# Tag Info

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Because you take $x-y=0$ as the unit point $(1,1)$, you have to reverse signs. If you were to take $x+y=0$ as the unit, then $(a,b)$ would be the coordinates of $ax+by=0$.

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The answer depends on what you mean by "bad in geometry." For many people the problem is very simple. Geometry isn't taught in schools the way it once was, so you can be very good at algebra and other things, but geometry can still seem foreign and strange. It might be good to learn geometry - at first - in normal school books from earlier times in order ...

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I'd recommend learning some non-Euclidean geometry as well. Pretty much master geometry with complex numbers and analytic geometry. The main idea at these mathematical competitions is to totally master your emotions and any nervousness you may have. Remember that you've trained well!

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See this blog post. It's true in general that the cohomology of a smooth hypersurface of degree $n$ in $\mathbb{CP}^d$ only depends on $d$ and $n$.

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Go up one dimension and into affine space. There $P$ corresponds to the line $L = (tb_0, tb_1,\ldots,tb_n)$ and $H$ still has the same equation. Now apply a linear transformation to $\Bbb A^{n+1}$ that moves $H$ to be orthogonal to $(1, 0,0,\ldots,0)$ (remember that $H$ originally is orthogonal to $(a_0, \ldots,a_n)$) (this might move $L$ to someplace ...

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To compute the degree of $\varphi(X)$ we need to cut it by generic hyperplanes $H_1,\cdots,H_k$, $k=\dim(X)$ and count the number of points in such generic intersection. Then notice that this is the same as intersecting $X$ with the pullback of the hyperplanes $\varphi^*H_i$, $i=1,\cdots,k$. So the computation of $\deg(\varphi(X))$ is reconduced to an ...

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Observe, in the Segre embedding $\mathbb{P}\times\mathbb{P}\rightarrow\mathbb{P}^3$, we have $$(x_0:x_1)\times(y_0:y_1)\mapsto(x_0y_0,x_0y_1,x_1y_0,x_1y_1).$$ If we choose the coordinates on $\mathbb{P}^3$ to be $(x,z,w,y)$, we see that $xy=zw$ on this surface.

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It depends on how exactly you define perspective projection, i.e. which properties are part of the definition and which are consequences thereof. But you can think about this geometrically. You have a line in space, and an eye point. Assuming that the line didn't pass through the eye, connecting the eye to every point on the line yields a bundle of lines, ...

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"Projection" can mean many things, and about the only feature they have in common is that it's about an (idempotent) transformation that moves points somewhere along straight lines. (And even that is not fully general -- although some "map projections" can be explained as moving points along straight lines from the surface of a sphere to an image plane, ...

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Here's a geometric proof. Let $O$ denote the origin. If $P$ and $Q$ are two points on the unit sphere, then $\sin(\angle POQ)$ is twice the area of triangle $\triangle POQ$. Thus, it suffices to prove that $$\text{area}(\triangle POQ) \,+\, \text{area}(\triangle QOR) \;\geq\; \text{area}(\triangle POR)$$ for any three points $P$, $Q$, and $R$ on the ...

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A probably-irrelevant remark: if you define the distance to be the sine of half the angle between the points (required to be in the same hemisphere!), then the triangle inequality is straightforward, because $2 \sin \frac{\theta}{2}$ is exactly the length of the chord from $P$ to $Q$, so the thing you're trying to prove becomes...the Euclidean triangle ...

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As I start writing, I'm not convinced it's even a $4$-polytope. It probably is, but I'll at least attempt a decomposition into $3$-faces and hopefully compute the Euler characteristic. I don't believe it's a well-known polytope, and I think the 'outer' facet is actually a truncated octahedron with hexagonal pyramids attached to each square face (there are ...

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The theorem you have quoted is true but only tells part of the story. An improved version is as follows. Let $U$ be a real $m\times n$ matrix with orthonormal columns, that is, its columns form an orthonormal basis of some subspace $W$ of ${\Bbb R}^m$. Then $UU^T$ is the matrix of the projection of ${\Bbb R}^m$ onto $W$. Comments The restriction to ...

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There is a projective transformation that takes any line $L$ to $\mathbb V(x)$. And there is a projective transformation that takes any nondegenerate conic $C$ to $\mathbb V(x^2 + y^2 + z^2)$. The first is just linear algebra, and the second can be proved using your result before and the fact that there is a unique conic passing through any $5$ general ...

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The whole point of the representation you're using for affine transformations is that you're viewing it as a subset of projective space. A line has been chosen at infinity, and the affine transformations are those projective transformations fixing this line. Therefore, abstractly, the use of the extra parameters is to describe where the line at infinity ...

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First, I think it may be a mistake to think about "translation", "rotation" and "scale", which is one particular decomposition of the affine group -- perhaps it's better to think about what transformations can be effected by affine maps. For affine maps: We can move any collection of three noncollinear points to any other collection of three points (which ...

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