Featured Post

We have just released an update to Maple, Maple 2018.2. This release includes improvements in a variety of areas, including code edit regions, Workbooks, and Physics, as well as support for macOS 10.14.

This update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2018.2 download page, where you can also find more details.

For MapleSim users, the update includes optimizations for handling large models, improvements to model import and export, updates to the hydraulics and pneumatics libraries, and more. For more details and download instructions, visit the MapleSim 2018.2 download page.

Featured Post


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.



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)]:
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]);
Area:=L->(1/2)*add(L[k, 1]*L[k+1, 2]-L[k, 2]*L[k+1, 1], k = 1 .. nops(L)-1);
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);
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

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

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

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.

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

writeto command issue

Maple 2018 asked by Christophe... 4805 November 15

Execution output.

Maple asked by Christian... 267 November 15