vv

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@acer  Nice. Unfortunately the blue point cannot be dragged over the boundary, so, the undefined value cannot be tested.

@mmcdara Here is a test using Explore. For older versions of Maple, use the previous Wind.

restart;
Wind:=proc(L::listlist, P::list)
  local p,u,k, x0:=P[1],y0:=P[2], t;
  if member(P,L) then return undefined fi;
  u:=seq(arctan(p[2]-y0,p[1]-x0),p=L); if L[1]<>L[-1] then u:=u,u[1] fi;
  add(piecewise(abs((t:=u[k+1]-u[k]))=Pi,undefined, t<-Pi,+1, t>Pi,-1, 0), k=1..nops([u])-1)
end:
K:=[[10,10], [10,90], [70,90], [70,50], [30,50], [30,30], [60,30], [60,80], [20,80], [20,70], [40,70], [40,40], [20,40], [20,60], [50,60], [50,10], [10,10]]:
T:=proc(x,y)  local w:=Wind(K, [x,y]);
  uses plots;
  display(plottools:-polygon(K),color=yellow,linestyle=solid,thickness=2, 
    pointplot(K, color=red, symbolsize=16, symbol=solidcircle),
    pointplot([x,y], color=blue, symbolsize=16, symbol=solidbox),
    view=[0..80, 0..100], title=typeset([x,y],'wind'=w));
end:
Explore(T(x,y), x=0..80, y=0..100);

@Rouben Rostamian  

alpha_max / Pi  is one of the real roots of the polynomial

9*t^6 - 54*t^5 + 48*t^4 + 168*t^3 - 336*t^2 + 192*t - 32

> 

restart;

> 

R1:=alpha/(2*Pi): h1:=sqrt(1-R1^2): V1:=Pi*R1^2*h1/3:

> 

V := V1 + eval(V1, alpha=2*Pi-alpha);

(1/24)*alpha^2*(-alpha^2/Pi^2+4)^(1/2)/Pi+(1/24)*(2*Pi-alpha)^2*(-(2*Pi-alpha)^2/Pi^2+4)^(1/2)/Pi

(1)
> 

#plot(V, alpha=0..Pi);

> 

plot(V, alpha=Pi/4..Pi);

 
> 

sol:=maximize(V, alpha=0..Pi, location):

> 

evalf(sol)

.4566405910, {[{alpha = 2.035839071}, .4566405910]}

(2)
> 

alpha__max:=eval(alpha, sol[2][][1]);

Pi-(1/3)*Pi*(29-4*43^(1/2)*cos((1/3)*arctan((9/1121)*191^(1/2)))-4*3^(1/2)*43^(1/2)*sin((1/3)*arctan((9/1121)*191^(1/2))))^(1/2)

(3)
> 

evalf(alpha__max*180/Pi); # degrees

116.6449865

(4)
> 

 

 

@nm A much larger maxslope:=100000:  is probably acceptable in most non-pathological cases.

@mmcdara The free and generating families are useful notions. Usually it is simpler to check these two, instead of a single condition. E.g., if a family F is free in a vector space E, then F is a basis for the vector subspace generated by F in E.

@Christopher2222  The strange fact is that evalb and select do not confirm numboccur.
numboccur(L, 1.) = nops(select(`=`, L, 1.));  # 2 = 3

@Carl Love Yes, of course.

Q:=eval([x,y], solve([x+2*y-1, 2*(x-6) - (y+3)])): # projection onto directrix
F:=2*[6,-3]-Q:
P:=(x-F[1])^2 + (y-F[2])^2 = (x+2*y-1)^2 / 5:
with(plots): display(implicitplot([P, x+2*y-1], x=-10..10,y=-10..10), pointplot([[6,-3], F], color=[red,blue]));

@JAMET You have already an answer. What did you not understand?

@mmcdara The unicity (uniqueness) of the linear combination makes the family free. So, the definitions are the same.

@Carl Love Vote up (especially for your comeback)!

@tomleslie The .eps is correct (with a few artifacts though). ghostscript converts it properly into a vector graphics pdf.
Maybe GIMP has an option to do that too.

@tomleslie Your file test2.pdf obviously contains a low resolution bitmap. Have you zoomed into it?

Both your graphics are bitmaps (not vectorized).
To obtain true vector graphics, convert the eps to pdf using e.g. ghostscript or latex.
Compare the results by zooming in:

from-eps.pdf

 

@Anthrazit Unlike sin or arctan, arctand does not check the (number of) arguments. You may use:

arctandeg := () -> 180/Pi*arctan(args);

 

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