Peaucellier's linkage questions
Correctly constructed sketches and scripts are available by clicking on the links in quotes or the diagrams shown. If you have trouble linking to Sketchpad, then go to [GSP Help].
- Sketch for Peaucellier's linkage: "link1.gsp"
Script for Peaucellier's linkage: "link1.gss"
- Script for inversion of a point: "link2.gss"
- Constructed inverse of a circle through the center of the
circle of inversion (the center of the circle of inversion was called A in the lab): "link3.gsp"
- Constructed inverse of a line not through the center of
the circle of inversion (the center of the circle of inversion was called A in the lab): "link4.gsp"
- Constructed circle of inversion for the linkage: "link5.gsp"
- Constructed shape traced by the point Q in the picture of
the linkage (this is the same picture as in the lab): "link6.gsp"
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For the questions, let A be the center of inversion, F be the point to invert, and F' be the inverse of point F. Then the definition of inversion tells that (AF)(AF') = r^2, where r is the radius of the circle of inversion.
- Where is the inverse of a point inside the circle?
The inverse is outside the circle of inversion.
If F is inside the circle of inversion, then AF < r. So AF' > r by the definition of inversion. Thus F' is on ray AF and outside of the circle of inversion.
- Where is the inverse of a point outside the circle?
The inverse is inside the circle of inversion.
If F is outside the circle of inversion, then AF > r. So AF' < r by the definition of inversion. Thus F' is on ray AF and inside of the circle of inversion.
- Are there points that are their own inverses?
All points on the circle of inversion are their own inverses.
For any point F on the circle of inversion, AF = r. So AF' = r by the definition of inversion. Since F' is on ray AF, point F' is identical to point F.
- Where is the inverse of the center of the circle?
The inverse is "at infinity".
As a point F (inside the circle of inversion) gets closer to A, AF approaches 0. By the definition, as AF approaches 0, AF' get larger - "approaches infinity".
Since the inverse such a circle is a line, invert two points on circle c (using link2.gss) and construct a line through those two image points.
We discovered that the inverse of such a line is a circle through
the center of inversion. We know from basic geometry that 3 points
uniquely determine a circle. So all we need to do is invert two points
on the line (using link2.gss) and construct a
circle through those two image points and the center of inversion.
How do we construct a circle through these 3 points? Where is its center? Go back to Problem 2 in the Geometer's Sketchpad Homework where you did this.
In Peaucellier's linkage, P and Q can be shown to be related by inversion through a circle with center T.
Verification:
The radius of the circle of inversion is (k^2 - m^2)^(1/2).
Proof of Special Cases: First Case: In the first picture, P and Q lie on top of each other. In this case, triangle TSR is isosceles and since P (and Q) is the midpoint of side RS, TP is an altitude. Thus triangle TPR is a right triangle. We know |TR| = k and |PR| = m. U
sing the Pythagorean theorem, By the definition of inversion, Second Case: In the second picture, R and S lie on top of each other. In this case, points T, P, R, S, and Q are collinear. Thus |TP| = k - m and |TQ| = k + m. By the definition of inversion, Proof of General Case: Using the Pythagorean theorem on right triangle TOR, we find that We have |TP| = |TO| - x and |TQ| = |TO| +
x. Combining these facts with the definition of inversion gives
Construction: A radius of the circle of inversion can be constructed using the Pythagorean theorem and the segments of lengths k and m. First construct two perpendicular lines. Call the point of where the perpendicular lines intersect point K. Construct a circle
centered at K with a radius equal to length m. Mark one point - named L - where this circle intersects one of the perpendicular lines. Now construct a circle centered at L with a radius equal to length k. Mark one point - named M - where this circle i
ntersects the other line. Connect points K, L, and M with segments. This forms a right triangle with sides of lengths m, k, and (k^2 - m^2)^(1/2).
Construct the circle of inversion using T as the center and MK as the length of the radius.
Start with an angle determined by three points A, B, and C, and two
segments AB and BC, shown in the above figure. Construct a circle of
inversion with center not collinear with AB or BC. Invert the points
and segments through this circle. Remember that the line segments will
invert to curved arcs.
The angle between two arcs meeting at a point is the angle formed by
their tangent lines at that point. This is illustrated in the above
figure. Using this, measure the original angle and the new angle, and
compare your answers. Your answer to this problem should reference a
Sketchpad sketch.
Inversion preserves angle measure!
Why do line segments invert to curved arcs? A line segment is part of a line, and the inverse of a line is a circle. Thus the segment is the arc on the circle with endpoints being the inverted endpoints of the segment. Note that the arc must not con
tain the center of inversion. Why? The inversion of the point at the center of inversion is "at infinity", certainly not on the line segment.
Construction: The sketch up to this point is available here.
[GSP Help]
Likewise construct arc B'C' and its tangent through B'. Now you can easily measure the angle between the tangents and angle ABC. When you move points A, B, and C around, the angles always remain equal.
A sample sketch2 up to this point is available here.
[GSP Help]
In this problem you create the first four circles in the picture shown
above. Starting with three tangent circles, two inside of the third,
here is a method of constructing a circle tangent to all three
circles. See the two figures below. Repeating this process gives the
beautiful picture of arbelos, known to the ancient Greeks.
As shown in the picture above, construct two tangent circles, c1 and c2. Label the point of tangency A. To construct circle c3 so that it is tangent to c1 and c2, construct the diameter AE of circle c1. This diameter intersects circle c2 at A and an
other point - call it C. Construct segment CE. Find the midpoint of CE - call it D. Now make a circle centered D with radius DE.
Referring to the labels in the figure, create a circle c5 with center
A, as shown in the figure below. Invert circles c1,c2, and c3 through
circle c5. What do you get? Relate this to the chart you made for the
lab.
The inversions of c1 and c2 are two parallel lines. The inversion of c3 is a circle which the parallel lines are tangent to.
Using the chart we made in Problem 3, we know the inversion of a circle which passes through A (center of inversion) will be lines not passing through A. The chart also states that the inversion of a circle not passing through A w
ill also be a circle not passing through A.
Create circle c6 tangent to the three inverses. Notice that by
inverting, you made it possible to do this construction.
Let c3' represent the inverion of circle c3. For c6 to be tangent to the parallel lines, it must be a translation of c3' in the same direction as the parallel lines. To do this, construct a line parallel to the two parallel lines that passes thr
ough the center of c3'. Select the two points where this line intersects c3'. Go to Transform / Mark Vector. Then select c3' and choose Transform / Translate... and it should translate by the marked vector to create circle c6.
Using your answer from the previous problem justify this statement:
inversion preserves tangency. Using this information, create circle
c4, tangent to c1, c2, and c3. This is the first step in constructing
the arbelos picture. For extra credit, create two more circles in
the arbelos picture.
See Dan Pedoe, Geometry, A Comprehensive Course page 89 for
more information.
When two objects are tangent at a point, their tangent lines at that point form a 0 degree angle. (They the same line.) In the previous problem, we found that inversion preserves angle measure. So inversion must preserve tang
ency.
Construct c4 by inverting c6. Since c6 was tangent to the inversions of c1, c2, and c3, then c4 must be tangent to c1, c2, and c3.
You can use a script to make this process of inverting circles easier. A sample script is available here.
[GSP Help]
Warning: This sample script is difficult to use.
A sketch of a more complete arbelos picture is here.
[GSP Help]
Shape
Inverse of the Shape
Circle through A
Line not through A
Circle not through A
Circle not through A
(distinct circles unless circle of inversion)
Line through A
Same line through A
Line not through Circle through A
By measuring the distance from T to P and the distance from T to Q. Multiplying these distances together gives a constant value, no matter how P and Q are moved around in the linkage. By the definition of inversion, this constant is
the square of the radius of inversion with point T being the center of inversion.
In Part 3B of the lab you were asked to investigate two pictures representing special cases.
Keep in mind that T, P, and Q are collinear. Draw in segments TQ and RS. Let O be the intersection point of segments TQ and RS. Since figure PRST is a rhombus, its diagonals are perpendicular bisectors of each other. Let x = |PO| = |QO|. Using the Pyt
hagorean theorem on right triangle POR, we find that
=
(k^2 - m^2 + x^2) - x^2 = k^2 - m^2.
To construct a segment the length of the radius, construct a right triangle with hypotenuse length k and one leg of length m. The other leg will necessarily be the length of the radius.
General Sketchpad questions on inversion.
It is assumed that you are familiar with Sketchpad. Thus, the
following questions do not give step by step help on the technique of
creating a sketch.
The tricky part of this construction is creating the arcs. Invert points A, B, and C (using link2.gss) and the lines containing the segments (done in Problem 5). Let A', B', and C' be the inversions of the poi
nts. The inversion of segment AB is the arc on the circle with endpoints A' and B'. To actually have only arc A'B' shown in Sketchpad, select the circle and points A' and B' in a COUNTERCLOCKWISE manner. Go to Construct / Arc On Circle. Construct the
tangent to the arc (circle) through B'. Then hide the circle.
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