Hot answers tagged

14

Intuitive interpretation: A curve has an infinity of points. So many that the amount counteracts the abscence of dimensions, like an $0\times\infty$ undeterminacy. More precisely, a curve still has no dimension transversally, but a finite (or infinite) dimension longitudinally. You can think of it as the limit of a necklace of pearls (points of finite ...


10

Euclid did say that that "A point is that which has no part". It is nothing more that a whisper of an indication that "You are here." You can't really see a point since there is nothing there. "...if point has no dimensions, i.e. in other words there is nothing, then how is it possible to draw any curve?" It isn't possible. Pick a ridiculously small ...


6

With $A$ the origin, rotate everything so that $AG$ lies on the $x$-axis. Drop a perpendicular from $H$ to meet the $x$-axis at $J$. Extend segment $CE$ to meet segment $HJ$ at point $K$. The key to the proof is the fact that that angle $\angle HEK$ has measure $18^\circ$, since each interior angle in a regular pentagon is $108^\circ$. Picture (not to ...


5

A curve is completely determined by two facts: Knowledge of all of the points lying on the curve Knowledge that the curve is drawn on the Euclidean plane When it's said that a curve is made out of points, one really means to include in the latter fact too, or something similar (e.g. a topology or a metric on the collection of points). There are more ...


5

You are confusing theory with parctise. A point does not exist in the real universe. Anything you draw will have some dimensions.


4

As pointed out by @mjqxxxx, this is possible. In fact, it is possible to pack two families of open balls of radii $\frac12, \frac13, \ldots$ into unit sphere. The picture below illustrate one possible configuration (with $n$ up to $300$): $\hspace1in$ In above picture, the two families are related to each other by a reflection with respect to origin. ...


4

Migrating my thoughts from comments below my answer to another question ... The square and triangle have little to do with the appearance of the golden ratio here. The essence of the construction is this: The figure has two salient characteristics: $O$ lies on the perpendicular bisector of edge $\overline{RS}$ of the regular pentagon; and, $B$ lies ...


4

You are confusing reality and the idealized world of maths. In an ideal world, a point has no dimensions. You could never actually draw a point. In reality, you can. You take a pencil, make a point on a piece of paper and say "that is the point". But the fact that you can see the point already means that is has some dimensions (the ink occupies a non-empty ...


3

A point in an n-dimensional Euclidean space is that point which consists of n coordinates which give a distinct location of that point. A function in Euclidean space essentially provides a rule for defining a set coordinates in space for which that function is true. You are correct in that the physical realization of a collection of 0-dimensional points ...


3

This is an algebraic proof using polynomial computations. What I like to think of as the “don't make me think” approach. As such it's pretty far away from what you already have. But seeing that there was no other answer in the past two days, I reckon that this proof is better than none, and that bumping the question may do more good than having it marked as ...


3

Inspired by a comment posted by OP and by the answer given by @Blue, here are three other ways to combine the regular pentagon, square, and equilateral triangle of equal sides in such a way as to divide a segment between vertices in the ratio $\phi:1$. The basic idea is simply to place the square so that it provides part of the perpendicular to one ...


2

Notice that if you draw the line $CB$, we have that the $\triangle SCB$ is equilateral. Thus, $\angle SBC = 60^\circ$. Now, the quadrilateral $DABC$ is a cyclic quadrilateral , thus the angle which is opposite to the $\angle SBC$ will be: $$\angle ADC = 180^\circ - 60^\circ = 120^\circ.$$


2

Computing the power of point $A$ with respect to $\bigcirc R$ in two ways, and the power of point $B$ with respect to $\bigcirc O$ in two ways: $$\begin{align} |\overline{AP}|^2 = |\overline{AO}| |\overline{AT}| &\qquad\to\qquad r \cdot 3 r = a^2 \\ |\overline{BP}| |\overline{BA}|\; = |\overline{BQ}| |\overline{BS}| &\qquad\to\qquad r \cdot 3 r = ...


2

This is not a construction of the golden ratio. It is division of a segment into parts whose ratio is golden, but the segment is incommensurable with the rest of the figure, so it is not a construction in the classical sense. It also assumes as given a regular pentagon, which is practically the same thing as the golden ratio. This is why in the figure can ...


2

Assuming you want to compute the $n$-dimensional volume of the intersection of the half-space with the ball. First of all, reduce the problem to a "standard form". Suppose $\mathbf e_1,\dots,\mathbf e_n$ is the orthonormal basis in $\mathbb R^n$, and $\mathbf e^1,\dots,\mathbf e^n$ is dual to it. Move and rotate the space that the half-space becomes ...


2

Look at this picture, for the distance $d$, you can use the formula $d=|w\cdot x_0+b|/|w|$. Now $r_1=\sqrt{r^2-d^2}$. Then the volume (the red part) is just the integration of the volume of the interface $S_1$ with radius $r_1$ to the interface with radius $r$.


2

As at least one other answer has mentioned, it takes an infinitude of points to make up a curve. (Or to make up any other interval, like the rather simple "curve" of the closed interval [0,1], for example.) At least one other answer has also mentioned the notion of "measure," and that's probably the more appropriate way to think about the small-ness of a ...


2

From your diagram C seems to be at $(4,4)$ G seems to be at $(4+2\sqrt3,4)$ H seems to be at $(4,-12+8\sqrt3 )$ Which makes $p=2\sqrt3$ and $q=16-8\sqrt3$ and $\dfrac{p}{q}=\dfrac34+\dfrac{\sqrt3}2 \approx 1.616025$ which is slightly less than the golden ratio $\dfrac12+\dfrac{\sqrt5}2 \approx 1.618034$


1

Let $F$ on the circle with diameter $AB$ be such that $FC$ is the angle bisector of $\angle AFB$. Since $(ACBD)$ is a harmonic range, $FD$ is the exterior angle bisector at $F$ of $\triangle AFB$. Consequently $\angle AFD=135^\circ$ and $FC\perp FD$. It follows that $\angle OAF+\angle PDF=45^\circ.$ Hence $$\angle FOP+\angle FPO=90^\circ,$$ and we have ...


1

If $K$ is the convex hull of a finite number of points $\{p_i\}_{i=1}^m$, $K^\circ$ can be (trivially) written as intersection of halfspaces: $$K^\circ = \bigcap_{i=1}^m \{x \in \mathbb{R}^n \mid x^\top p_i \le 1\}.$$ This helps to get some feeling for small numbers $m$.


1

Some other answers have already mentioned the distinction between the ideal geometric world (which we can describe by mathematical rules) and the real world (which our ideal world is intended to approximate). However, an interesting point is that Euclid never talked about dimension but merely said: [Euclid Book 1 Def 1] σημειον εστιν ου μερος ουθεν ...


1

This article by Prof. Lawrence Spector helped me a lot and I think it addresses exactly what your confusion is. I'm not sure what the policy is on just posting links, but I don't think i would do a great job paraphrasing what he wrote, so here goes: http://www.themathpage.com/aCalc/apoint.htm I would also highly recommend reading on the same site: ...


1

(Converting comment to answer, as requested.) If the side of each sub-square has length $2$, then the radius of the circle is $\sqrt{10} = \sqrt{2}\cdot\sqrt{5}$. Once you have $\sqrt{5}$ floating around, it's not too difficult to find $\phi$.


1

The ratio is $\sqrt{\dfrac{2}{3 - \sqrt{5}}}$. Consider the right triangle HIC. We may calculate the length of the blue segment, which is $r\sqrt{3}$, and the angle ICH, which is $\pi/6$. Consider the triangle FDC. From the Law of Sines, we may find angle CFD, which is $\sin^{-1}(1/4)$. Since the sum of the interior angles is $\pi$, we find angle FDC, ...


1

This might be easier to visualize if you identify a point $(x,y)$ in $\mathbb{RP}^2$ with line through $(x,y,1)$ in $\mathbb R^3$ (less the origin). I.e., the projective plane is the plane $z=1$ in $\mathbb R^3$ with points at infinity and the line at infinity added. The $w$-coordinate of the point’s homogeneous representation is just a $z$-coordinate in ...


1

An algebraic curve, with affine polynomial equation $p(x,y)=0$, $\deg p=d$, has a corresponding projective curve with polynomial homogeneous equation $P(X,Y,W)=0$, defined as $$P(X,Y,W)=W^d p\Bigl(\frac XW,\frac YW\Bigr)$$ In practice this means that to, say, the affine cubic curve $y^2=x(x^2+1)$, there corresponds the projective cubic curve ...


1

Part of the value of thinking about adding points "at infinity" to the Euclidean plane (so constructing the projective plane) and working with the projective plane is that in the projective plane there is symmetry between points and lines. Since each pair of points determines a line, each pair of lines should determine a point. For lines $L$ and $M$ that are ...



Only top voted, non community-wiki answers of a minimum length are eligible