Two Islands

Two islands have the same form of a convex pentagon. The smaller is the reduction of the other one with a coefficient equal to 75%. [That is, every line on one island is exactly 75% the length of one found on the other island.] Strangely, their respective maps on two tracing papers are in such a way that 4 sides of the smaller can be placed exactly along 4 sides of the other one [when rotated appropriately]. Find a possible representation of the two islands.

Source: Philippe Fondanaiche, citing Elisabeth Busser and Gilles Cohen POLE 1999

Solutions were received from Richard Winterstein and Philippe Fondanaiche. Thanks also to Denis Borris for lots of analysis on this problem and for forwarding the question to Kevin Brown, who solved it and created his Cutting Self-Similar Pentagons page to show his solution.

Richard's solution: 

  Let me describe the notation I used.  The larger pentagon P is composed of 
the five lines a, b, c, d, and e, occurring in that order preceding 
clockwise. The smaller pentagon Q has corresponding sides v, w, x, y, and z. 
  Since P ~ Q with a coefficient of k (= 3/4), v=ka, w=kb, etc.
  In solving this problem, I began with the observation that for four lines 
of the smaller pentagon (Q) to lie on top of four lines of the larger 
pentagon (P), at least two adjacent sides of Q had to be equal in length to 
two adjacent sides of P.  It's also necessary for four of the interior 
angles of P (and Q, by similarity) to be equal.  Let these angles be equal 
to angle A, and call the fifth angle of the pentagons B.
  Proceeding beyond these assumptions requires one to make some assumptions 
about the form of P and Q and the appearance of them when overlaid.  My 
technique was to make one set of assumptions, then analyze the problem to 
see if contradictions would arise.  Once a contradiction appeared, it was 
possible to refine the assumptions for a next attempt at a solution.  On the 
third go around, I was making the following assumptions.
  (1) The four equal interior angles (A) = 120 deg.  It follows that B=60 
deg (int. angles sum to 540 deg -- a pentagon may be represented as three 
triangles).  The angle B is formed by the sides a and b of P, and, 
similarly, v and w of Q.
  (2) Q aligns with P when Q is rotated one side in a clockwise manner.  
Side w lies atop c, x atop d, y atop e, and z atop a.  Side v does not lie 
atop side b.
  (3) Side x = d, and side y = e.  Side w is slightly smaller than side c, 
differing by a small quantity s.
  (4) Consider for a moment a pentagon R with sides v, c+s, x, y, z+s, lying 
on top of Q.  (R is the sum of Q and the narrow parallelogram s by v.)  The 
difference between P and R is an equilateral triangle T with sides b, v, and 
(a-e-s).

  Since v = ka, b=ka and (a-e-s)=ka.  With this set of relationships, the 
information derived or given from the problem statement, and the preceding 
assumptions, one is able to define all the sides of P and Q in terms of a 
and s.  (To save space I won't give the steps used to obtain these 
equations.)
    P:      a = a
            b = ka
            c = (k^2)(a) + s
            d = (k)[(k^2)(a) + s]
            e = (k^2)[(k^2)(a) + s]

    Q:      ( kP )

  From examination of the equilateral triangle T we found an expression 
relating a and s.  It is a = ka + s + e, or:

            a = ka + s + (k^2)[(k^2)(a) + s]

Solving for s yields s = (13/1456)a . (approx. 0.0089257a )

  Substituting this value for s into the equations for the sides of P and Q 
allows us to determine them exactly in terms of a.  Thus all sides and 
angles of P and Q are known.

---------

For unit a, the sides of P, to five decimal places, are 1, 0.75, 0.57143, 
0.42857, 0.32143.  The corresponding sides of Q are 0.75, 0.5625, 0.42857, 
0.32143, and 0.24107.  To verify my solution I checked these values using 
trigonometry.
Philippe's solution: 
TWO ISLANDS

There are two ways to represent the two islands:

1) The 2 pentagons have 2 common sides and all the angles defined by the 5 
sides are equal to 3*pi/5=108°
    Let consider a regular pentagon abcde of side 1. The 2 pentagons ABCDE 
     (the smaller) and A'B'C'D'E' (the greater) are built as follows:
    A on a and B on b.
    C on bc such as BC=4/3
    E on ae such as AE=3/4
    The parallel to cd from C and the parallel to de from E cut themselves at 
     D such as CD=(11-2*sqrt(5))/12 and DE=(29+3*sqrt(5))/24
    A' on b and B' on C such as A'B'= 4/3 = 4*AB/3
    C' on the line CD such as B'C' = 16/9 = 4*BC/3
    E' on a such as A'E' = 1 = 4*AE/3
    The parallel to ed from C' cuts the line ae at D'. It is easy to check 
     that C'D' = (11-2*sqrt(5))/9 = 4*CD/3 and D'E' = (29+3*sqrt(5))/18 = 4*DE/3

2) The 2 pentagons have 3 common sides.    
    The smaller pentagon is defined by:
    AB = 1
    BC = 4/3 = 1,3333333
    CD = 16/9 = 1,777777
    DE = 64/27 = 2,370370370...
    AE = a = 3,1442...
    angle(ABC) = angle(BCD) = angle(CDE) = angle(BAE) = 120,168..° =180 - alpha
    angle(AED) = 4*alpha - 180° = 59,328..°
    The greater pentagon is defined by:
    A' on B
    B' on C such as A'B' = BC = 4/3 = 4*AB/3
    C' on D such as B'C' = CD = 16/9 = 4*BC/3
    D' on E such as C'D' = DE = 64/27 = 4*CD/3
    The circle of center D' and radius 256/81 cut the line BA at E' such as:
    D'E' = 256/81 = 4*DE/3
    A'E' = 4,192266666... = 4*AE/3
    angle(A'B'C') = angle(B'C'D') = angle(C'D'E') = angle(E'A'B') = 120,168..°
    angle(A'E'D') = 59,328..°
    
    Nota:The angle alpha and the side AE = a are determined with the relation 
     of the sines in the triangle AEE':
    a/sin(angle(AE'E)) = (256/81) / sin(angle(E'AE)) = (4*a/3 - 1) / sin(angle(AEE'))
    where angle(EAE') = alpha, angle(AEE') = 360° - 5*alpha, angle(AE'E) = 4*alpha-180°
    Therefore alpha is determined by the equation:
     sin(alpha) = 256/81*[sin(5*alpha) - 4*sin(4*alpha)/3]
    A calculation by successive approximations give the solution alpha = 59,832..°
    Then a = -256*sin(4*alpha)/[81*sin(alpha)] = 3,1442....
     and we infer that A'E' = 4*a/3 = 4,19226666....
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