积极 AI Logo 积极 AI

e与π在24边形星罗中的展示

e与π在24边形星罗中的展示
几何画板及绘图内容加载中，请稍等……
2022-04-26 04:31
圆函数三角函数解析几何
这个范例展示了e和π在24边形星罗中的应用。它涉及指数函数、圆周率和旋转等核心数学概念。通过使用GeoGebra命令如Point、Intersect、Circle、Sequence等，可以构造出这些数学概念的具体图形。适合高中数学教学和学习。

GeoGebra 命令及参数

  1. Point(yAxis) → A
  2. Intersect(xAxis, yAxis) → B
  3. Circle(A, B) → c
  4. Sequence((k, 0), k, 1, 4) → P, {(1, 0), (2, 0), (3, 0), (4, 0)}
  5. Sequence(x = k, k, 1, 4) → L, {x = 1, x = 2, x = 3, x = 4}
  6. Sequence(Circle[((2 * cos((k * π / 12))), 2 + (2 * sin((k * π / 12)))), 2], k, 0, 23) → K, {(x - 2)² + (y - 2)² = 4, (x - 1.9318516525781366)² + (y - 2.5176380902050415)² = 4, (x - 1.7320508075688774)² + (y - 3)² = 4, (x - 1.4142135623730951)² + (y - 3.414213562373095)² = 4, (x - 1.0000000000000002)² + (y - 3.732050807568877)² = 4, (x - 0.5176380902050419)² + (y - 3.9318516525781364)² = 4, x² + (y - 4)² = 4, (x + 0.5176380902050413)² + (y - 3.9318516525781364)² = 4, (x + 0.9999999999999996)² + (y - 3.7320508075688776)² = 4, (x + 1.414213562373095)² + (y - 3.414213562373095)² = 4, (x + 1.732050807568877)² + (y - 3.000000000000001)² = 4, (x + 1.9318516525781364)² + (y - 2.517638090205042)² = 4, (x + 2)² + (y - 2)² = 4, (x + 1.9318516525781368)² + (y - 1.4823619097949594)² = 4, (x + 1.7320508075688776)² + (y - 1.0000000000000004)² = 4, (x + 1.4142135623730958)² + (y - 0.5857864376269057)² = 4, (x + 1.0000000000000009)² + (y - 0.26794919243112303)² = 4, (x + 0.517638090205043)² + (y - 0.06814834742186382)² = 4, x² + y² = 4, (x - 0.5176380902050406)² + (y - 0.06814834742186315)² = 4, (x - 0.9999999999999987)² + (y - 0.2679491924311219)² = 4, (x - 1.4142135623730947)² + (y - 0.5857864376269046)² = 4, (x - 1.7320508075688767)² + (y - 0.9999999999999991)² = 4, (x - 1.9318516525781362)² + (y - 1.4823619097949567)² = 4}
  7. Sequence((tan((k * π / 12)) * x) + 2, k, 0, 23) → St, {(tan((0 * π / 12)) * x) + 2, (tan((1 * π / 12)) * x) + 2, (tan((2 * π / 12)) * x) + 2, (tan((3 * π / 12)) * x) + 2, (tan((4 * π / 12)) * x) + 2, (tan((5 * π / 12)) * x) + 2, (tan((6 * π / 12)) * x) + 2, (tan((7 * π / 12)) * x) + 2, (tan((8 * π / 12)) * x) + 2, (tan((9 * π / 12)) * x) + 2, (tan((10 * π / 12)) * x) + 2, (tan((11 * π / 12)) * x) + 2, (tan((12 * π / 12)) * x) + 2, (tan((13 * π / 12)) * x) + 2, (tan((14 * π / 12)) * x) + 2, (tan((15 * π / 12)) * x) + 2, (tan((16 * π / 12)) * x) + 2, (tan((17 * π / 12)) * x) + 2, (tan((18 * π / 12)) * x) + 2, (tan((19 * π / 12)) * x) + 2, (tan((20 * π / 12)) * x) + 2, (tan((21 * π / 12)) * x) + 2, (tan((22 * π / 12)) * x) + 2, (tan((23 * π / 12)) * x) + 2}
  8. Line(B, xAxis) → f
  9. OrthogonalLine(B, f) → g
  10. Sequence(Text[k, P(k) + (-0.10, -0.6)], k, 1, 4) → T, {"1", "2", "3", "4"}
  11. Intersect(St(2, x), gl1) → S1
  12. Intersect(St(4, x), gl1) → S2
  13. Intersect(St(24, x), gl1) → S3
  14. Intersect(St(22, x), gl1) → S4
  15. Intersect(K(3), gl2) → T1, U1
  16. Intersect(K(23), gl2) → T2, U2

AI 绘图提示语参考

绘制一个24边星形玫瑰花，使用e和π作为关键参数。首先画出x轴和y轴，然后画一个圆心在y轴上的圆。接着在x轴上标出四个点，分别代表1到4。利用这些点和π/12的倍数，构造24个圆，每个圆的半径为2，中心点在x轴上以π/12为间隔旋转。然后画出24条通过原点的直线，斜率分别为tan(kπ/12)，k从0到23。接着画出x轴和y轴的交点，以及与x轴垂直的直线。在每个圆与垂直直线的交点处标出数字1到4。最后，找到特定直线与垂直直线的交点，连接这些点形成星形玫瑰花。
以上提示语为根据原始绘图命令整理而来，仅供参考。