All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
5 equilaterals (Posted on 2021-09-10) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Jer    
Rating: 5.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
solution(s) | Comment 1 of 7
The way to do this, given below, it to build five equilateral triangles along a line segment in such a way that the apices of the triangles form the arc of a circle. The first two apices, together with the start of the first triangles base at the start of the segment, determine the center of the circle and its radius. The rest of the triangles need to be constructed so that each apex lies on the circle so defined.  In general, the end of the last (5th) base will not lie exactly on the circle, but will be too far or fall short. The size of the first triangle is given as 1, but the size of the second can be varied, and with adjustments, the end of the last base (the end of the sement) can be made to lie exactly on the circle, and at that point you have a solution, where the segment is a chord of the circle.

A choice of the size of the second triangle (WCX) determines the remainder.

Let A be at the origin and W, X, Y, Z and G be on the x-axis. The size of WCX determines where point C is. The perpendicular bisectors meet where the center of the circle must be located. Each subsequent triangle must be big enough for its apex to be on the circle thus found. Some choices of the size of WCX will result in the 5th triangle extending beyond the intersection of the circle with the x-axis, and others will not even extend as far as that point. Having a program to vary the size of WCX will allow homing in on the point where the final triangle ends at the circumference of the circle (twice the x coordinate of the center of the circle), and the intervening points give us the answers.

The first version of the below program had ranges for ratio1, the size of the second triangle. Eventually narrowing the range found the second triangle should have sides equal to 7.8630543116374....  

clc
A=[0 0]; B=[1/2 sqrt(3)/2];
W=[1 0];

 ratio1=7.8630543116374
    
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; 
x2=Pt2(1); y2=Pt2(2);
Eq2=y==y2+m2*x; 


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])

ratio1 =
           7.8630543116374
           
                       center of circle
                  x                        y                      radius
          20.2367132985458         -11.2506591685365          23.6732308091374
          
B =
                       0.5         0.866025403784439
C =
           4.9315271558187          6.80960478521475
X =
           8.8630543116374                         0
D =
          15.7921873844231          12.0016105344706
Y =
          22.7213204572087                         0
E =
          28.9369913479624          10.7658577859123
Z =
          35.1526622387161                         0
F =
          37.8130444179039          4.60791710190395
G =
          40.4734265970916                         0
          
          
                                   2 versions of G
                                   differ by
           7.8630543116374       1.4210854715202e-14


7.8630543116374 is indeed the size of each side of WCX. The sides of XDY are  22.7213204572087 - 8.8630543116374 = 13.8582661455713; of YEZ, 35.1526622387161 - 22.7213204572087 = 12.4313417815074; of ZFG, 40.4734265970916 - 35.1526622387161 = 5.32076435837551.

Of course 40.4734265970916 is the length of the whole chord of this circle of radius 23.6732308091374, so it's not quite a diameter.

The set of triangles is not symmetrical, and there should be a reverse version where the initial ratio is 12.4313417815074/5.32076435837551 = 2.33638269695951, but of course making a miniature version where the first is shrunk form 5.32... to the 1 of the puzzle, but something goes wrong with the calculation and it doesn't come out right.

So:    ABW            WCX              XDY               YEZ                 ZFG
        1     7.8630543116374   13.8582661455713   12.4313417815074    5.32076435837551
  or    1     2.3363826969595   2.60456303120373    1.4778054020115     .18794292185217
  
The bottom numbers are proportional (reduced by a factor of  5.32076435837551) to the top numbers but in reverse order.  

  Posted by Charlie on 2021-09-10 19:55:57
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information