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4

Yes, if the $n+1$ points are in general position, which simply means that the $n+1$ points must not lie in a hyperplane. We can proceed by induction: If $x_0,\ldots, x_{n}$ are our $n+1$ points in general position, then any $n$ of them, for example $x_0,\ldots, x_{n-1}$, certainly lie in a common $(n-1)$-dimensional hyperplane $H$. We can identify $H$ with $...


4

The length is $2AD+DE$. Obviously $DE=2$, so we have to find $AD$. Consider the triangle $ADF$. $DF=1,\angle AFD=60^o$, so $AF=\sqrt3DF=\sqrt3$. Hence $AB=2+2\sqrt3$.


3

Hagen von Eitzen's answer gives a neat theoretical approach of this problem. However, I would like to expose a constructive and computational way to find the radius and center of the $(n-1)$-sphere determined by $n+1$ suitable points in $\mathbb{R}^n$. Let $n$ be an integer greater than $1$ and let say $x_i:=(x_{i,j})_{j\in\{1,\cdots,n\}},i\in\{0,\cdots,n\}$...


3

As mentioned, the limiting case is when the small circle touches the larger circle internally. Then, we have the figure below:- Applying Pythagoras theorem to the red triangle, we have $OM^2 = 1 – (o.5L)^2$. Applying Pythagoras theorem to the green triangle, we have $(R + d)^2 = (1 – R)^2 – OM^2$ Eliminating $OM^2$ from the two equations, we get $d$ in ...


2

We can see: José Ferreirós' review of C.K. Raju, Cultural Foundations of Mathematics: The Nature of Mathematical Proof and the Transmission of the Calculus from India to Europe in the 16th c. CE, in Philosophia Mathematica Volume 17, Issue 3: In his interest to revise traditional historiography and oppose proofcentred mathematics, Raju devotes a lot ...


2

Why not just apply a circular inversion? If we have $p_0,p_1,\ldots,p_n\in\mathbb{R}^n$ in general position, we may consider $q_1,q_2,\ldots,q_n$ as the images of $p_1,p_2,\ldots,p_n$ under a circular inversion with respect to a unit hypersphere centered at $p_0$. There is a hyperplane $\pi$ through $q_1,q_2,\ldots,q_n$, and by applying the same circular ...


1

Assume $x_0 = 0$ for simplicity and let $x_i' = \frac{x_i}{|x_i|^2}$ be the images of $x_i$'s under an inversion centered at $x_0$. By a well-known property of inversions, $x_0,\ldots,x_{n+1}$ lie on an affine $n-1$-plane or an $n-1$-sphere if and only if $x_1',\ldots,x_{n+1}'$ lie on an affine $n-1$-plane. When the latter is expressed using the ...


1

The articles by Raju have a conspirational flavor. The history of Indian mathematics is still an uncharted territory. There are more informative unbiased articles, for instance, there are much deeper and less biased studies I've read: A. Seidenberg, “The Origin of Mathematics,” Archive for History of Exact Sciences 18, 301-342 (1978). S.C.Kak, “Science in ...


1

An interesting approach may be the following one: you may construct the Nagel point $N$ as $3G-2I$. You have the $BC$-line, i.e. the perpendicular to $IQ_a$ through $Q_a$. You may assume that some $P_a\in BC$ is the contact point of the $A$-excircle, then: $A$ lies on $NP_a$; The midpoint of $M_a$ of $P_a Q_a$ is also the midpoint of $BC$; $A$ lies on $M_a ...


1

Examining the sign of $\vec{A_{1}A_{2}} \times \vec{A_{2}A_{3}} \cdot \vec{A_{1}A_{4}}$ (the scalar triple product) should give you chirality.


1

Here's how I think you solve the problem. Label the center of the unit circle $O$, the midpoint $A$ of the chord, a point $B$ where the chord meets the unit circle, and pick a point $T$ on the line tangent to the unit circle at point $B$. After you've moved the tiny circle, call its new center $A^\prime$. You want to find the length $\overline{AA^\prime}$. ...


1

This is NOT a solution. I just want to share some of my findings. Construction: 1) Extend BO to cut the red circle at D; 2) DA cut the red circle at E and BM extended at F; 3) OE is joined. By midpoint theorem, we have 1) OMI // DEFA; 2) BJ = JE; BM = MF. All angles marked with the same color are equal. OJMI is the line of centers of the 4 circles and ...



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