \equiv \left ( \frac{2 - \pi L^2}{2\sqrt{\pi}}, 0\right )$

$\text{Now, I am going to find the area BCE} \\ A_{BCE} = \int^{a}_{\frac{1}{\sqrt{\pi}} - L} \sqrt{L^2 - \left (x - \frac{1}{\sqrt{\pi}} \right )^2} \, dx \\ \text{Let } x - \frac{1}{\sqrt{\pi}} = L\sin \theta \Rightarrow x = L\sin \theta + \frac{1}{\sqrt{\pi}} \\ \frac{dx}{d\theta} = L\cos\theta \\ \therefore dx = L\cos\theta \, d\theta \\ x = \frac{1}{\sqrt{\pi}} - \frac{\sqrt{\pi}L^2}{2}, \theta = -\arcsin \left (\frac{\sqrt{\pi}L}{2} \right ) = b \\ x = \frac{1}{\sqrt{\pi}} - L; \theta = -\frac{\pi}{2} \\ \therefore A_{BCE} = \int^{b}_{-\frac{\pi}{2}} \sqrt{L^2 - L^2\sin^2 \theta} L\cos \theta \, d\theta \\ = L^2 \int^{b}_{-\frac{\pi}{2}} \cos^2 \theta \, d\theta \\ = L^2 \left [\frac{\theta}{2} + \frac{1}{4}\sin 2\theta \right ]^{b}_{-\frac{\pi}{2}} \\ = L^2 \left [\frac{-\arcsin \left (\frac{\sqrt{\pi L}}{2} \right )}{2} - \frac{1}{4}\sin\left (2\arcsin \left ( \frac{\pi L}{2}\right ) \right ) + \frac{\pi}{4} \right ]$

$\text{Now for the area AEC } \\ A_{AEC} = \int^{\frac{1}{\sqrt{\pi}}}_{a} \sqrt{\frac{1}{\pi} - x^2} \, dx \\ \text{Let } x = \frac{1}{\sqrt{\pi}}\sin \theta \\ \therefore \frac{dx}{d\theta} = \frac{1}{\sqrt{\pi}}\cos \theta \\ dx = \frac{1}{\sqrt{\pi}}\cos \theta \, d\theta \\ x = \frac{1}{\sqrt{\pi}}, \theta = \frac{\pi}{2} \\ x = \frac{1}{\sqrt{\pi}} - \frac{\sqrt{\pi}L^2}{2}, \theta = \arcsin \left (1 - \frac{\pi L^2}{2} \right ) = c \\ \therefore A_{AEC} = \int^{\frac{\pi}{2}}_{c} \sqrt{\frac{1}{\pi} - \frac{1}{\pi}\sin^2 \theta} \frac{1}{\sqrt{\pi}}\cos \theta \, d\theta \\ = \frac{1}{\pi}\int^{\frac{\pi}{2}}_{c} \cos^2 \theta \, d\theta \\ = \frac{1}{\pi}\left [\frac{\theta}{2} + \frac{1}{4}\sin 2\theta \right ]^{\frac{\pi}{2}}_{c} \\ = \frac{1}{\pi} \left [\frac{\pi}{4} - \frac{\arcsin \left (1 - \frac{\pi L^2}{2} \right )}{4} - \frac{1}{4}\sin \left (2\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right ) \right ]$

$\text{The two terms which are annoying are } - \frac{1}{4}\sin\left (2\arcsin \left ( \frac{\pi L}{2}\right ) \right ) \text{ and }- \frac{1}{4}\sin \left (2\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right ) \\ - \frac{1}{4}\sin\left (2\arcsin \left ( \frac{\pi L}{2}\right ) \right ) = -\frac{1}{2}\sin\left (\arcsin\left (\frac{\pi L}{2} \right ) \right )\cos \left (\arcsin \left (\frac{\pi L}{2} \right ) \right ) \\ = -\frac{\pi L}{4}\cos \left (\arcsin\left (\frac{\pi L}{2} \right ) \right ) \\ \text{We let } \theta = \arcsin\left (\frac{\pi L}{2} \right ) \text{ and it simplifies to } \\ \frac{-\pi L}{4}\cos \theta = \frac{-\pi L \sqrt{4 - \pi^2 L^2 }}{8} \\ \text{The other expression is quite complicated, being } \frac{1}{4}\sin \left (2\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right ) = \frac{\left (2 - \pi L^2\right )\sqrt{12 + 4\pi L^2 - \pi^2 L^4}}{4}$

Now, I can keep on going. Or someone can realise that this will involve differentiating very complicated expressions to apply Newton's method

Also, could someone please tell me if they find any mistakes? Thanks

Aquawhite · Jan 10, 2010

Woah, that's an amazing solution

The amazing thing is, I understood almost all of that

... however I doubt I would have been able to solve so far if it were in an exam in the HSC.

shaon0 · Jan 11, 2010

If you have the Coordinates of C and where the larger circle meets the x-axis near B. Could you just use A=1/2 r^2(@-sin@) in both segments by constructing BD,BC,AD,AC?

jet · Jan 11, 2010

Yes you could on second thought, though it would probably still be a bastard to solve. Will complete tomorrow.

Trebla · Jan 11, 2010

shaon0 said:
If you have the Coordinates of C and where the larger circle meets the x-axis near B. Could you just use A=1/2 r^2(@-sin@) in both segments by constructing BD,BC,AD,AC?

But how would you eliminate @? That method involves an angle and its sine hence might still give a transcendental equation.

shaon0 · Jan 11, 2010

Trebla said:
But how would you eliminate @? That method involves an angle and its sine hence might still give a transcendental equation.

I don't know, as I haven't worked on the problem yet (and don't plan to). I was just suggesting another method to calculate the areas as a iterative process wouldn't have to be used in the method. @ would just be atan(y/x) where y,x are co-ords of C, i think.

buchanan · Jan 11, 2010

The solution to jetblack2007's final equation is $L=\sqrt{2/\pi}$

Trebla · Jan 11, 2010

buchanan said:
The solution to jetblack2007's final equation is $L=\sqrt{2/\pi}$

From inspection? haha

cutemouse · Jan 11, 2010

Could've also used Mathematica

untouchablecuz · Jan 11, 2010

great work

i did exactly the same thing, props on actually evaluating the integrals

for completion:

$\\\frac{1}{4}\sin \left (2\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right ) = \frac{\left (2 - \pi L^2\right )\sqrt{12 + 4\pi L^2 - \pi^2 L^4}}{4}\\\sin \left (2\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right ) = \left (2 - \pi L^2\right )\sqrt{12 + 4\pi L^2 - \pi^2 L^4}\\\2\sin \left (\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right )\cos\left (\arcsin \left (1 - \frac{\pi L^2}{2} \right ) \right )=\left (2 - \pi L^2\right )\sqrt{12 + 4\pi L^2 - \pi^2 L^4}\\2\left (1 - \frac{\pi L^2}{2} \right )\sqrt{1-\left (1 - \frac{\pi L^2}{2} \right )^2}=\left (2 - \pi L^2\right )\sqrt{12 + 4\pi L^2 - \pi^2 L^4}\\2\left (\frac{2-\pi L^2}{2} \right )\sqrt{1-\left (1 - \frac{\pi L^2}{2} \right )^2}=\left (2 - \pi L^2\right )\sqrt{12 + 4\pi L^2 - \pi^2 L^4}\\\sqrt{1-\left (1 - \frac{\pi L^2}{2} \right )^2}=\sqrt{12 + 4\pi L^2 - \pi^2 L^4}\\1-\left (1 - \frac{\pi L^2}{2} \right )^2=12 + 4\pi L^2 - \pi^2 L^4\\$Simplify and solve to find that, $L=\sqrt{\frac{2}{\pi}}$

buchanan · Jan 11, 2010

Actually, his final equation is incorrect.

The correct equation is ${1\over2}L\sqrt{\pi(4-\pi L^2)}-({\pi L^2\over2}-1)\cos^{-1}({\pi L^2\over2}-1)={\pi\over2}$

And when I put that one into wolframalpha, I get L=0.653742534637021....

Subsequent check of the sum of the areas of the 2 segments with this value of L indeed get's a value of exactly 0.5 for the area.

And this answer is NOT $\sqrt{2\over\pi}$

untouchablecuz · Jan 11, 2010

buchanan said:
Actually, his final equation is incorrect.

The correct equation is ${1\over2}L\sqrt{\pi(4-\pi L^2)}-({\pi L^2\over2}-1)\cos^{-1}({\pi L^2\over2}-1)={\pi\over2}$

And when I put that one into wolframalpha, I get L=0.653742534637021....

Subsequent check of the sum of the areas of the 2 segments with this value of L indeed get's a value of exactly 0.5 for the area.

And this answer is NOT $\sqrt{2\over\pi}$

o rly

buchanan · Jan 11, 2010

Yeah. rly.

My solution is attached.

untouchablecuz · Jan 11, 2010

buchanan said:
Yeah. rly.

My solution is attached.

nice work : )

Maths question! (1 Viewer)

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