Simplification.mw

restart;

L:=[]:

 for A  from .1 by .01 to 3 do

  for B  from .1 by .01 to 3 do

  l1:=(exp(B)-A)-1:

  l2:=(exp(B)-A/2)-1.5:

  if l1>0 and l2>0 then L:=[op(L), [A,B]]: break: end if:

  end do:

 end do:

L; 

In total there are 192 solutions:

with(combinat):

S:=permute(9): L:=[]:

for a in S do

if  a[9] = add(a[2*k-1]/a[2*k], k= 1..4) then L:=[op(L), a]: fi:

od:

nops(L);

                                               192

The first solution in the list  L  is Carl's solution, the last is

It is interesting that  there is no solution in which all the fractions are irreducible.

There are several solutions.

solve({c2+vc = .34, 

c2+c3 = .519,

t8+c3-c4-vc = .132,

t1+t5-c2+0.0435=0., 

t1-t2-c2+c3-1.334*vc = 0.,

t2+t6-c3-c4+0.0435+1.334*vc=0.,

-112.5*t1+112.5*t5+300*c2^2+1000*vc = 38.25, 

112.5*t1-112.5*t2-300*c2^2+300*c3-300*c3^2-2000*vc+23.4 = 0.,

-112.8*t8+112.5*t2-112.5*t6+300*c4^2-300*c3+300*c3^2+1000*vc+14.85 = 0.}, 

{t1,t2,t5,t6,t8,c2,c3,c4,vc});

{c2 = .3099822380, c3 = .2090177620, c4 = .2431877898e-1, t1 = .2247720450, t2 = .8376387448e-1, t5 = .4171019302e-1, t6 = .6602897199e-1, t8 = -.2268122102e-1, vc = .3001776198e-1}, {c2 = 3.167226095, c3 = -2.648226095, c4 = .2431877898e-1, t1 = 2.201508587, t2 = .1575760075, t5 = .9222175084, t6 = .9465362873, t8 = -.2268122102e-1, vc = -2.827226095}, {c2 = .3099822380, c3 = .2090177620, c4 = .7266812210, t1 = .2247720450, t2 = .8376387448e-1, t5 = .4171019302e-1, t6 = .7683914140, t8 = .6796812210, vc = .3001776198e-1}, {c2 = 3.167226095, c3 = -2.648226095, c4 = .7266812210, t1 = 2.201508587, t2 = .1575760075, t5 = .9222175084, t6 = 1.648898729, t8 = .6796812210, vc = -2.827226095}

The original code has been executed in Maple 12 (classical worksheet). Found mistake in Dist2 (bracket in the wrong place).
  
Here's another variant of the code:

M1 := [-2, 3, -4]: M2 := [4, -1, 3]:

A := [-t+3, 2*t-4, t+2]:

B := [s+4, 2*s, 2*s-1]:

Dist1 := sqrt((M1[1]-A[1])^2+(M1[2]-A[2])^2+(M1[3]-A[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2+(B[3]-A[3])^2)+sqrt((B[1]-M2[1])^2+(B[2]-M2[2])^2+(B[3]-M2[3])^2):

Dist2 := sqrt((M1[1]-B[1])^2+(M1[2]-B[2])^2+(M1[3]-B[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2+(B[3]-A[3])^2)+sqrt((A[1]-M2[1])^2+(A[2]-M2[2])^2+(A[3]-M2[3])^2):

Optimization[Minimize](Dist1); Optimization[Minimize](Dist2);

min(%[1], %%[1]);

[17.8369577415605676,

[s = 0.602330835119798569, t = 1.52243814242405584]]

[14.4132655077310190,

[s = -0.189757117490280680, t = 0.845165089700380400]]

14.4132655077310190

This is quite simply:

M1:=[-2,3,-4]:  M2:=[4,-1,3]:

A:=[-t+3,2*t-4,t+2]:  B:=[s+4,2*s,2*s-1]:

Dist1:=sqrt((M1[1]-A[1])^2+(M1[2]-A[2])^2+(M1[3]-A[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2+(B[3]-A[3])^2)+sqrt((B[1]-M2[1])^2+(B[2]-M2[2])^2+(B[3]-M2[3])^2):

Dist2:=sqrt((M1[1]-B[1])^2+(M1[2]-B[2])^2+(M1[3]-B[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2)+(B[3]-A[3])^2+sqrt((A[1]-M2[1])^2+(A[2]-M2[2])^2+(A[3]-M2[3])^2):

evalf(minimize(Dist1));  evalf(minimize(Dist2));

min(%, %%); 

                                                                         21.59170144

                                                                         16.24881940

                                                                         16.24881940

1) If you consider the function y=psi(x)  in the real range, then you should write surd(1-x, 3)  instead  (1-x)^(1/3). Else for  (1-x)^(1/3)  if  x>1 , you get complex values.

2) By the code  q:=y->fsolve(y=psi(x), x)  you are looking the function inverse to the function y=psi(x) . For correct solution of the problem the function y=psi(x) should be monotonic in the corresponding range. Such ranges simpliest can be found using the plot of the function. Function  y=psi(x)  in the entire domain does not have an inverse function.

V:=1.8e-5:  R:=8.314:  T:=298:  k:=0.841:  G:=1e6:

psi:=unapply((ln(x)-x+k*(1-x)^2+(V*G)/(R*T)*(surd(1-x,3)-(1-x)/2))*(R*T/V), x);

plot(psi(x), x=0..3, title="The plot of function y=psi(x)");

q:=y->fsolve(psi(x)=y, x):

plot('q'(y), y=-.9e9..-.1.4e9);

If this is important to you, then you can specify  the corresponding function, called P

P:=unapply(n!/(n-r)!, n,r);
 P(10,6);  P(n,r);

                        n!
P := (n, r) -> --------
                     (n - r)!

151200

   n!
--------
(n - r)!

It is easy to see that the product of the members of which are equidistant from the ends is 4, for example

simplify((sqrt(3)+tan(Pi/180))*(sqrt(3)+tan(29*Pi/180)));
                                           4

So, we obtain

mul(simplify((sqrt(3)+tan(Pi*k/180))*(sqrt(3)+tan((30-k)*Pi/180))), k=1..14)*simplify(sqrt(3)+tan(15*Pi/180));
                                     536870912

z:=(x,y)->5/(1+x^2+y^2):

A:=plot3d(z(x,y), x=-4..7, y=-4..4, numpoints=3000):

x:=t->2+3*cos(t): y:=t->-1+2*sin(t):

N:=100:

B:=seq(plots[spacecurve]([x(t),y(t),z(x(t), y(t))], t=0..2*Pi*k/N, color=black, thickness=3), k=0..N):

bug:=seq(plottools[sphere]([x(2*Pi*k/N),y(2*Pi*k/N),z(x(2*Pi*k/N), y(2*Pi*k/N))], 0.15, style=surface, color=red),  k=0..N):

C:=seq(plots[display](A, B[k], bug[k]), k=1..N+1):

plots[display](C, insequence=true, axes=normal, view=[-4.7..7.7, -4.7..4.7, -1..5.7]);

Your task is similar to the problem of wandering drunk sailor. Do a keyword search drunken sailor.

Vector(10,[seq(sin(i), i=-0.5..-0.1, 0.1), seq(sin(i), i=0.1..0.5, 0.1)]);

or

Vector(10, [seq(sin(i), i in {seq(-0.5..0.5, 0.1)} minus {0.})]);

subs(a=`3`, sqrt(a)/a);

To find the center of rotation is sufficient to use two points:

restart; with(geometry):

point(X, x0, y0), point(A, 4, -2), point(B, 5, -4):   #  X - the center of rotation

L:=coordinates(rotation(A1, A, alpha, 'counterclockwise', X)):

M:=coordinates(rotation(B1, B, alpha, 'counterclockwise', X)):

solve({L[1]=4, L[2]=2, M[1]=6, M[2]=3});

You can use  seq  command.

Example:

X:=[seq(i, i=0..20)]:
Y:=[seq((-1)^i*x, i=0..19)]:
y:=piecewise(seq(op([X[i]<x and x<=X[i+1], Y[i]]), i=1..20)):
plot(y, x=0..20);