Peaucelleir's Linkage

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

1. Peaucellier's linkage.

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2. A script to construct the inverse of a point on the circle.

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3.The Definintion of Inversion (AF)(AF')=r^2

  1. The inverse of a point inside a circle is a point outside of a circle.
  2. The inverse of a point outside the circle is a point inside the circle.
  3. A point on the circle is it's own inverse.
  4. The inverse of the center of the circle is infinity.

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4. Table 2C

Shape inverted Inverse of shape
Circle through A Line
Circle not through A Circle
Line through A Line through A
Line not through A A Circle

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5. Constructing the inverse of a circle

  1. Construct Circle 1 (c1) with A as it's center.
  2. Construct Circle 2 (c2) with center B, and Radius AB.
  3. Construct a Point P on c2.
  4. Construct Ray AP.
  5. Label Point R at the intersection of RayAP and c1
  6. Construct Segment AP
  7. Choose a point on c1, label it T. Make segment AT.
  8. Select AP, and then AT, and mark the ratio.
  9. Choose A as your Center of Dilation, and dilate P by the marked ratio above.
  10. Label this point P'.
  11. Trace P'
  12. Move P around c2 to get the inverse of c2.

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6. The inverse of a line not through the center of the circle of inversion.

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  1. Construct c1 with A as center.
  2. Construct line j, not intersecting c1.
  3. Choose P on j, to be inverted.
  4. Construct Ray AP
  5. Mark R as the intersection of AP and c1.
  6. Construct Segment AP
  7. Make a radius segment from A to a point on c1.
  8. Select radius segment, hold down Shift key and select Segment AP. Mark ratio.
  9. Select A to be your Center for Dilation.
  10. Dilate R by marked ratio.
  11. Name dilated point P', trace this point.
  12. Move P along the line to get the inverse of line j.

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7. P and Q are related by inversion.

The definition of inversion through circle of radius r centered at A: The inverse F' of a point F is such that A,F, and F' are collinear, and (AF)(AF')=r^2. Using the sketch above, a line can be drawn through T,P, and Q. (They are collinear.) Calculations show that (TP)(TQ)=(TU)^2. Thus, P is the inverse of Q, TU is the radius, and T is the center of the circle of inversion.

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8. Peaucellier's Linkage with the circle of inversion.

The radius of the circle can be found when P lies on top of Q. We know that at that point, TP=TQ=the radius of the circle. When constructing this line,notice the right triangle with leg m and hyp k. Using the Pythagorean theorm, r^2=k^2-m^2, so r=The square root of (k^2-m^2)

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9. A sketch of the inverse of the circular path of P.

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