Collinearity First, it must be proven that points , , are
collinear. This may be easily seen by observing that the linkage is mirror-symmetric about line , so point must fall on that line. More formally, triangles and are congruent because side is congruent to itself, side is congruent to side , and side is congruent to side . Therefore, angles and are equal. Next, triangles and are congruent, since sides and are congruent, side is congruent to itself, and sides and are congruent. Therefore, angles and are equal. Finally, because they form a complete circle, we have : \angle OBA + \angle ABD + \angle DBC + \angle CBO = 360^\circ but, due to the congruences, and , thus :\begin{align} & 2 \times \angle OBA + 2 \times \angle DBA = 360^\circ \\ & \angle OBA + \angle DBA = 180^\circ \end{align} therefore points , , and are collinear.
Inverse points Let point be the intersection of lines and . Then, since is a
rhombus, is the
midpoint of both line segments and . Therefore, length = length . Triangle is congruent to triangle , because side is congruent to side , side is congruent to itself, and side is congruent to side . Therefore, angle = angle . But since , then , , and . Let: :\begin{align} & x = \ell_{BP} = \ell_{PD} \\ & y = \ell_{OB} \\ & h = \ell_{AP} \end{align} Then: :\ell_{OB}\cdot \ell_{OD}=y(y+2x)=y^2+2xy :{\ell_{OA}}^2 = (y + x)^2 + h^2 (due to the
Pythagorean theorem) :{\ell_{OA}}^2 = y^2 + 2xy + x^2 + h^2(same expression expanded) :{\ell_{AD}}^2 = x^2 + h^2 (Pythagorean theorem) :{\ell_{OA}}^2 - {\ell_{AD}}^2 = y^2 + 2xy = \ell_{OB} \cdot \ell_{OD} Since and are both fixed lengths, then the product of and is a constant: :\ell_{OB}\cdot \ell_{OD} = k^2 and since points , , are collinear, then is the inverse of with respect to the circle with center and radius .
Inversive geometry Thus, by the properties of
inversive geometry, since the figure traced by point is the inverse of the figure traced by point , if traces a circle passing through the center of inversion , then is constrained to trace a straight line. But if traces a straight line not passing through , then must trace an arc of a circle passing through .
Q.E.D. A typical driver Peaucellier–Lipkin linkages (PLLs) may have several inversions. A typical example is shown in the opposite figure, in which a rocker-slider four-bar serves as the input driver. To be precise, the slider acts as the input, which in turn drives the right grounded link of the PLL, thus driving the entire PLL.
Historical notes Sylvester (
Collected Works, Vol. 3, Paper 2) 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.’” ==Cultural references==