Polygons with 2 diagonals of fixed length (part two)

Question

In this question of mine

Polygons with two diagonals of fixed length

I've presented the following particular polygon $P$

and I've asked the following question: is it possible to shorten one or more sides or blue diagonals of $P$ in a continuous process in such a way that both the following conditions are satisfied:

• by shortening said sides or blue diagonals of PP no other side or blue diagonal of $P$ gets longer
• by shortening said sides or blue diagonals of $P$ the lengths the two red diagonals remain constant

User Larry B. proved that the answer is "no" using a linear system of inequalities concerning the lengths of sides and diagonals.

Now I want to understand if the same is true for any polygon which "looks as $P$": by that I mean any polygon with 8 vertexes and with 4 convex angles and 4 concave angles in alternating order and such that the two red diagonals stay inside the polygon. Clearly this will make things more complicated than in my previous question..

Answer 1

Edit: Note this answer is invalid. Included here only for reference.

This problem is trivially solved through the Truss approach, as indicated in one of the answers of the original question. This problem is repetitive in systems.

This kind of graphs is nothing more than a certain number of nodes, each one with 2 degrees of freedom (DOF), all of which must be solved.

• The polygon has N=8 vertex, and we are in a 2D space, hence we have 2*N DOF available,

• For limit the traslation, we must set one node as the origin (0,0), fixing 2 DOF. In R dimensions, we fix R DOF by doing this,

• For limit the rotation, we must restrict the rotation of the whole graph. Hence the rotation of the set is only one number, fixing thus 1 DOF. In R dimensions, rotations have R-1 parameters (DOFs).

• Hence, we have 2N-3=13 DOFs remaining. Hence, the topology of the graph - that is, the relation between the nodes - must add the required equations for defining the system.

• There are 8 black bars. Assume their distance fixed and we fix 1 new DOF per each node, fixing 8 DOFs new in total,

• There are 2 red bars, restricting 1 new DOF at 2 nodes, fixing 2 new DOFs in total,

• Now add 3 blue bars, restricting 1 new DOFs at 3 additional nodes not connected by the previous bars, fixing 3 new DOFs in total.

With that, all the 2N DOFs of the system got well constrained, and the quadratic system is well determined. Note that the way we added the constraints is the way we added the equations of the quadratic system, for not adding a degeneracy -i.e. adding all the diagonals in a single node, and letting nodes disconnected.

Thus, we only have to check if the quadratic matrix is diagonalizable.

Geometrically in 2D, it is equivalent to say that each node or shape is connected to a triangular shape which "stiff" them. In N dimensions this is equivalent to say that each node or shape is connected to a N-simplex shape (tetrahedrons).

Fixing 3 blue diagonals at any part of the graph will make the system of equation well-defined. Adding more than 3 diagonals will make the system over-determined, that is, the additional diagonals are linearly dependent constraints from the above.

If we allow the distance to be altered according some function of the metric: i.e. f(x)=kx^2, the quadratic system add an additional term.

Answer 2

In here i am including the assumptions of the problem, please got me corrected if i am failing here?:

• Graph with 8 nodes, 4 outer and 4 inner,
• Red links are fixed,
• Blue and Black links can decrease in magnitude, can not increase,
• Red and Blue links are interior to the shape (that is the only difference between blue and black links?).

Lets add the following less trivial assumptions:

• Each node must have at least 3 links and/or constraints,
• a total of 2N degrees of freedoms must be fixed,
• 2 constraints must be added for fixing the (2D) position,
• 1 constraint must be added for fixing the (2D-1) rotation,
• hence a total of 2N-3=13 constraints shall be included.

Optimization through Calculus of Variations - Karush Kuhn Tucker

Each virtual displacement over any node should not increase any node, respecting all the constraints:

$$ J=\frac{1}{2}\sum_{i=1}^N \sum_{j in I(i)} d_{ij}^2 $$

Where $d_i$ is the distance (strength) of every link measured from the $x_i$ node onto every other node, where the $i(j)$ overloaded notation is the $j$th node connected to $i$, $I$ is the set of all indexes connected to the $i$ node, and $f$ Is the current distance constraint.

$$ F_{ij}=f_{ij}-d_{ij}^2 \ge 0, i=1:N, j \ \text{in} \ I(i) $$

The constraints on the red links are expressed as:

$$ G_{kj}=d_{kj}-g_{kj}=0, (k,k(j)) \ \text{is red} $$

Thus the optimal conditions came from the Karush Kuhn Tucker (https://en.m.wikipedia.org/wiki/Karush–Kuhn–Tucker_conditions) conditions which are no more than the Euler Lagrange multiplier extended for the constrained case.

These conditions seek for the optimal solution of every term on $F$ and every term on $G$, through the solution of the generalized gradient $\nabla$ form expressions - the KKT conditions:

$$ \nabla J_i+ \sum_{ij} \mu_{ij} \nabla F_{ij}+ \sum_{kj} \lambda_{kj} \nabla G_{kj}=0 $$

The $\mu_{ij}\ge 0$ and $\lambda_{kj}$ are the KKT multipliers of the problem. Note that the gradients express the direction of increase of the distance, and we are interested in the opposite of the gradient for minimizing them.

All the previous is only written as an introduction for the KKT, quite standard without the artificial indexes for the nodes included:

$$ \nabla J_j+ \sum_{j} \mu_{j} \nabla F_{j}+ \sum_{j} \lambda_{j} \nabla G_{j} $$

This is simply the sum of each gradient of the strength functionals for each link, included the constraints, when applicable (when there is no red link attached, $\lambda_j$=0).

Also, the gradients of the expressions are nothing more than the directions vectors of the links, toward the increase direction of the length: the line itself.

For the given geometry, it is straightforward to prove that it is always possible to find multipliers for making the satellite expression equal to zero.

For the outer nodes, it is always possible to find positive numbers satisfying the condition. Note the arrangement of the expression for maintaining all multipliers positive. Hence the black links and the red must not be $/pi$ or 0. This holds because the red direction is always between the black directions -inside the polygon:

$$ \mu_{i1}\nabla F_{i1}+\mu_{i2}\nabla F_{i2}=\lambda_i\nabla G_i, i=1:4 $$

For the inner nodes, the expression is similar, though including the fact that the third direction will always be LD with the previous. The arrange in here again take into account the geometry positioning a pair of gradients at the opposite side of the node. Hence again, the angle between the black links $acos(\nabla F_{i1},\nabla F_{i2})$ must not be $\pi$ or 0. It is always possible to find the positive directions, because there is always a node in the polygon which have an angle greater than $pi/2$, with both black rods at the node. If not, then the polygon is flat, and several links passes through the indicated node, which introduces a degeneracy. Thus, for the inner nodes, we can always find three links and positive multipliers such that all directions turns into a zero sum, satisfying again the KKT condition:

$$ \mu_{i1}\nabla F_{i1}+\mu_{i2}\nabla F_{i2}+\mu_{i3}\nabla F_{i3}, i=1:4 $$

This proves that for the given geometry and constraints, it is not possible to minimize the functional of distances, and in particular, not any single distance.