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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 ...

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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.

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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 $\overline{AA^\prime}$. ...

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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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