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Hyperbolic Octahedron

Hyperbolic Octahedron
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2023-03-15 20:55
Coordinate GeometrySolid GeometryGeometric Construction
In collections geometriadinamica.es
This example demonstrates the construction of a hyperbolic octahedron. It involves key concepts in hyperbolic geometry, such as hyperbolic polygons. Using GeoGebra commands like Segment, Point, OrthogonalLine, etc., the hyperbolic octahedron's faces can be constructed. Suitable for high school or university geometry courses, particularly in solid and hyperbolic geometry.

GeoGebra Commands & Parameters

  1. Segment(Q3, Q4) → p2
  2. Point(p2) → P3
  3. Segment(P3, Q4) → p4
  4. Point(p4) → P4
  5. Segment(Q1, Q2) → p1
  6. Point(p1) → P1
  7. Segment(P1, Q2) → p3
  8. Point(p3) → P2
  9. OrthogonalLine(C, xAxis) → zz2
  10. Circle(C, 0.62) → zz0
  11. Intersect(zz0, zz2, 1) → ZZ4
  12. Intersect(zz0, zz2, 2) → ZZ3
  13. Ellipse(ZZ3, ZZ4, 0.65) → zElipse2
  14. Point(zElipse2) → Z
  15. Intersect(zElipse2, zz2) → ZZ7, ZZ8
  16. Angle(ZZ8, C, Z) → zAngInc
  17. OrthogonalLine(C, yAxis) → zz1
  18. Intersect(zz0, zz1, 2) → ZZ2
  19. Intersect(zz0, zz1, 1) → ZZ1
  20. Ellipse(ZZ1, ZZ2, 0.65) → zElipse1
  21. Point(zElipse1) → XY
  22. Intersect(zElipse1, zz1) → ZZ5, ZZ6
  23. Angle(ZZ6, C, XY) → zAngRot
  24. Segment(Q5, Q6) → p5
  25. Point(p5) → P5
  26. Segment(ZZ5, ZZ6) → zz4
  27. Segment(ZZ7, ZZ8) → zz3
  28. Intersect(zElipse2, zElipse1, 1) → zPunto1
  29. Intersect(zElipse2, zElipse1, 4) → zPunto2
  30. Segment(zPunto1, zPunto2) → zSeg2
  31. CurveCartesian(((1.5 * r) * ((((cos(u) * cos(α)))^(3) * sin(zAngRot)) + (((sin(u) * cos(α)))^(3) * cos(zAngRot)))), ((1.5 * r) * ((((-((cos(u) * cos(α)))^(3)) * cos(zAngRot)) * sin(zAngInc)) + ((((sin(u) * cos(α)))^(3) * sin(zAngRot)) * sin(zAngInc)) + ((sin(α))^(3) * cos(zAngInc)))), u, u1, u2) → curva2
  32. CurveCartesian((((1.5 * r) * (cos(v))^(3)) * sin(zAngRot)), ((1.5 * r) * ((((-(cos(v))^(3)) * cos(zAngRot)) * sin(zAngInc)) + ((sin(v))^(3) * cos(zAngInc)))), v, v1, v2) → curva1
  33. CurveCartesian(((1.5 * r) * ((((cos(β) * cos(v)))^(3) * sin(zAngRot)) + (((sin(β) * cos(v)))^(3) * cos(zAngRot)))), ((1.5 * r) * ((((-((cos(β) * cos(v)))^(3)) * cos(zAngRot)) * sin(zAngInc)) + ((((sin(β) * cos(v)))^(3) * sin(zAngRot)) * sin(zAngInc)) + ((sin(v))^(3) * cos(zAngInc)))), v, v1, v2) → a1
  34. Sequence(CurveCartesian[((1.5 * r) * ((((cos(u) * cos(v)))^(3) * sin(zAngRot)) + (((sin(u) * cos(v)))^(3) * cos(zAngRot)))), ((1.5 * r) * ((((-((cos(u) * cos(v)))^(3)) * cos(zAngRot)) * sin(zAngInc)) + ((((sin(u) * cos(v)))^(3) * sin(zAngRot)) * sin(zAngInc)) + ((sin(v))^(3) * cos(zAngInc)))), v, v1, v2], u, u1, u2, (2 * paso)) → list2
  35. Sequence(CurveCartesian[((1.5 * r) * ((((cos(u) * cos(v)))^(3) * sin(zAngRot)) + (((sin(u) * cos(v)))^(3) * cos(zAngRot)))), ((1.5 * r) * ((((-((cos(u) * cos(v)))^(3)) * cos(zAngRot)) * sin(zAngInc)) + ((((sin(u) * cos(v)))^(3) * sin(zAngRot)) * sin(zAngInc)) + ((sin(v))^(3) * cos(zAngInc)))), u, u1, u2], v, v1, v2, paso) → list1

AI Drawing Prompt Reference

Draw a hyperbolic octahedron, showcasing its faces made of hyperbolic polygons. Start by drawing two line segments and creating points on them. Then, use orthogonal lines to construct circles and ellipses from the points. Intersect the ellipses with the lines to find new points, and draw segments and curves connecting them to form the intricate structure of the hyperbolic octahedron, highlighting key concepts in hyperbolic geometry.
The prompt above was compiled from the original drawing commands and is for reference only.