Hot answers tagged

5

Great question! Let's start by looking at all the data we need to make sense of $\pi$ in the first place. You may have heard of metric spaces before. These are one of the more basic ways of generalizing geometry: a metric space is just a pair $(X, d)$: a collection $X$ of "points", together with a "distance" function $d$, satisfying some obvious rules. For ...


5

The hyperplane will intersect "all positive space" iff its normal vector is neither in "all positive space" nor in "all negative space". This gives a probability of $1-2^{1-n}$.


4

Identify the euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$. Let $T$ be the position of the treasure. If the direction of rotation is the one given in the picture, then $$\begin{cases} M &= A - i (A-P)\\ N &= B + i (B-P) \end{cases} \implies T = \frac12(M+N) = \frac{A+B}{2} - i\frac{A-B}{2}$$ This means in order to get to ...


4

As the diagram suggests, any starting point and the endpoints ($A^\prime$ and $B^\prime)$ of the routes through $A$ and $B$ determine pairs of congruent triangles with the (other) vertices $X$ and $Y$ of the square with diagonal $\overline{AB}$. The key midpoint property then becomes clear (and nicely related to the distance from starting point to ...


4

Let $PC$ be the side length of the given square and $PA$, with $\angle CPA=90^\circ$, the given side length of the desired rectangle. Let $O$ be the intersection of the bisector of $AC$ with $\overleftrightarrow{AP}$. The circle around $O$ through $A$ (and $C$) intersects $AP$ in a second point $B$. Then $PB$ is the other side length of the desired ...


4

Consider a triangle $ABC$ with $\angle A = 10^\circ$, $\angle B = 150^\circ$, and $\angle C=20^\circ$. Let $O$ be the circumcenter of the triangle $ABC$. Then $$\angle AOC = 360^\circ - 2 \angle B = 60^\circ,$$ so triangle $AOC$ is equilateral. Let $S$ be the circumcenter of the triangle $OBC$. Then $$\angle BSC = 2\angle BOC = 4\angle BAC = 40^\circ$$ ...


3

This doesn't directly answer your question, but should help bridge the gap between Euclidean geometry and curvature, and so may be of interest nonetheless. In the sense you're asking, "curvature" (meaning Gaussian or intrinsic curvature) can be defined using angular defect in small triangles. More precisely, if $\triangle ABC$ has "straight lines" (geodesic ...


3

First of all, you have to decide which axioms of euclidean geometry you want to keep in the Riemannian setting. The most natural thing to do is to work with simply connected complete noncompact Riemannian surfaces. (Cf. Andrew Hwang's answer.) The Riemannian notion of a line is a complete geodesic. You probably also want to assume that two distinct lines ...


3

Here are diagrams of what's going on: You can work out how long the line at the bottom is by comparing the big triangle with the little triangle. They are similar triangles.


2

Consider the complex number $r=\cos \theta + i \sin \theta$, such that $|r|=1$. For a complex number $z=x+iy$ we have: $$ rz=(\cos \theta + i \sin \theta)(x+iy)=x\cos \theta -y\sin \theta+i(x \sin \theta+ y \cos \theta) $$ If we interpret $z$ as the vector ${\bf z}=(x,y)^T$ we can write this result as: $$ {\bf z'}= r\begin{bmatrix} \cos \theta & -\sin ...


2

Any distance (mathematically speaking) could be useful a priori. It strongly depends on what you consider to be "close". For example the infinite norm could be appropiate if your readings are "maximum strengh of each muscle" and you would like to see whether any of these 4 muscles can reach 100 or close to 100 (Newton?) of strengh or not, to know whether to ...


2

Here is a counterexample. Let $n=4$ and let $K$ consist of the rows of the following array: $$\begin{matrix} -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \\ \end{matrix}$$ Note that any isometry $D\to D$ which fixes the first $3$ points must be the identity (fixing the first ...


2

I think you are about one step away from constructing a counterexample. Consider your last figure. The interiors of regions $E_1$, $E_2$, and $E_4$ are pairwise disjoint convex sets whose union is a proper subset of $C$. The only convex subsets of $C$ that contain $E_1$ and do not intersect either $E_2$ or $E_4$ are subsets of the closure of $E_1$. That ...


2

The answer is that he will have to dig in two places. The places are found by bisecting the line $AB$, and on the perpendicular bisector of $AB$ marking of the distances $AB/2$ from the midpoint of $AB$ in either direction. The proof: By rescaling and rotating, we can place $A$ and $B$ at $(-1,0)$ and $(1,0)$ in a cartesian coordinate system. Now the ...


2

It doesn't matter. He can put down a new pole wherever he wants and follow the same instructions. He will still find the treasure.


2

An easy way to see this is to use Rodrigues's rotation formula: $$ g{\bf v}=\cos \alpha \,{\bf v}+(1-\cos \alpha)({\bf u}\cdot {\bf v}){\bf u}+\sin \alpha \,{\bf u}\times {\bf v} $$ where ${\bf u}$ is the axis of rotation. Now it can be shown that conjugating $g$ by $h$ gives $$ hgh^{-1}{\bf v}=\cos \alpha \,{\bf v}+(1-\cos \alpha)((h{\bf u})\cdot {\bf ...


2

Do you know the equation for a circle with center $(a,b)$ and radius $4$? You are being asked to evaluate $a,b$. You are given two properties of the circle to do so. The first one, tangent to the $x$ axis, gives you $b$ The other, you need to solve the equation of the circle simultaneously with the equation of the line. For most values of $a$ you will ...


2

Hint: Since the radius is $r=4$ and the circle touch the$x$ axis, its center is a point with coordinates $C=(\alpha,4)$. So its equation is: $$ (x-\alpha)^2+(y-4)^2=16 $$ with $\alpha>0$ to be in the first quadrant. You want that it is tangent to the line $4x-3y=0$ and this means that the system: $$ \begin{cases} (x-\alpha)^2+(y-4)^2=16\\ 4x-3y=0 ...


2

Suppose we consider a coordinate system centred at the point $A$. Now, suppose we shrink the entire figure, keeping point $A$ fixed such that the distance between $A$ and $B$ becomes equal to $1$ inch. Hence, we can say that $B$ moves to a point $B_1$ on $AB$ such that $AB_1=1$ inch. Also, suppose that in this process, $G$ shifts to $G_1$ and $H$ shifts to ...


2

Apply the triangle inequality $\|a+b\| \leq \|a\| + \|b\|$, with $a=x-z$ and $b=z-y$. $$ \|x-y\| = \| x -z + z - y \| = \| (x -z) + (z - y) \| \leq \| (x -z) \| + \|(z - y) \| < 2+3=5 \\ \implies \|x-y\| < 5.$$


1

I like projective geometry, so I'll be explaining using terms from there. Your first parabola can be written as $$(x,y,1)\cdot\begin{pmatrix}0&0&2b\\0&-1&0\\2b&0&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$ Its axis is the $x$ axis, evidenced by the fact that $(1,0,0)$ is the only point at infinity (i.e. last ...


1

No the distance between two midpoints is the same as a side, which is not what you want. The easiest is probably to start by calculating the area of the parallelogram (by taking the length of the cross product of two adjacent sides in 3D, or as a determinant in 2D) and then divide by the length of the side you want to have as base.


1

If by opposing sides you mean parallel sides, then no, it isn't. That will be the length of the side adjacent to the one you are referring to. One way would be to find the area of the parallelogram using the coordinates and equate the area to the product of the base and the height. The length of the base can again be found using the coordinates of two ...


1

This is one approach. I don't pretend it's the nicest. Let $u = \tan\alpha$. Then $$ \tan2\alpha = \frac{2u}{1-u^2} $$ $$ \tan3\alpha = \frac{u(3-u^2)}{1-3u^2} $$ Since $\tan3\alpha = 2\tan2\alpha$, we have $$ \frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2} $$ $$ (1-u^2)(3-u^2) = 4(1-3u^2) $$ $$ 3-4u^2+u^4 = 4-12u^2 $$ $$ 1-u^4 = 8u^2 $$ Now let $y = BE$. ...


1

(1) The brown dotted line is the extension of $BE$. (2) The green dotted line ($AC’$) is the angle bisector of $∠BAE$ such that $\beta_1 = \beta_2 = \beta$. Another obvious fact is $\beta ‘ = 2\beta$. (3) The blue dotted line is the perpendicular bisector of $AB$ cutting $AB$ and $AC$ at $P$ and $Q$ respectively. By intercept theorem, $AQ = QC$. By ...


1

You're essentially just intersecting lines. My preferred means to deal with this is projective geometry and homogeneous coordinates. Then all of this boils down to cross products. So here is a crash course. Represent a point $(x,y)$ in the plane by a vector $(x,y,1)\in\mathbb R^3$. or any multiple thereof. So $(2x,2y,2)$ is the same point, just written ...


1

No matter how small we try to make a point, it still has some size/dimension. It sounds a bit like you are talking about drawing a point, but that is not what we're doing when we imagine a point in geometry. A point is an idealized, primitive notion. It does not have any physical size to speak of. how can these infinitude of points add up to give ...


1

You have the right idea, you should be considering it as $\mathbb{R}^2$. The line $L(z_1,z_2)$ can be defined as the image of the straight-line path between its endpoints: $$ f : [0,1] \rightarrow \mathbb{C} \\ f(t) = z_1 + t(z_2 - z_1) \\ = tz_2 + (1-t)z_1 $$ So any point on the line can be represented as $f(t)$ for some $t$. The step you're looking at is ...


1

Let $ABCD$ be the given quadrilateral with perpendicular diagonals meeting at $E$, whose vertices are the centers of the four circles passing through $E$ mentioned in the question. Let then $PQRS$ be the quadrilateral whose vertices are the reflections of $E$ across the sides of $ABCD$ and whose sides are the chords mentioned in the question (see picture ...


1

I will present a construction in two steps. There are a number of exceptional cases to be considered, but they should be ignored on first reading. We will assume $A, B, C$ are not aligned. (Otherwise, the only solution is to let $K = M$ be the midpoint of $BC$.) First note that there are two rotations $\rho$ which carry line $AB$ to line $AC$ and which ...



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