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# Creating Conway's circle fails?

Posted 1 year ago
 Hello everyone, I have a triangle with vertices (0,0), (5,1), and (4,7). At this moment, I am trying to find "Measure" on these lines the other segments, which need to calculate for this the lengths of sides and identify those points as being on the line and the added length to be as requested. Draw the triangle with the extended lengths. Here is my code. Clear[v1, v2, v3, l1, l2, l3, y, x, b, a] {v1, v2, v3} = {{0, 0}, {5, 1}, {4, 7}}; a = Graphics[{ {Blue, PointSize[Large], Point[{v1, v2, v3}]}, {EdgeForm[Gray], LightBlue, Triangle[{v1, v2, v3}]}, {Purple, PointSize[Large]}}, Axes -> True]; sv2v3 = -6 x + 31; sv1v2 = 0.2 x; sv1v3 = 1.75 x; b = Plot[{sv2v3, sv1v2, sv1v3}, {x, -10, 20}, PlotRange -> {-10, 20}, PlotLegends -> "Expressions"]; showt = Show[a, b]  Then I tried to find the length for this triangle so that I can create the Conways's circle v1v2 = EuclideanDistance[{v1}, {v2}]; dv2v3 = EuclideanDistance[{v2}, {v3}]; dv1v3 = EuclideanDistance[{v1}, {v3}]; pts1 = {x, y} /. Solve[EuclideanDistance[{x, y}, {5, 1}] == dv1v3 && y == -6 x + 31 && x > 5 && y <= 0, {x, y}, Reals] // N pts2 = {x, y} /. Solve[EuclideanDistance[{x, y}, {5, 1}] == dv1v2 && y == -6 x + 31 && x >= 0 && y > 0, {x, y}, Reals] // N pts3 = {x, y} /. Solve[EuclideanDistance[{x, y}, {4, 7}] == dv1v2 && y == 1.75 x && x >= 4 && y >= 7, {x, y}, Reals] // N pts4 = {x, y} /. Solve[EuclideanDistance[{x, y}, {0, 0}] == dv1v2 && y == 1.75 x && x <= 0 && y <= 0, {x, y}, Reals] // N pts5 = {x, y} /. Solve[EuclideanDistance[{x, y}, {0, 0}] == dv2v3 && y == 0.2 x && x <= 0 && y <= 0, {x, y}, Reals] // N pts6 = {x, y} /. Solve[EuclideanDistance[{x, y}, {0, 0}] == dv1v3 && y == 0.2 x && x > 5 && y > 1, {x, y}, Reals] // N  Not sure if I did something wrong. Please help me. Thank you.
 Hi chiao-yinOne problem is that dv1v2 is not computed so any Solve that uses dv1v2 fails.