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PostPosted: Fri, 13 Jan 2012 07:06:46 UTC 
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Image

Attempt at a solution:

i) circle has equation (x+c)^2 + (y-c)^2 = (2c)^2
hyperbola has equation y=1/x
Substituting y=1/x into the equation for the circle gives the equation x^4 + 2cx^3 -2cx + 1 = 0

ii)The equation has roots x=alpha, alpha, beta and beta because this is where the two curves touch. It is a quartic equation therefore the two points of contact must be double roots.

iii) Not sure how to do part 3 - help would be greatly appreciated!


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PostPosted: Fri, 13 Jan 2012 07:14:49 UTC 
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simonsmith wrote:
Image

Attempt at a solution:

i) circle has equation (x+c)^2 + (y-c)^2 = (2c)^2
hyperbola has equation y=1/x
Substituting y=1/x into the equation for the circle gives the equation x^4 + 2cx^3 -2cx + 1 = 0

ii)The equation has roots x=alpha, alpha, beta and beta because this is where the two curves touch. It is a quartic equation therefore the two points of contact must be double roots.

iii) Not sure how to do part 3 - help would be greatly appreciated!


What image?

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PostPosted: Fri, 13 Jan 2012 07:20:27 UTC 
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simonsmith wrote:
https://8501882917160242609-a-1802744773732722657-s-sites.googlegroups.com/site/patelltd/maths-problem/SCAN0341.JPG?attachauth=ANoY7coff_52MToLpPd4CHWDJjs_E-bH_KqDDcnxn4zoWq23Ke6BgITovrq7xh503N0ZhgFyzrAUWFoX3jYLiYxEYYJPdznOoJBJEPW-flaSQEGByA-wotUuttRcYSH3jTXKJ5ICsuBGrkAIW3QE1CnlrM3l4nHGnbrcLSfS2Qn6K9Esr4DX_WvJ3Tg_RItvgL3TADthIuK8UV4lFZ6TJ9bMOjUc8mHkSQ%3D%3D&attredirects=0

Attempt at a solution:

i) circle has equation (x+c)^2 + (y-c)^2 = (2c)^2
hyperbola has equation y=1/x
Substituting y=1/x into the equation for the circle gives the equation x^4 + 2cx^3 -2cx + 1 = 0

ii)The equation has roots x=alpha, alpha, beta and beta because this is where the two curves touch. It is a quartic equation therefore the two points of contact must be double roots.

iii) Not sure how to do part 3 - help would be greatly appreciated!


Start by writing down the Viete's formulas.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 13 Jan 2012 07:22:10 UTC 
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There's a name for a result that obvious?

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PostPosted: Fri, 13 Jan 2012 10:53:55 UTC 
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Shadow wrote:
There's a name for a result that obvious?


I usually refuse to call it Viete's formula, but only as "symmetry of roots". However, that is the name given in the wiki article.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 13 Jan 2012 20:43:48 UTC 
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outermeasure wrote:
Shadow wrote:
There's a name for a result that obvious?


I usually refuse to call it Viete's formula, but only as "symmetry of roots". However, that is the name given in the wiki article.


OK, that's fair then.

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