The vicious cycle (2)

Since these 3 questions need to be solved simultaneously, im putting all $3$ solutions here:

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Question $1$ : Find the minimum value of $x^{2}+P_{3}x+1$ . $x$ ranges over the reals.

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The minimum occurs when $x=\frac{-P_{3}}{2}$ from the formula $\frac{-b}{2a}$ . There are many ways to prove it. See this

So, the minimum value of $x^{2}+P_{3}x+1$ is:

$x^{2}+P_{3}x+1 = \frac{{P_{3}}^{2}}{4} - \frac{{P_{3}}^{2}}{2} + 1 = 1 - \frac{{P_{3}}^{2}}{4}$

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Question $2$ : This problem

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This problem is very interesting. It can be solved with geometry.

Observe that ${x}^{2} + {y}^{2} = {r}^{2}$ is the equation of a circle and that $3x+4y=P_{1}$ is the equation of a line.

This makes the problem a whole lot easier and you can easily derive that $P_{2} = \frac{{P_{1}}^{2}}{25}$ Take a look at this if you don't understand.

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Question $3$ : Let a real $x$ be such that $\sec x=\sqrt {P_{2}}.$

Find $AB$ where

$A=\frac {\sin x}{1-\sin x}$

$B=\frac {\sin x}{1+\sin x}$

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First thing first, let's find $AB$ :

$AB=\frac { \sin x }{ 1-\sin x } \times \frac { \sin x }{ 1+\sin x } =\frac { \sin ^{ 2 }{ x } }{ 1-\sin ^{ 2 }{ x } }$

Now, lets find $\sin(x)$ :

$\frac { 1 }{ \cos { (x) } } =\sqrt { { P }_{ 2 } } \\ \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } =1\\ \sin x=\sqrt { 1-\cos ^{ 2 }{ x } } =\sqrt { 1-\frac { 1 }{ { P }_{ 2 } } }$

So therefore:

$AB=\frac { \sin ^{ 2 }{ x } }{ 1-\sin ^{ 2 }{ x } } ={ P }_{ 2 }-1$

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Now for the most exciting part, solving for the $3$ question simultaneously:

We have here $3$ equations:

${ P }_{ 1 }=1-\frac { { { P }_{ 3 } }^{ 2 } }{ 4 }$

${ P }_{ 2 }=\frac { { { P }_{ 1 } }^{ 2 } }{ 25 }$

${ P }_{ 3 }={ P }_{ 2 }-1$

Solving them gives:

$\boxed{\boxed{\boxed{\begin{matrix} { P }_{ 1 }=-15 \\ { P }_{ 2 }=9 \\ { P }_{ 3 }=8 \end{matrix}}}}$

Since the last step needs quartics, here is a method (Newton's method) to bash it through quickly and very accurately.

Since $\space (3x+4y)^2 \le (3^2+4^2)(x^2+y^2) \quad \Rightarrow P_1^2 \le 25 (x^2+y^2)$

Therefore, $\space P_2 = min(x^2+y^2) = \dfrac {P_1^2}{25}\quad \Rightarrow P_1^2 = 25P_2$ .

Go to solution of Problem 1 .

From Problem 1 , $\space P_3 = 8$ .

From Problem 3 , $\space P_2=1+P_3 = 1+8 = \boxed{9}$