# Recent Changes

### Saturday, April 9

1. Lecture 12 edited \chapter{Spherical Geometry} ... word: measuring the the Earth. The The geodesics on a sphe…
\chapter{Spherical Geometry}
...
word: measuring thethe Earth. The
The geodesics on a sphere are parts of great circles, i.e. the circles which are the intersections of the sphere with a plane passing through the center of the sphere. For instance the shortest way to fly from a point on the equator of the Earth to another point on the equator is along the Equator. We won't prove it in this lecture, however.
Instead we will define the spherical geometry in the way Klein would: the space we will be dealing with will be the sphere of radius $R$ centered at the origin of an Euclidean space and the group of isometries of spherical geometry will be the group of rotations around the origin.
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Now we would like to study the simplest figures on the sphere. Triangles were the simplest figures on a plane, but on a sphere we have a figure which is even simpler - a slice. A slice is the figure formed by two lines on the sphere. Note that on a sphere any two lines intersect at two points which are opposite to each other. The only parameter that a slice has is the angle formed by the two lines defining the slice (recall that angle between any two curves is defined as the angle between the tangent lines to these curves at the point of intersection; alternatively we can define an angle between two great circles on the sphere as the angle between the two planes that contain these great circles).
Let's consider a slice of angle $\alpha$ and compute it's area. It's pretty obvious that the area of such a slice should be proportional to the angle $\alpha$. Now if we take $\alpha=\pi$, the slice is half-sphere, and hence it should have area $2\pi R^2$. Hence we get that the area of a slice of angle $\alpha$ is $2\alpha R^2$.
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slices: $4\pi R^2+4A=4\alphaR^2+4A=4\alpha R^2+4\beta R^2+4\gamma The result we got is very different from the sort of results we have in Euclidean geometry - in Euclidean geometry the sum of angles of a triangle is always$\pi$. In spherical geometry, however, the difference between the sum of angles and$\pi$is proportional to the area of triangle! The statement that the sum of angles of planar triangle is$\pi$follows in fact from what we proved for the spheres. Imagine a huge sphere of radius$R$lying on a plane and imagine a triangle drawn on the plane. The spherical triangle obtained from it by central projection with center at the center of the sphere from the plane to the sphere is very close to the planar triangle, when$R$is big enough. But then the difference of sum of angles of the spherical triangle and$\pi$is equal to the area divided by$R^2$, so as$R$tends to infinity, the spherical triangle approaches the planar one, the area approaches the area of planar triangle and thus the difference of sum of angles and$\pi$approaches zero. The formula we proved could be used in principle to measure the radius of the Earth without using anything which doesn't lie on its surface. Indeed, if we could draw a big triangle on the surface of our sphere, measure its area accurately and measure its angles accurately, we would be able to find the radius of Earth. ... To deal with duality in concrete terms, we can use the notion of cross-product of vectors. Recall that the cross-product of two vectors$v$and$w$is the vector$v\times w$which is orthogonal to$v$and$w$and whose length is equal to the area of parallelogram spanned by$v$and$w$. The direction of this vector can be determined by a right- or left- hand rule, depending on the orientation chosen for the ambient three-dimensional space. In fact if we compute the cross-product of$v$and$w$in the other order, we get the same result, except pointing in opposite direction:$w\times v=- v\times w$. Other properties of cross-product include bilinearity: ... (also$(v_1+v_2)\times w=v_1\timesw=v_1\times w +
$v\times (\lambda w)=(\lambda v) \times w = \lambda (v\times w)$
and the Jacobi identity
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Let's apply this idea to proving that three medians in a triangle are concurrent. To prove this statement it is enough to show that the points dual to the medians are collinear. Now the midpoint of the segment $AB$ can be expressed as $\frac{A+B}{|A+B|}$ - since vectors $A$ and $B$ have equal length, the direction $A+B$ in plane through the origin, $A$ and $B$ which points towards the midpoint between $A$ and $B$.
Thus the point dual to the median from vertex $C$, i.e. to the line connecting $C$ with the midpoint of $AB$ is proportional to $(A+B)\times C$. Similarly the points dual to other medians are proportional to $(B+C)\times A$ and $(C+A)\times B$ respectively. Now notice that antisymmetry of the cross-product implies that the sum $(A+B)\times C+(B+C)\times A+(C+A)\times B$ is equal to zero. This means that the origin and the points $(A+B)\times C$, $(B+C)\times A$ and $(C+A)\times B$ are coplanar, which implies the collinearity (on the sphere) of the points dual to the medians.
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proportional to $(B\times$(B\times C)\times A$\section{Bonus: a Funny Proof of Pascal's Theorem} (to be moved to the section on Pascal's theorem) Let's go back to Pascal's theorem and try to prove it without mentioning the notion of cross-ratio. Our strategy this time will be as follows: ... quadric where$L_1$L_1 L_3 L_5 ... so that$L_1$L_1 L_3 L_5 First we present a very simple proof that relies, however, on knowledge of complex analysis. Consider all the equations we have as equations on complex numbers that define figures in complex plane. Also make also all the equations homogeneous and consider them in the complex projective plane. Then the quadric becomes isomorphic to the Riemann sphere - this follows from the fact that there exists a rational parametrization of the quadric and that the quadric is smooth. The function$\frac{L_1 L_3 L_5}{L_2 L_4 L_6}$is in fact a function on the complex projective plane, since the numerator and the denominator are homogeneous of the same degree (3). Finally, when restricted to the quadric, this function doesn't have any poles - the zeros of the denominator get canceled with the zeros of the numerator (they both are at the vertices of the hexagon). Thus Liouville's theorem implies that the function must be constant:$\frac{L_1 L_3 L_5}{L_2 L_4 L_6}=\lambda\$.
While this proof is quite transparent, we would like to give an alternative proof that uses considerations with real numbers only. This way we will have a real (in the sense of real numbers) proof of Pascal's theorem, which the readers unfamiliar with complex analysis will be able to understand.
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11:55 am

### Thursday, March 29

1. home edited ... {_Chapter5.pdf} {_Chapter6.pdf} {_Chapter7.pdf} {problems_on_proj_geom.pdf} {problem…
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{_Chapter5.pdf}
{_Chapter6.pdf}
{_Chapter7.pdf}
{problems_on_proj_geom.pdf}
{problems_on_duality_and_sph.pdf}
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7:25 pm
7:24 pm
3. home edited ... {_Chapter6.pdf} {problems_on_proj_geom.pdf} {problems_on_duality_and_sph.pdf} Lecture …
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{_Chapter6.pdf}
{problems_on_proj_geom.pdf}
{problems_on_duality_and_sph.pdf}
Lecture 10 comment
Problems for preparation for quiz 3:
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7:08 pm

### Tuesday, March 20

1. home edited ... {_Chapter5.pdf} {_Chapter6.pdf} {problems_for_quiz4.pdf} (these are problems on projecti…
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{_Chapter5.pdf}
{_Chapter6.pdf}
{problems_for_quiz4.pdf} (these are problems on projective geometry --- I am not sure whether they are for quiz 4 or not){problems_on_proj_geom.pdf}
Lecture 10 comment
Problems for preparation for quiz 3:
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10:30 am
2. 10:30 am
3. home edited ... {_Chapter5.pdf} {_Chapter6.pdf} {problems_for_quiz4.pdf} (these are problems on projec…
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{_Chapter5.pdf}
{_Chapter6.pdf}
{problems_for_quiz4.pdf} (these are problems on projective geometry --- I am not sure whether they are for quiz 4 or not)
Lecture 10 comment
Problems for preparation for quiz 3:
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10:29 am
4. home edited ... {_Chapter5.pdf} {_Chapter6.pdf} {problems_for_quiz4.pdf} Lecture 10 comment Problems …
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{_Chapter5.pdf}
{_Chapter6.pdf}
{problems_for_quiz4.pdf}
Lecture 10 comment
Problems for preparation for quiz 3:
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10:28 am
5. 10:28 am