Kitonum

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Here is a classic puzzle:
You are camping, and have an 8-liter bucket which is full of fresh water. You need to share this water fairly into exactly two portions (4 + 4 liters). But you only have two empty buckets: a 5-liter and a 3-liter. Divide the 8 liters in half in as short a time as possible.

This is not an easy task and requires at least 7 transfusions to solve it. 

The procedure  Pouring  solves a similar problem for the general case. Given n buckets of known volume and the amount of water in each bucket is known. Buckets can be partially filled, be full or be empty (of course the case when all is empty or all is full is excluded). With each individual transfusion, the bucket from which it is poured must be completely free, or the bucket into which it is poured must be completely filled. It is forbidden to pour water anywhere other than the indicated buckets.

Formal parameters of the procedure: BucketVolumes  is a list of bucket volumes,  WaterVolumes  is a list of water volumes in these buckets. The procedure returns all possible states that can occur during transfusions in the form of a tree (the initial state  WaterVolumes  is its root).

restart;
Pouring:=proc(BucketVolumes::list(And(positive,{integer,float,fraction})),WaterVolumes::list(And(nonnegative,{integer,float,fraction})), Output:=graph)
local S, W, n, N, OneStep, j, v, H, G;
uses ListTools, GraphTheory;

n:=nops(BucketVolumes); 
if nops(WaterVolumes)<>n then error "The lists should be the same length" fi;
if n<2 then error "Must have at least 2 buckets" fi;
if not `or`(op(WaterVolumes>~0)) then error "There must be at least one non-empty bucket" fi;
if BucketVolumes=WaterVolumes then error "At least one bucket should not be full" fi;
if not `and`(seq(WaterVolumes[i]<=BucketVolumes[i], i=1..n)) then error "The amount of water in each bucket cannot exceed its volume" fi;
W:=[[WaterVolumes]];

OneStep:=proc(W)
local w, k, i, v, V, k1, v0;
global L;
L:=convert(Flatten(W,1), set);
k1:=0; 
for w in W do
k:=0; v:='v';
for i from 1 to n do
for j from 1 to n do
if i<>j and w[-1][i]<>0 and w[-1][j]<BucketVolumes[j] then k:=k+1; v[k]:=subsop(i=`if`(w[-1][i]<=BucketVolumes[j]-w[-1][j],0,w[-1][i]-(BucketVolumes[j]-w[-1][j])),j=`if`(w[-1][i]<=BucketVolumes[j]-w[-1][j],w[-1][j]+w[-1][i],BucketVolumes[j]),w[-1]); fi;
od; 
od; 
v:=convert(v,list);
if `and`(seq(v0 in L, v0=v)) then k1:=k1+1; V[k1]:=w else 
for v0 in v do  
if not (v0 in L) then k1:=k1+1; V[k1]:=[op(w),v0] fi;
od;
fi;
L:=L union convert(v,set);
od;
convert(V,list);
end proc:

S[0]:={};
for N from 1 do
H[N]:=(OneStep@@N)(W);
S[N]:=L;
if S[N-1]=S[N] then break fi;
od;
if Output=set then return L else
if Output=trails then interface(rtablesize=infinity);
return <H[N-1]> else
G:=Graph(seq(Trail(map(t->t[2..-2],convert~(h,string))),h=H[N-1]));
DrawGraph(G, style=tree, root=convert(WaterVolumes,string)[2..-2], stylesheet = [vertexcolor = "Yellow", vertexfont=[TIMES,20]], size=[800,500])  fi; fi;

end proc: 

Examples of use:

Here is the solution to the original puzzle above. We see that at least 7 transfusions are  
required to get equal volumes (4 + 4) in two buckets

Pouring([8,5,3], [8,0,0]);
           

 

 With an increase in the number of buckets, the number of solutions is extremely 
 increased. Here is the solution to the problem: is it possible to equalize the amount of water (7+7+7+7) in the following example? 

Pouring([14,10,9,9],[14,10,4,0]);
S:=Pouring([14,10,9,9],[14,10,4,0], set);
is([7,7,7,7] in S);
nops(S);
         

 

 

Download Pouring.mw

 

 

Yesterday, I accidentally discovered a nasty bug in a fairly simple example:

restart;
Expr:=a*sin(x)+b*cos(x);
maximize(Expr, x=0..2*Pi);
minimize(Expr, x=0..2*Pi);
                                    

I am sure the correct answers are  sqrt(a^2+b^2)  and  -sqrt(a^2+b^2)  for any real values  a  and  b .  It is easy to prove in many ways. The simplest method does not require any calculations and can be done in the mind. We will consider  Expr  as the scalar product (or the dot product) of two vectors  <a, b>  and  <sin(x), cos(x)>, one of which is a unit vector. Then it is obvious that the maximum of this scalar product is reached if the vectors are codirectional and equals to the length of the first vector, that is, sqrt(a^2+b^2).

Bugs in these commands were noted by users and earlier (see search by keywords bug, maximize, minimize) but unfortunately are still not fixed. 

In this post an interesting geometric problem is solved: for an arbitrary convex polygon, find a straight line that divides both the area and the perimeter in half. The following results on this problem are well known:
1. For any convex polygon there is such a straight line.
2. For any convex polygon into which a circle can be inscribed, in particular for any triangle, the desired line must pass through the center of the inscribed circle.
3. For a triangle, the number of solutions can be 1, 2, or 3.
4. If the polygon has symmetry with respect to a point, then any straight line passing through this point is a solution.

The procedure called  InHalf  (the code below) symbolically solves this problem. The formal parameter of the procedure is the list of coordinates of the vertices of a convex polygon (vertices must be passed opposite or clockwise). The procedure returns all solutions in the form of a list of pairs of points, lying on the perimeter of the polygon, that are the ends of segments that implement the desired dividing.


 

restart;

InHalf:=proc(V::listlist)
local L, n, a, b, M, N, i, j, P, Q, L1, L2, Area, Area1, Area2, Perimeter, Perimeter1, Perimeter2, sol, m, k, Sol;
uses LinearAlgebra, ListTools;
L:=map(convert,[V[],V[1]],rational); n:=nops(L)-1;
a:=<(V[2]-V[1])[1],(V[2]-V[1])[2],0>; b:=<(V[n]-V[1])[1],(V[n]-V[1])[2],0>;
if is(CrossProduct(a,b)[3]<0) then L:=Reverse(L) fi;
M:=[seq([L[i],L[i+1]], i=1..n)]:
N:=0;
for i from 1 to n-1 do
for j from i+1 to n do
P:=map(t->t*(1-s),M[i,1])+map(t->t*s,M[i,2]); Q:=map(s->s*(1-t),M[j,1])+map(s->s*t,M[j,2]);
L1:=[P,L[i+1..j][],Q,P];
L2:=[Q,L[j+1..-1][],L[1..i][],P,Q];
Area:=L->(1/2)*add(L[k, 1]*L[k+1, 2]-L[k, 2]*L[k+1, 1], k = 1 .. nops(L)-1);
Area1:=Area(L1);
Area2:=Area(L2);
Perimeter:=L->add(sqrt((L[k,1]-L[k+1,1])^2+(L[k,2]-L[k+1,2])^2), k=1..nops(L)-2);
Perimeter1:=Perimeter(L1);
Perimeter2:=Perimeter(L2);
sol:=[solve({Area1=Area2,Perimeter1=Perimeter2,s>=0,s<1,t>=0,t<1}, {s,t}, explicit)] assuming real;
if sol<>[] then m:=nops(sol);
for k from 1 to m do
N:=N+1; if nops(sol[k])=2 then Sol[N]:=simplify(eval([P,Q],sol[k])) else Sol[N]:=simplify(eval([P,Q],s=t)) fi;
od; fi;
od; od;
Sol:=convert(Sol, list);
`if`(indets(Sol)={},Sol,op([Sol,t>=0 and t<1]));
end proc:  


Examples of use

# For the Pythagorean triangle with sides 3, 4, 5, we have a unique solution

L:=[[4,3],[4,0],[0,0]]:
P:=InHalf(L);
plots:-display(plot([L[],L[1]], color=green, thickness=3), plot(P,  color=red), scaling=constrained);

[[[8/5-(2/5)*6^(1/2), 6/5-(3/10)*6^(1/2)], [4, (1/2)*6^(1/2)]]]

 

 

# For an isosceles right triangle, there are 3 solutions. We see that all the cuts pass through the center of the inscribed circle

L:=[[0,0],[4,0],[4,4]]:
InHalf(L);
P:=InHalf(L);
r:=(4+4-4*sqrt(2))/2: a:=4-r: b:=r:
plots:-display(plot([L[],L[1]], color=green, thickness=3), plot(P,  color=red), plot([r*cos(t)+a,r*sin(t)+b, t=0..2*Pi], color=blue), scaling=constrained);

[[[2*2^(1/2), 0], [2*2^(1/2), 2*2^(1/2)]], [[4, 0], [2, 2]], [[4, -2*2^(1/2)+4], [-2*2^(1/2)+4, -2*2^(1/2)+4]]]

 

[[[2*2^(1/2), 0], [2*2^(1/2), 2*2^(1/2)]], [[4, 0], [2, 2]], [[4, -2*2^(1/2)+4], [-2*2^(1/2)+4, -2*2^(1/2)+4]]]

 

 

# There are 3 solutions for the quadrilateral below

L:=[[0,0],[4.5,0],[4,3],[0,2]]:
P:=InHalf(L);
plots:-display(plot([L[],L[1]], color=green, thickness=3), plot(P,  color=red), scaling=constrained);

[[[(1/44844)*6^(1/2)*(17^(1/2)-13/2)*37^(3/4)*(2*17^(1/2)+13)^(1/2)*((1836*37^(1/2)-6956)*17^(1/2)+7995*37^(1/2)-56425)^(1/2)-(1/4)*17^(1/2)-(1/8)*37^(1/2)+23/8, 0], [(6^(1/2)*37^(1/4)*((1836*37^(1/2)-6956)*17^(1/2)+7995*37^(1/2)-56425)^(1/2)*(2*17^(1/2)+13)^(1/2)+(-156*37^(1/2)+7770)*17^(1/2)-711*37^(1/2)+50505)/(1776*17^(1/2)+11544), (-6^(1/2)*37^(1/4)*((1836*37^(1/2)-6956)*17^(1/2)+7995*37^(1/2)-56425)^(1/2)*(2*17^(1/2)+13)^(1/2)+(156*37^(1/2)+222)*17^(1/2)+711*37^(1/2)+1443)/(296*17^(1/2)+1924)]], [[(1/90576)*17^(3/4)*(37^(1/2)+37)^(1/2)*(37^(1/2)-37)*((1461*37^(1/2)+29415)*17^(1/2)-986*37^(1/2)-149702)^(1/2)+(5/4)*17^(1/2)+(1/8)*37^(1/2)-27/8, 0], [(17^(1/4)*((1461*37^(1/2)+29415)*17^(1/2)-986*37^(1/2)-149702)^(1/2)*(37^(1/2)+37)^(1/2)+(37*17^(1/2)-51)*37^(1/2)+703*17^(1/2)-1887)/(17*37^(1/2)+629), (17^(1/4)*((1461*37^(1/2)+29415)*17^(1/2)-986*37^(1/2)-149702)^(1/2)*(37^(1/2)+37)^(1/2)+(37*17^(1/2)+85)*37^(1/2)+703*17^(1/2)+3145)/(68*37^(1/2)+2516)]], [[-(1/90576)*17^(3/4)*(37^(1/2)+37)^(1/2)*(37^(1/2)-37)*((1461*37^(1/2)+29415)*17^(1/2)-986*37^(1/2)-149702)^(1/2)+(5/4)*17^(1/2)+(1/8)*37^(1/2)-27/8, 0], [(-17^(1/4)*((1461*37^(1/2)+29415)*17^(1/2)-986*37^(1/2)-149702)^(1/2)*(37^(1/2)+37)^(1/2)+(37*17^(1/2)-51)*37^(1/2)+703*17^(1/2)-1887)/(17*37^(1/2)+629), (-17^(1/4)*((1461*37^(1/2)+29415)*17^(1/2)-986*37^(1/2)-149702)^(1/2)*(37^(1/2)+37)^(1/2)+(37*17^(1/2)+85)*37^(1/2)+703*17^(1/2)+3145)/(68*37^(1/2)+2516)]]]

 

 

# There are infinitely many solutions for a polygon with a center of symmetry. Any cut through the center solves the problem. The picture shows 2 solutions.

L:=[[1,0],[1+2*sqrt(3),2],[2*sqrt(3),sqrt(3)+2],[0,sqrt(3)]]:
P:=InHalf(L);
plots:-display(plot([L[],L[1]], color=green, thickness=3), plot(eval(P[1],t=1/3),  color=red), scaling=constrained);

[[[2*3^(1/2)*t+1, 2*t], [-2*3^(1/2)*(t-1), 3^(1/2)-2*t+2]], [[2*3^(1/2)-t+1, 3^(1/2)*t+2], [t, -3^(1/2)*(t-1)]]], 0 <= t and t < 1

 

 

 


 

Download In_Half.mw

Recently I looked through an interesting book D. Wells "The Penquin book of Curious and Interesting Geometry" and came across this result, which I did not know about before: starting with a given quadrilateral , construct a square on each side. Van Aubel's theorem states that the two line segments between the centers of opposite squares are of equal lengths and are at right angles to one another. See the picture below:

                                  

It is interesting that this is true not only for a convex quadrilateral, but for arbitrary one and even self-intersecting one. This post is devoted to proving this result in Maple. The proof was very short and simple. Note that the coordinates of points are given not by lists, but by vectors. This is convenient when using  LinearAlgebra:-DotProduct  and  LinearAlgebra:-Norm  commands.

The code of the proof (the notation of the points on the picture coincide with their names in the code):

restart;
with(LinearAlgebra):
assign(seq(P||i=<x[i],y[i]>, i=1..4)):
P||5:=P||1:
assign(seq(Q||i=(P||i+P||(i+1))/2+<0,1; -1,0>.(P||(i+1)-P||i)/2, i=1..4)):
expand(DotProduct((Q||3-Q||1),(Q||4-Q||2), conjugate=false));
is(Norm(Q||3-Q||1, 2)=Norm(Q||4-Q||2, 2));

The output:

                                                      0
                                                    true

 

VA.mw

This post is devoted to the rigorous proof of Miquel's five circles theorem, which I learned about from this question. The proof is essentially very simple and takes only 15 lines of code. The figure below, in which all the labels coincide with the corresponding names in the code, illustrates the basic ideas of the code. First, we symbolically define common points of intersection of blue circles with a red unit circle  (these parameters  s1 .. s5  are the polar coordinates of these points). All other parameters of this configuration can be expressed through them. Then we find the centers  M  and  N  of two circles. Then we find the coordinates of the point  K  from the condition that  CK  is perpendicular to  MN . Then we find the point  and using the result obtained, we easily find the coordinates  of all the points  A1 .. A5. Then we find the coordinates of the point   P  as the point of intersection of the lines  A1A2  and  A3A4 . Finally, we verify that the point  P  lies on a circle with center at the point  N , which completes the proof.

                      

 

Below - the code of the proof. Note that the code does not use any special (in particular geometric) packages, only commands from the Maple kernel. I usually try any geometric problems to solve in this style, it is more reliable,  and often shorter.

restart;
t1:=s1/2+s2/2: t2:=s2/2+s3/2:
M:=[cos(t1),sin(t1)]: N:=[cos(t2),sin(t2)]:
C:=[cos(s2),sin(s2)]: K:=(1-t)*~M+t*~N:
CK:=K-C: MN:=M-N:
t0:=simplify(solve(CK[1]*MN[1]+CK[2]*MN[2]=0, t)):
E:=combine(simplify(C+2*eval(CK,t=t0))):
s0:=s5-2*Pi: s6:=s1+2*Pi:
assign(seq(A||i=eval(E,[s2=s||i,s1=s||(i-1),s3=s||(i+1)]), i=1..5)):
Dist:=(p,q)->sqrt((p[1]-q[1])^2+(p[2]-q[2])^2):
LineEq:=(P,Q)->(y-P[2])*(Q[1]-P[1])=(x-P[1])*(Q[2]-P[2]):
Line1:=LineEq(A1,A2):
Line2:=LineEq(A3,A4):
P:=combine(simplify(solve({Line1,Line2},[x,y])))[]:
Circle:=(x-N[1])^2+(y-N[2])^2-Dist(N,C)^2:
is(eval(Circle, P)=0);  
# The final result

                                                                    true


It may seem that this proof is easy to repeat manually. But this is not so. Maple brilliantly coped with very cumbersome trigonometric transformations. Look at the coordinates of point  , expressed through the initial parameters  s1 .. s5 :

simplify(eval([x,y], P));  # The coordinates of the point  P

  

  

 

ProofMiquel.mw

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