Solution
The locus of D is an exact straight line segment. This is a classic construction called the Peaucellier-Lipkin linkage.
I found it interesting since a few years earlier, Chebyshev (Lipkin was a student of Chebyshev) found an elegant linkage — Chebyshev's "lambda linkage" — that yields a curve that is close to but not exactly equal to a straight line. Almost a century earlier, James Watt found a linkage that was again approximate, but quite simple. More information on these can be found on Wikipedia:
http://en.wikipedia.org/wiki/Peaucellier-Lipkin_linkage
http://en.wikipedia.org/wiki/Watt's_linkage
http://en.wikipedia.org/wiki/Chebyshev_linkage
There are some demonstrations/animations of these also at:
http://demonstrations.wolfram.com/ChebyshevsLambdaMechanism/
http://demonstrations.wolfram.com/PeaucelliersInversor/
http://demonstrations.wolfram.com/InversiveGeometryIIThePeaucellierInversorMechanism/
And there is also a Russian project on mathematical animations that has some high quality animations. The Peaucellier-Lipkin linkage is here: http://www.etudes.ru/en/etudes/inversor/. And Chebyshev's is at http://en.tcheb.ru/1.
Now, the proof that the locus of the given problem is a straight line is
best done by circular inversion. A proof can be found at the Wikipedia link,
which also has this nice quote about the impression the linkage made on
Kelvin:
Sylvester writes that when he showed a model to Kelvin, he "nursed it as if
it had been his own child, and when a motion was made to relieve him of it,
replied 'No! I have not had nearly enough of it — it is the most beautiful thing I have ever seen in my life.'"
Sandy Norman (Univ. of Texas at San Antonio) worked out a few things. Recall
that one arm had length d and the other d/2. A first question is: Which
straight lines occur as d varies? The answer is that these vertical lines
fill a circle of radius 3 centered at (2, 0), with the exclusion of a circle of radius 1/3 centered at -2/3:
Now the given values have a ratio of 1/2 in the two arms. The linkage works the same way for other ratios. Sandy worked out that if the ratio of the smaller arm to the larger is q, then the radius of the disk representing the locus of segments is simply (1 + q)/(1 − q). When q is 1/2, this is 3. And the excluded disk, with leftmost point at (-1, 0), has radius (1 − q)/(1 + q).
I received solutions to the stated problem from Sandy, Piotr Zielinksi, YaJie Zhang, and Joseph DeVincentis. The most elegant way to solve this would be to use circular inversion, the details of which are easily found at the given link. We finish by including the solution found by DeVincentis.
Since 2/3 < d/2 < 2, and OB is limited to OA − AB = d/2 at the minimum and OA + AB = 3d/2 at the maximum, it is always possible for B to reach the point (1, 0) where it is shown in the diagram (although for small d, this makes B the outer point on the rhombus, and D is inside the unit circle, at least for part of the locus). But for any d, B is not able to go all the way around the circle; it is restricted to the parts of the circle which are at least d/2 away from O. When OB = d/2, D is forced to its farthest possible position from O, at a distance 3d/2 from O.
The diagonals of the rhombus are perpendicular (call their intersection E), and O lies on BD due to symmetry.
Given the length of OB = b, since OB is a chord on the unit circle, the angle between OB and the X axis can be calculated as arccos(b/2). Since OBD are collinear, this angle also partly determines the position of D. It would
be more typical to vary the angle and solve for b, but since we are actually given limits on b, it makes sense to vary b and solve for the angle in this case.
If we take the distance OB = b as given, then we have all the lengths of the sides of the triangle OAB, and we can solve for angle AOB with the law of cosines.
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AB²
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=
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OA² + OB² − 2 OA OB cos AOB
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or
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d²/4
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=
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d² + b² − 2 d b cos AOB
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cos AOB
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=
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((3/4)d² + b²)/(2db)
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Use this to solve for the legs of the right triangle AOE, which has the same angle as AOB.
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OE / OA
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=
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cos AOB
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=
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((3/4)d² + b²)/(2db)
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=
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OE / d
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OE
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=
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((3/4)d² + b²)/(2b)
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=
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(3/8b)d² + b/2
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By symmetry, OE is the average of OB and OD. OB = b, so
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OD
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=
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2(3/8b)d²
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=
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(3/4b)d²
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Now, we know that the cosine of the angle between OD and the horizontal is b/2. Let (x, y) be the coordinates of D. Then
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b/2
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=
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x/((3/4b)d²)
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=
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4bx/(3d²)
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or
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4bx
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=
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3d² b/2
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And x = (3/8)d², a constant, regardless of b.
Its limits occur when b is as small as possible, specifically d/2, which means OD = (3/2d)d² = (3/2)d. Then
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OD²
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=
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x² + y²
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or
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(9/4)d²
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=
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(9/64)d4 + y²
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So,
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y²
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=
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(9/4)d² − (9/64)d4
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=
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(9/4)d² (1 − d²/16)
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So y = (3/2)d √(1 − d²/16).
So the locus of d is a vertical line segment between ...
(0, (3/2)d √(1 − d²/16))
... and ...
(0, −(3/2)d √(1 − d²/16))
[Back to Problem 1194]
© Copyright 2014 Stan Wagon. Reproduced with permission.