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How do you form an equilateral triangle using taxicab geometry? - Answers
I may be wrong but I do not believe you can because the metric in the vertical or horizontal directions is different from that in the other ("slanted") directions.Answer:Taxicab geometry relies entirely on movements in the x and y directions (like a taxi cab on city streets which form squares). As a consequence the "shortest" distance between two points is the sum of the x and y components (Pythagorus doesn't work)An equilateral triangle requires all angles to be 60o with no 90o angles and all sides of equal length.But then you might ask "If I have two points on the grid, an x value and a y value where y= x√2 and I travel from x1=0 to y (corresponding to x2) for any non-zero value of x (x2) then back to the x axis at x3= 2x1 then back to x1 wouldn't the outer perimeter of my trip form an equilateral triangle?" Afraid not, the lines formed between x1, y, and x3 would each be stepped not straight and equal to (Δx+Δx√2) not 2Δx as in a true equilateral triangle. (Δx in this case is the distance from x1 to x2)
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How do you form an equilateral triangle using taxicab geometry? - Answers
I may be wrong but I do not believe you can because the metric in the vertical or horizontal directions is different from that in the other ("slanted") directions.Answer:Taxicab geometry relies entirely on movements in the x and y directions (like a taxi cab on city streets which form squares). As a consequence the "shortest" distance between two points is the sum of the x and y components (Pythagorus doesn't work)An equilateral triangle requires all angles to be 60o with no 90o angles and all sides of equal length.But then you might ask "If I have two points on the grid, an x value and a y value where y= x√2 and I travel from x1=0 to y (corresponding to x2) for any non-zero value of x (x2) then back to the x axis at x3= 2x1 then back to x1 wouldn't the outer perimeter of my trip form an equilateral triangle?" Afraid not, the lines formed between x1, y, and x3 would each be stepped not straight and equal to (Δx+Δx√2) not 2Δx as in a true equilateral triangle. (Δx in this case is the distance from x1 to x2)
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How do you form an equilateral triangle using taxicab geometry? - Answers
I may be wrong but I do not believe you can because the metric in the vertical or horizontal directions is different from that in the other ("slanted") directions.Answer:Taxicab geometry relies entirely on movements in the x and y directions (like a taxi cab on city streets which form squares). As a consequence the "shortest" distance between two points is the sum of the x and y components (Pythagorus doesn't work)An equilateral triangle requires all angles to be 60o with no 90o angles and all sides of equal length.But then you might ask "If I have two points on the grid, an x value and a y value where y= x√2 and I travel from x1=0 to y (corresponding to x2) for any non-zero value of x (x2) then back to the x axis at x3= 2x1 then back to x1 wouldn't the outer perimeter of my trip form an equilateral triangle?" Afraid not, the lines formed between x1, y, and x3 would each be stepped not straight and equal to (Δx+Δx√2) not 2Δx as in a true equilateral triangle. (Δx in this case is the distance from x1 to x2)
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- og:descriptionI may be wrong but I do not believe you can because the metric in the vertical or horizontal directions is different from that in the other ("slanted") directions.Answer:Taxicab geometry relies entirely on movements in the x and y directions (like a taxi cab on city streets which form squares). As a consequence the "shortest" distance between two points is the sum of the x and y components (Pythagorus doesn't work)An equilateral triangle requires all angles to be 60o with no 90o angles and all sides of equal length.But then you might ask "If I have two points on the grid, an x value and a y value where y= x√2 and I travel from x1=0 to y (corresponding to x2) for any non-zero value of x (x2) then back to the x axis at x3= 2x1 then back to x1 wouldn't the outer perimeter of my trip form an equilateral triangle?" Afraid not, the lines formed between x1, y, and x3 would each be stepped not straight and equal to (Δx+Δx√2) not 2Δx as in a true equilateral triangle. (Δx in this case is the distance from x1 to x2)
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