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Hint: Following your partial attempt, the (Apollonius) circle with diameter $GD$ passes through $K$ because of the right-angle, and then you can angle-chase using circle properties to finish.

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Yes, we need the field to be algebraically closed. Then it is clear. $B(x,y,0,0)$ is a homogenous polynomial in $x,y$. Such a polynomial certainly admits a root other than $(x=0,y=0)$.

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$\mathbb{R}^2$ Fix $z$, let $x$ run around the unit circle, and forget $y$ for now. The locus of projections of $z$ on $x$ is the circle with diameter 1 containing the origin and $z$. The same locus for $y$. Since the diameter is $||z||=1$, any two points are no farther away from each other than 1. QED. $\mathbb{R}^n$ The projection $z_{xy}$ of $z$ ...

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First of all, the map is clearly not a map from $\mathbb C^3$ to $\mathbb C^2$, but a map from $C^3\setminus\{(0,t_2,t_3)|t_2,t_3\in\mathbb C\}$ to $\mathbb C^2$. The map is clearly undefined on $(0,1,1)$ for example. Now, if $\phi(t_1,t_2,t_3)$ is defined, then you can find the kernel using this hint: $$\frac{t_2}{t_1} = 0 \iff t_2 = 0$$

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This is a proof by homogeneous coordinates, projective geometry and computer algebra. I start with the first case. First off, observe that your whole setup is invariant under projective transformations. So without loss of generality you may assume that the original conic is the unit circle, and you may further assume that $P$ is at position $[1:0:1]$. Then ...

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HINT You want to reflect lines $x= \pm 0.5, y= \pm 0,5$ about $y = m x , m= \tan \theta .$ Instead of that consider four among the eight rotated lines $$\pm x \cos \theta \pm y \sin \theta = 0.5 ; \, \pm x \sin \theta \pm y \cos\theta = 0.5$$

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Let $ABCD$ be your square. Let's call $G$ and $H$ the points where $l$ intersects, respectively, the sides $AB$ and $CD$. We have the two trapezia $GBCH$ and $AGHD$ that both have the same area (half of the original square, so $1/2$). We now want to reflect $AGDH$ around $l$, let's call $L$ and $K$ the images of $A$ and $B$ after the reflection. The new ...

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You can tweak $U$ and $V$ so that $A$ becomes $\left[\begin{array}{ccc}a&a&a\\a&a&b\\a&b&c\end{array}\right]$. $(a=0.9595767,b=-0.4841173,c=0.0808466)$ Then $A^TA=A^2$ has eigenvalues $1,1,4$, so $A$ has eigenvalues $\pm1,\pm1,\pm2$ The choice $1,-1,2$ gives $2a+c=2,a(a-b)^2=2$ and I think: $2ac-b^2-a^2=-1$. I think it turns into a ...

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This is a svd decomposition, and it's not unique, because two singular values are equal. That means that U and V are not unique, and probably not in the most symmetric form. The same degeneracy may be present in the solution itself, although the absolute value in the condition on component may then push this into an awkward angle. The fact that the same ...

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Many of these are proved using the universal hypersurface. Let $P$ the projective space of all degree $d$ forms (since the equation and any non-zero constant multiple give the same variety) and consider $Z\subset P\times\mathbb{P}^n$ the universal hypersurface defined in the obvious way - these are pairs $(f,p)$ with $f(p)=0$. Consider $T\subset Z$, defined ...

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I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in $\mathbb{P}^{5}$ by $[z_{0}:\ldots:z_{5}]$ and the coordinates in $\mathbb{P}^{2}$ by $[x_{0}:x_{1}: x_{2}]$. Claim The Veronese surface is cut out by three quadrics: $$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$ $$C_{2}: ... 0 They're collinear if the matrix of their coordinates has rank <3, or if the determinant of their coordinates is 0. Now it is easy to check this using row reduction:$$\begin{bmatrix}1&0&3\\3&6&3\\2&8&-2 \end{bmatrix}\to \begin{bmatrix}1&0&3\\0&6&-6\\0&8&-8 \end{bmatrix}\quad\begin{matrix}R_2 \leftarrow ...

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Hint $$3A-C=B$$ This means you can express $C$ as linear combination of $B$ and $A$. Thus they can not span a three dimensional space.  So you can skip $C$. Now set the origin to $A$ ($A\to 0$) you only have $B$ ($B\to B-A$) showing in one direction. Thus all points are collinear. Because they all lie on one line (With the origin $A$ and from this ...

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I think it’ll help if we restrict our view to the finite plane, i.e. setting $Z=1$, using the equivalence $(x,y,z)\leftrightarrow(x/z,y/z,1)$, which is good except on the “line at infinity” given by $Z=0$. You’ve done that in your exposition. Now, to modify slightly what you’ve said, the given map $\phi$ has the effect of sending $(x,y,1)$ to ...

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Here are some recommendations from my reading list: Linear Geometry by Gruenberg and Weir Projective and Polar Spaces by Peter Cameron Projective Geometry: From Foundations to Applications by Albrecht Beutelspacher and Ute Rosenbaum Outline of Projective Geometry by L. E. Garner

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What definition of irreducible are you using? The one I have in mind, which is also the definition given in Hartshorne, is: Definition: A nonempty subset of a topological space $Y \subseteq X$ is irreducible if it cannot be expressed as a union $Y = Y_1 \cup Y_2$ of two proper subsets each of which is closed in $Y$. Thus, we say that a reducible space ...

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$f$ is homogeneous and the denominator is never zero, so it gives a well-defined function on $\mathbb{P}^n$. It is clear that a matrix $A$ in the image of $f$ has trace $1$, is symmetric, and the fact that $A^2 = A$ is an easy algebraic verification. In the coordinate patches given here, away from, for example, $x_{n+1} = 0$, the map is given locally as ...

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The smooth and topological answers should agree so I will just concentrate on the topological case. Let me restrict attention to the complex case although the real case looks similar. The classifying space of $\mathbb{P}^{n-1}$ bundles is $BPGL_n(\mathbb{C})$. There is a natural functor from rank $n$ complex vector bundles to $\mathbb{P}^{n-1}$-bundles and ...

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The Cartesian equation of the circle of radius $r$ and center $(a, b)$ in the $(x, y)$-plane is $$(x - a)^{2} + (y - b)^{2} = r^{2}. \tag{1}$$ To "homogenize" in projective coordinates $(w, x, y)$, multiply out, then multiply each monomial by the appropriate power of $w$ to make each term have the same total degree in $(w, x, y)$. Here, (1) becomes  ...

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Suggestion: Get a copy of Hartshorne's little book on Projective Geometry. It'll answer questions like this. Your question seems to me "How do we know that any two lines have the same number of points?" (since whether we call this "$n$" or "$n+1$" really doesn't matter). Here's a quick proof, starting with Hartshorne's four axioms for projective planes. ...

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The equations which define $P_3$ have coefficients in $k$, so $Gal(\bar k/k)$ preserves the locus of these equations i.e it fixes the coordinates of $P_3$ since it is the unique point of this locus.

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If you already know that the fundamental group is $\mathbb Z_2$ then there is little to prove except that the homotopy class of $s$ is nontrivial, i.e., that a projective line cannot be continuously contracted to a point. On the other hand the whole point of the exercise seems to be to prove that all closed curves are homotopic to either a point or a ...

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