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