Expert Help with High School Geometry

Calculate angle between two vectors, given their rotation w.r.t. a third vector.
I have three vectors, $\stackrel{\to <!--\; \to \; -->}{a},\stackrel{\to <!--\; \to \; -->}{b},$ and $\stackrel{\to <!--\; \to \; -->}{c}$ in n-dimensional space. I know the coordinates of all three vectors and their dot products. Both $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$ are rotated away from $\stackrel{\to <!--\; \to \; -->}{c}$ by an angle $\mathrm{\alpha <!--\; \alpha \; -->}$, in their own respective directions, obtaining ${\stackrel{\to <!--\; \to \; -->}{a}}^{\prime }$ and ${\stackrel{\to <!--\; \to \; -->}{b}}^{\prime }$ . What is the angle between ${\stackrel{\to <!--\; \to \; -->}{a}}^{\prime }$ and ${\stackrel{\to <!--\; \to \; -->}{b}}^{\prime }$ ?

Estimating the parameters of gaussians to fit a lot of samples can be do with Exceptation Maximization, for instance if we want to fit two gaussian on points, to have the clusters $a$ and $b$. (1)
$b_i=P(b|x_i)=\frac{P(x_i|b)P(b)}{P(x_i|b)P(b)+P(x_i|b)P(a)}$
Here $P(b)$ is the prior that depicts the overall importance of the $b$ cluster.
This prior is then updated for the next step, according on how many the points belongs to the $b$ cluster:
$P(b)=\frac{1}{n}\underset{i}{\sum <!--\; \sum \; -->}{b}_i$
However, what is the value of the prior $P(b)$ on the first iteration of the algorithm?

Maximize
$f(x,y,z)=xy+z^2,$
while $2x-<!--\; -\; -->y=0$ and $x+z=0$. Lagrange doesnt seem to work.

Consider the problem for some vectors $v,m\in <!--\; \in \; -->{\mathbb{R}}^n:$

$f(v)=(v^{T}m{)}^2$

w.r.t $\Vert <!--\; \Vert \; -->v{\Vert <!--\; \Vert \; -->}^2=1$

I want to maximize f

If I consider the lagrangian, I get:

$L(v)=(v^{T}m{)}^2+\mathrm{\lambda <!--\; \lambda \; -->}(1-<!--\; -\; -->\Vert <!--\; \Vert \; -->v{\Vert <!--\; \Vert \; -->}^2)$

Taking derivative, I get: $2m{m}^{T}v-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}2v=0$ Therefore $m{m}^{T}v=\mathrm{\lambda <!--\; \lambda \; -->}v(\ast <!--\; \ast \; -->)$

Multiplying by $v^T$ from left, I end up with

$(v^{T}m{)}^2=\mathrm{\lambda <!--\; \lambda \; -->}$

If I put that in(*), I do cannot simplfy that.

Is there a trick I can apply?

Let $(c_1,c_2)$ be a fixed point in ${\mathbb{R}}^2$. How to maximize $|x_1{x}_2-<!--\; -\; -->c_1{c}_2|$ subject to the condition that $(x_1-<!--\; -\; -->c_1{)}^2+(x_2-<!--\; -\; -->c_2{)}^2<1$. i.e.
$\underset{(x_1,x_2)\in <!--\; \in \; -->{\mathbb{R}}^2:(x_1-<!--\; -\; -->c_1{)}^2+(x_2-<!--\; -\; -->c_2{)}^2<1}{sup}|x_1{x}_2-<!--\; -\; -->c_1{c}_2|=?$
Note : This is a generalization of the problem of maximizing $|x_1{x}_2|$ subject to the condition $x_1^2+x_2^2<1$. I'm stuck with it. Any help would be much appreciated.

Let us start with these two equations of two lines:
$x+y=4$
$x-<!--\; -\; -->y=2$
They intersect at $(x,y)=(3,1)$
Let us now translate (move) both lines so that they intersect at (0, 0). We need to move both lines by -3 along x-axis and by -1 along y-axis. So the equations of the lines become.
$(x+3)+(y+1)=4$
$(x+3)-<!--\; -\; -->(y+1)=2$
This is equivalent to
$x+y=0$
$x-<!--\; -\; -->y=0$
Why do the RHS become 0 for both equations? This happens no matter which two intersecting lines we begin with. What is the geometrical interpretation of this?

I always saw the algorithm transform its objective function by log. I wonder this is because logarithm function is monotonically increased (logx) to do maximization, or decreased (−logx) to do minimization? So for MM algorithm, any functions should be transformed into log?

Are there also other reasons that log transformation is often used? I wonder log is always work or we have some mathematical operations that can replace log for this purpose?

I'm facing this problem:
$\underset{x\in <!--\; \in \; -->{\mathbb{R}}_{+}^3}{min}max\{\frac{\underset{i=1}{\overset{3}{\sum <!--\; \sum \; -->}}{x}_i^2-<!--\; -\; -->2x_1{x}_3}{{\left(\underset{i=1}{\overset{3}{\sum <!--\; \sum \; -->}}{x}_i\right)}^2},\frac{\underset{i=1}{\overset{3}{\sum <!--\; \sum \; -->}}{x}_i^2+2(x_1{x}_3-<!--\; -\; -->x_1{x}_2+x_2{x}_3)}{{\left(\underset{i=1}{\overset{3}{\sum <!--\; \sum \; -->}}{x}_i\right)}^2}\}$
I don't know how to deal with inner max and choose one of two!

I'm trying to use $max(A,B)\ge <!--\; \ge \; -->\frac{1}{2}(A+B)$! Do you have any idea?

Non-geometric Proof of Pythagorean Theorem
Is there a purely algebraic proof for the Pythagorean theorem that doesn't rely on a geometric representation? Just algebra/calculus. I want to TRULY understand the WHY of how it is true. I know it works and I know the geometric proofs.

"Pythagorean theorem" for projection onto convex set
I'm going through the book on online convex optimization by Hazan, and in the first chapter I saw this assertion (which Hazan calls the "pythagorean theorem"):
Let $K\subset <!--\; \subset \; -->{\mathbb{R}}^d$ be a convex set, $y\in <!--\; \in \; -->{\mathbb{R}}^d$, and $x={\mathrm{\Pi <!--\; \Pi \; -->}}_K(y)$. Then for any $z\in <!--\; \in \; -->K$ we have:
$\Vert <!--\; \Vert \; -->y-<!--\; -\; -->z\Vert <!--\; \Vert \; -->\ge <!--\; \ge \; -->\Vert <!--\; \Vert \; -->x-<!--\; -\; -->z\Vert <!--\; \Vert \; -->.$
It is presented without proof - what is a proof for this? Also, how does it relate to the pythagorean theorem?

Let $N=\{1,\dots <!--\; \dots \; -->,n\}$. For a given $(r_1,\dots <!--\; \dots \; -->,r_n)\in <!--\; \in \; -->{\mathbb{R}}_{++}^n$. I need to solve
$\begin{array}{r}\underset{(k_1,\dots <!--\; \dots \; -->,k_n)\in <!--\; \in \; -->{\mathbb{R}}_{++}^n}{max}\underset{i\in <!--\; \in \; -->N}{\prod <!--\; \prod \; -->}[\underset{j\in <!--\; \in \; -->N}{\sum <!--\; \sum \; -->}(r_j-<!--\; -\; -->k_j)+r_i\ln <!--\; \; -->\left(\frac{k_i}{r_i}\right)].\end{array}$
I could not come up with a closed form solution of the maximizers by using first order conditions, because of the log. Is it possible to tell, that there does not exist a closed form solution and we thus have to maximize the product numerically.

Find the coordinates of $\stackrel{\to <!--\; \to \; -->}{v}$ if $\stackrel{\to <!--\; \to \; -->}{v}=2\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->3\stackrel{\to <!--\; \to \; -->}{b}+4\stackrel{\to <!--\; \to \; -->}{c}$ and $\stackrel{\to <!--\; \to \; -->}{a}(4;1),\stackrel{\to <!--\; \to \; -->}{b}(1;2)$ and $\stackrel{\to <!--\; \to \; -->}{c}(2;7)$

How does one maximize the following equation? $(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)(a-<!--\; -\; -->b)$ given that $-<!--\; -\; -->n\le <!--\; \le \; -->c<b<a\le <!--\; \le \; -->n$ for some value of $n$
So far, I have attempted to do this using AM-GM and got $\frac{64}{27}{n}^3$ but I don't think this is correct as some of the values of a, b, and c can be negative and when I tried to find the actual values which satisfied the max value found it gave me a contradiction.

Consider the vectors:
$a_1=\left(\begin{array}{c}0\\ -<!--\; -\; -->1\\ 1\\ 0\end{array}\right),a_2=\left(\begin{array}{c}0\\ 0\\ -<!--\; -\; -->1\\ 1\end{array}\right),a_3=\left(\begin{array}{c}2\\ 0\\ 0\\ 1\end{array}\right)$
Find a single vector $p$ which maximizes $p{a}_i$ for $i=1,2,3$.

To put this in context this is an economics profit max problem where p is a price and each component of the above vectors represents the quantity of the good.

I honestly have no idea how to find this p vector. It doesn't even seem possible to me that a single vector can maximize these three vectors.

I'm struggling with the following question for long. I tried to apply isoperimetric inequality $4\mathrm{\pi <!--\; \pi \; -->}A\le <!--\; \le \; -->L^2$, but my attempt has been unsuccessful. Could anyone give me a hint?

Let $AB$ be a segment of straight line and let $l>$ length of $AB$. Show that the curve $C$ joining $A$ and $B$, with length $l$, and such that together with $AB$ bounds the largest possible area is an arc of a circle passing through $A$ and $B$.

Suppose I want to solve the following problem
$\underset{f(\cdot <!--\; \cdot \; -->)}{max}\int <!--\; \int \; -->f(x)g(x)d\mathrm{\mu <!--\; \mu \; -->}(x)$
where the maximization is over measurable functions $f:X\to <!--\; \to \; -->[0,1]$, $\mathrm{\mu <!--\; \mu \; -->}$ is a finite measure and $g$ is measurable.
Can someone come up with an example such that the problem does NOT have a solution? I imagine one can try to maximize point by point and obtain something that is not measurable and then the problem would not have a solution?

I have a function $f=-<!--\; -\; -->20p\cdot <!--\; \cdot \; -->q+9p+9q$. Player 1 chooses $p$ and player 2 chooses $q$. Both $p$ and $q$ are in the inclusive interval [0,1]. Player 1 wants to maximize $f$ while player 2 wants to minimize $f$.

Player 1 goes first, what is the most optimal value of $p$ he should choose knowing that player 2 will choose a $q$ in response to player 1's choice of $p$?

This seems to be some sort of minimization-maximization problem, but I am unsure how to solve it. I was thinking about approaching this from a calculus perspective by taking the partial derivative of $f$ with respect to $p$, but it doesn't seem I get an intuition by doing this, and it seems that $p$ and $q$ are a function of each other. How should I approach solving this problem analytically?

Solve the following NLP:
$\{\begin{array}{cc}min& -<!--\; -\; -->3x+y-<!--\; -\; -->z^2\\ s.t& g(x,y,z)=x+y+z\le <!--\; \le \; -->0\\ & h(x,y,z)=-<!--\; -\; -->x+2y+z^{2}z=0\end{array}$

My attempt
Using kkt conditions, we have 2 possibles situations:
1) If $g<0$: $\mathrm{\nabla <!--\; \nabla \; -->}f+\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\nabla <!--\; \nabla \; -->}h=0,h=0$
$\{\begin{array}{ccc}-<!--\; -\; -->3-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}& =& 0\\ -<!--\; -\; -->3y^2+1+2\mathrm{\lambda <!--\; \lambda \; -->}& =& 0\\ 2\mathrm{\lambda <!--\; \lambda \; -->}z& =& 0\\ -<!--\; -\; -->x+2y+z^2& =& 0\end{array}$
From first and second, we see it is impossible
**2)**$g=0$: $\mathrm{\nabla <!--\; \nabla \; -->}f+\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\nabla <!--\; \nabla \; -->}h+\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nabla <!--\; \nabla \; -->}g=0,\mathrm{\mu <!--\; \mu \; -->}\ge <!--\; \ge \; -->0$
$\{\begin{array}{ccc}-<!--\; -\; -->3-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}+\mathrm{\mu <!--\; \mu \; -->}& =& 0\\ -<!--\; -\; -->3y^2+1+2\mathrm{\lambda <!--\; \lambda \; -->}+\mathrm{\mu <!--\; \mu \; -->}& =& 0\\ 2\mathrm{\lambda <!--\; \lambda \; -->}z+\mathrm{\mu <!--\; \mu \; -->}& =& 0\\ -<!--\; -\; -->x+2y+z^2& =& 0\\ x+y+z& =& 0\end{array}$
I couldnt solve this last one. Any idea? I've tried to put y,z im function of $\mathrm{\mu <!--\; \mu \; -->}$ and use the last 2 eq., but didnt work (it became complicated).
Thanks in advance!

The Pythagorean theorem and Hilbert axioms
Can one state and prove the Pythagorean theorem using Hilbert's axioms of geometry, without any reference to arithmetic?
Edit: Here is a possible motivation for this question (and in particular for the "state" part of this question). It is known that the theory of Euclidean geometry is complete. Every true statement in this theory is provable.
On the other hand, it is known that the axioms of (Peano) arithmetic cannot be proven to be consistent. So, basically, I ask if there is a reasonable theory which is known to be consistent and complete, and in which the Pythagorean theorem can be stated and proved.
In summary, I guess I am asking - can we be sure that the Pythagorean theorem is true? :)

Find the maximum of
$x_2+y_1-<!--\; -\; -->x_1-<!--\; -\; -->y_2$
if $(x_1-<!--\; -\; -->1/2{)}^2+(y_1-<!--\; -\; -->1/2{)}^2=1/4$ and $(x_2-<!--\; -\; -->1/2{)}^2+(y_2-<!--\; -\; -->1/2{)}^2=1/4$ and $x_1{y}_2=y_1{x}_2$ and all variables are positive.

Is there a way to do this with inequalities? What about Lagrange Multipliers?