math.answers.com/math-and-arithmetic/How_to_arrange_four_points_at_equidistance

Preview meta tags from the math.answers.com website.

Linked Hostnames

8

Thumbnail

Search Engine Appearance

Google

https://math.answers.com/math-and-arithmetic/How_to_arrange_four_points_at_equidistance

How to arrange four points at equidistance? - Answers

You have to think 3D!!! A triangular based pyramid does the trick if all the edges are of equal length. More generally, you can arrange n+1 points equidistantly (Euclidean distance) in an n dimensional space, but it is simple to do it in an n+1 dimensional space. First an alogrithm for the n+1 dimensional space followed by an algorithm for an n dimensional space: For both algorithms: Define e_i as a unit vector in the direction of the ith axis for 1<= i <= n+1. #1: n+1 dimensions Place the ith point at e_i. You are done since the distance between any two points is just sqrt(2). #2: n dimensions Place the first point at the origin Place the second point at e_1. For the (k+1)th point (with k= 2 through n) iterate the following procedure: 1) Let X_k be the center of mass of the first k points (this will be a point only involving the first k-1 coordinates). 2) Locate the next point at X_k + sqrt(1- X_k ^2)*e_k. This means the first k-1 coordinates are the coordinates of the center of mass, X_k. The kth coordinate is computed by first taking the sum of the squares of the coordinates of X_k. Subtracting this value from 1 then taking the square root yields the value of the kth coordinate. When you have located the n+1 point, you are done. The proof of the equidistance is simple by induction. Assume that the first r points are equidistant (which is obvious for r=2 since a set two points must be equidistant). Then the distance that separates the first r points from each other must be 1 because that is the distance between the first two. Now consider the r+1 point. It's distance from all other points must be equal because all other points are symmetric about the center of mass and it sits directly "above" the center of mass in one higher dimension. We chose its distance above the center of mass to ensure that its distance from the origin is exactly 1 so its distance from each of the first r points is exactly 1. Since we know that the first r points are all exactly 1 apart from each other, we have showed that the first r+1 points are equidistant. This completes the induction proof.



Bing

How to arrange four points at equidistance? - Answers

https://math.answers.com/math-and-arithmetic/How_to_arrange_four_points_at_equidistance

You have to think 3D!!! A triangular based pyramid does the trick if all the edges are of equal length. More generally, you can arrange n+1 points equidistantly (Euclidean distance) in an n dimensional space, but it is simple to do it in an n+1 dimensional space. First an alogrithm for the n+1 dimensional space followed by an algorithm for an n dimensional space: For both algorithms: Define e_i as a unit vector in the direction of the ith axis for 1<= i <= n+1. #1: n+1 dimensions Place the ith point at e_i. You are done since the distance between any two points is just sqrt(2). #2: n dimensions Place the first point at the origin Place the second point at e_1. For the (k+1)th point (with k= 2 through n) iterate the following procedure: 1) Let X_k be the center of mass of the first k points (this will be a point only involving the first k-1 coordinates). 2) Locate the next point at X_k + sqrt(1- X_k ^2)*e_k. This means the first k-1 coordinates are the coordinates of the center of mass, X_k. The kth coordinate is computed by first taking the sum of the squares of the coordinates of X_k. Subtracting this value from 1 then taking the square root yields the value of the kth coordinate. When you have located the n+1 point, you are done. The proof of the equidistance is simple by induction. Assume that the first r points are equidistant (which is obvious for r=2 since a set two points must be equidistant). Then the distance that separates the first r points from each other must be 1 because that is the distance between the first two. Now consider the r+1 point. It's distance from all other points must be equal because all other points are symmetric about the center of mass and it sits directly "above" the center of mass in one higher dimension. We chose its distance above the center of mass to ensure that its distance from the origin is exactly 1 so its distance from each of the first r points is exactly 1. Since we know that the first r points are all exactly 1 apart from each other, we have showed that the first r+1 points are equidistant. This completes the induction proof.



DuckDuckGo

https://math.answers.com/math-and-arithmetic/How_to_arrange_four_points_at_equidistance

How to arrange four points at equidistance? - Answers

You have to think 3D!!! A triangular based pyramid does the trick if all the edges are of equal length. More generally, you can arrange n+1 points equidistantly (Euclidean distance) in an n dimensional space, but it is simple to do it in an n+1 dimensional space. First an alogrithm for the n+1 dimensional space followed by an algorithm for an n dimensional space: For both algorithms: Define e_i as a unit vector in the direction of the ith axis for 1<= i <= n+1. #1: n+1 dimensions Place the ith point at e_i. You are done since the distance between any two points is just sqrt(2). #2: n dimensions Place the first point at the origin Place the second point at e_1. For the (k+1)th point (with k= 2 through n) iterate the following procedure: 1) Let X_k be the center of mass of the first k points (this will be a point only involving the first k-1 coordinates). 2) Locate the next point at X_k + sqrt(1- X_k ^2)*e_k. This means the first k-1 coordinates are the coordinates of the center of mass, X_k. The kth coordinate is computed by first taking the sum of the squares of the coordinates of X_k. Subtracting this value from 1 then taking the square root yields the value of the kth coordinate. When you have located the n+1 point, you are done. The proof of the equidistance is simple by induction. Assume that the first r points are equidistant (which is obvious for r=2 since a set two points must be equidistant). Then the distance that separates the first r points from each other must be 1 because that is the distance between the first two. Now consider the r+1 point. It's distance from all other points must be equal because all other points are symmetric about the center of mass and it sits directly "above" the center of mass in one higher dimension. We chose its distance above the center of mass to ensure that its distance from the origin is exactly 1 so its distance from each of the first r points is exactly 1. Since we know that the first r points are all exactly 1 apart from each other, we have showed that the first r+1 points are equidistant. This completes the induction proof.

  • General Meta Tags

    22
    • title
      How to arrange four points at equidistance? - Answers
    • charset
      utf-8
    • Content-Type
      text/html; charset=utf-8
    • viewport
      minimum-scale=1, initial-scale=1, width=device-width, shrink-to-fit=no
    • X-UA-Compatible
      IE=edge,chrome=1
  • Open Graph Meta Tags

    7
    • og:image
      https://st.answers.com/html_test_assets/Answers_Blue.jpeg
    • og:image:width
      900
    • og:image:height
      900
    • og:site_name
      Answers
    • og:description
      You have to think 3D!!! A triangular based pyramid does the trick if all the edges are of equal length. More generally, you can arrange n+1 points equidistantly (Euclidean distance) in an n dimensional space, but it is simple to do it in an n+1 dimensional space. First an alogrithm for the n+1 dimensional space followed by an algorithm for an n dimensional space: For both algorithms: Define e_i as a unit vector in the direction of the ith axis for 1<= i <= n+1. #1: n+1 dimensions Place the ith point at e_i. You are done since the distance between any two points is just sqrt(2). #2: n dimensions Place the first point at the origin Place the second point at e_1. For the (k+1)th point (with k= 2 through n) iterate the following procedure: 1) Let X_k be the center of mass of the first k points (this will be a point only involving the first k-1 coordinates). 2) Locate the next point at X_k + sqrt(1- X_k ^2)*e_k. This means the first k-1 coordinates are the coordinates of the center of mass, X_k. The kth coordinate is computed by first taking the sum of the squares of the coordinates of X_k. Subtracting this value from 1 then taking the square root yields the value of the kth coordinate. When you have located the n+1 point, you are done. The proof of the equidistance is simple by induction. Assume that the first r points are equidistant (which is obvious for r=2 since a set two points must be equidistant). Then the distance that separates the first r points from each other must be 1 because that is the distance between the first two. Now consider the r+1 point. It's distance from all other points must be equal because all other points are symmetric about the center of mass and it sits directly "above" the center of mass in one higher dimension. We chose its distance above the center of mass to ensure that its distance from the origin is exactly 1 so its distance from each of the first r points is exactly 1. Since we know that the first r points are all exactly 1 apart from each other, we have showed that the first r+1 points are equidistant. This completes the induction proof.
  • Twitter Meta Tags

    1
    • twitter:card
      summary_large_image
  • Link Tags

    16
    • alternate
      https://www.answers.com/feed.rss
    • apple-touch-icon
      /icons/180x180.png
    • canonical
      https://math.answers.com/math-and-arithmetic/How_to_arrange_four_points_at_equidistance
    • icon
      /favicon.svg
    • icon
      /icons/16x16.png

Links

58