A, B, C, D, E, F, G are points in order along the circumference of a circle.
W, X, Y, Z are points in order along chord AG.
ABW, WCX, XDY, YEZ, ZFG are equilateral triangles.
If AB=1, find the side lengths of the other four triangles.
(In reply to
solution(s) by Charlie)
The unique solution is in fact symmetrical. I had neglected to subract the initial x in the linear equation formulae.
clc
A=[0 0]; B=[1/2 sqrt(3)/2];
W=[1 0];
ratio1=3.6855439326708
X=[1+ratio1 0];
C=(W+X)/2 + [0 ratio1*sqrt(3)/2];
Pt1=(A+B)/2; Pt2=(B+C)/2;
m1=(B(1)-A(1))/(A(2)-B(2));
m2=(C(1)-B(1))/(B(2)-C(2));
syms x y
x1=Pt1(1); y1=Pt1(2);
Eq1=y==y1+m1*(x-x1); % corrected line (subtract x1)
x2=Pt2(1); y2=Pt2(2);
Eq2=y==y2+m2*(x-x2); % corrected line (subtract x2)
solve([Eq1 Eq2],[x y]);
ctr=eval([ans.x ans.y]);
ctrx=ctr(1); ctry=ctr(2);
radsq=(C(1)-ctrx)^2 + (C(2)-ctry)^2;
disp([ctrx ctry sqrt(radsq)])
B
C
X
start=X(1);
Eq1=(x-ctrx)^2 + (y-ctry)^2==radsq;
Eq2=x==start+y/sqrt(3);
solve([Eq1 Eq2],[x y]);
D=eval([ans.x ans.y]);
D=D(2,:)
Y=[D(1)*2-start 0]
start=Y(1);
Eq1=(x-ctrx)^2 + (y-ctry)^2==radsq;
Eq2=x==start+y/sqrt(3);
solve([Eq1 Eq2],[x y]);
E=eval([ans.x ans.y]);
E=E(2,:)
Z=[E(1)*2-start 0]
start=Z(1);
Eq1=(x-ctrx)^2 + (y-ctry)^2==radsq;
Eq2=x==start+y/sqrt(3);
solve([Eq1 Eq2],[x y]);
F=eval([ans.x ans.y]);
F=F(2,:)
G=[F(1)*2-start 0]
disp([ratio1 G(1)-2*ctrx])
center of circle
x y radius
ratio1 =
3.6855439326708
7.29161703982328 -3.63246679158001 8.14631782140697
B =
0.5 0.866025403784439
C =
2.8427719663354 3.19177467245652
X =
4.6855439326708 0
D =
7.29161703982327 4.51385102982696
Y =
9.89769014697573 0
E =
11.7404621133111 3.19177467245653
Z =
13.5832340796465 0
F =
14.0832340796465 0.866025403784447
G =
14.5832340796466 0
3.6855439326708 0
>>
3.6855439326708 is the size of each side of WCX and of XDY is 9.89769014697573 - 4.6855439326708 = 5.21214621430493, which is the largest, central, triangle. The remaining two are mirror images of the first two.
The whole chord is 14.5832340796466 units long.
The center of the circle, when the horizontal chord begins at the origin, is at (7.29161703982328, -3.63246679158001) and its radius is 8.14631782140697, so the chord is not quite the diameter.
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Posted by Charlie
on 2021-09-12 12:03:37 |
re: solution(s) --- Correction!
