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8

If you have a 2 by 2 invertible matrix $M$ over $k$, then $M$ induces an isomorphism of $\mathbb P^1$ with itself. If you have a general point $[a:b]$ in $\mathbb P^1$, write down a matrix that sends $[a:b]$ to $[1:0]$ so that you have $\mathbb P^1 - [a:b] \cong \mathbb P^1 - [1:0]$. Then apply your isomorphism above. By the way, your isomorphism above is ...


4

Yes, the projective line minus any point is the affine line. I'll suppose the underlying field is $\mathbb F$, and that we are defining $\mathbb P$ to be the set of one dimensional subspaces of $\mathbb F^2$. The affine line is just the field $\mathbb A=\mathbb F$. Let me know if your definitions are substantially different. For any point $p'\in\mathbb P$ ...


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$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$In the projective line, an arbitrary quotient $Z/W$ of complex numbers (not both zero) makes sense: If $W \neq 0$, then $z = Z/W$ has its ordinary meaning as a complex number. If $W = 0 \neq Z$, we write $Z/W = \infty$. It's straightforward to check that some arithmetic operations with $\infty$ can ...


1

In projective geometry, a circle is just a special case of a conic, namely one passing through the ideal circle points $(1,\pm i,0)$. So if you allow for complex transformations, then you can take a pair of conics and map two points of intersection to these special points to obtain a configuration with two circles. For more than two conics, though, this only ...


1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$tl; dr: There's a hyperboloid model of the projective plane that is easily shown (using elementary linear algebra, see sketch below) to be homogeneous and isotropic under its isometry group. That's enough to guarantee its Gaussian curvature is constant. (In fact, every ...


1

That happens because 1° in longitude corresponds to different lengths at different latitudes. A not 100% accurate but fast remedy could be as follows: divide the longitude of each point by the cosine of the corresponding latitude before computing the orthogonal vector, and perform the reverse transformation on the final points. This corresponds to using a ...


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$\newcommand{\dd}{\partial}$No claim of elegance, but Cartesian coordinates handle both questions, and the answers are "yes": Up to translation, a general ellipsoid can be written in the form $$ Ax^{2} + By^{2} + Cz^{2} + 2(Dxy + Exz + Fyz) = 1 \tag{1} $$ for some positive-definite coefficient matrix $$ \left[\begin{array}{@{}ccc@{}} A & D & E \\ D &...


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Menelaus' theorem for $\triangle BCE$ and transversal $DA$ gives $ \frac{EF}{FC}\; \frac{CD}{DB}\; \frac{BA}{AE} = 1 = \frac{EF}{FC}\; \frac{2}{5}\; \frac{3+4}{4}$ thus $\frac{EF}{FC} = \frac{10}{7}$. Permute $5 \leftrightarrows 3, 2 \leftrightarrows 4$ to get $\frac{DF}{FA} = \frac{3}{14}$.


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After joining BF, label the areas as shown. For example, p = area of triangle BEF = [BEF]. The rule:- “$\triangle$s having the same altitude, ratio of their areas is proportional to the ratio of their bases.” For example, $\dfrac {p}{z} = \dfrac {3}{4}$. This means $p = \dfrac {3}{4}z$. Do the same to $\dfrac {q}{x} = …$ to get $q = ? x$. Therefore, $[...



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