Originally posted by HolyT
If the problem is valid as stated, then the total area (relative to area x) of an irregular pentagon that meets the given conditions is the same as that of a regular pentagon, which definitely meets the given conditions. I have found the area of the regular pentagon. It is 5x/4*[cos(54° )]^2. If there are any such irregular pentagons and the problem is vali posting it in another forum: http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/648.shtml
I am bored with this puzzle now. I am giving the solution. The solution for the general case runs as follows:-
Let ABCDE be the irregular pentagon satisfying the stated property. Join all the internal diagonals; a smaller inverted pentagon PQRST is created inside ABCDE. Let P be the tip opposite to A, Q opposite to B &c..&c.. A 5-pronged star is created inside the pentagon having its tips on the 5 corners of the pentagon. Let the area of the bigger pentagon ABCDE be w and that of the smaller inner pentagon PQRST be w'.
There are 5 triangles with their base on one of the sides of the internal pentagon and vertex on one of the tips of the outer pentagon (viz. triangles ASR, BTS, CPT, DQP and REP). Let their areas be a1, a2, a3, a4 and a5 respectively.These triangles form the five arms of the internal 'penta'-star.
Likewise 5 inwardly directed triangles are created which have their base on one of the sides of the outer pentagon and their vertices on the tips of the internal pentagon (viz. triangles PCD, QDE, REA, SAB and TBC). Let their areas be b1, b2, b3, b4 and b5 respectively.
According to the given condition,
a1+b3+b4 = a2+b4+b5 = a3+b5+b1 = a4+b1+b2 = a5+b2+b3 = x.
................................. .................eqns.(1)
It has been shown that the str. lines BE, CD are parallel; similarly str. lines CA, DE are parallel &c..&c..These results directly lead to triangles CBS and DRE being congruent; similarly triamgles DCT and ESA are congruent, Triangles EDP and ATB are congruent &c. &c..{Preacher had also gotten up to this result.}
This leads to:->
a2+b5 = a5+b2; a3+b1 = a1+b3; a4+b2 = a2+b4;
a5+b3 = a3+b5; a1+b4 = a4+b1.
...........................................eqns.(2)
Using eqns. (1) and (2) we get:-
a1=a2=a3=a4=a5=a(say).
b1=b2=b3=b4=b5=b(say). Let b/a = c (say).
Thus x = a+2b; w' = x-2a = 2b-a; w = 3x+a+b = 4a+7b....eqns.(3)
Hence w/x = (4+7c)/(1+2c)...............................................eqn(4).
The ratio c can be determined from simple geometry as follows:-
Join SD. Now, area of triangle SCD = area of triangle BCD = x;
[on account of having common base BC and equal height].
Hence area of triangle STD = x-a-b = b.
Now c = b/a = (Ar. tr.BTC)/(Ar. tr. BTS) = CT/TS...............eqn.(5)
Also, CT/TS = (Ar. tr.CTD)/(Ar. tr. STD) = (a+b)/b =1+1/c....eqn(6)
From eqns.(5) and (6), c = 1+1/c ; This leads to the quadratic eqn:-
c^2 - c - 1 = 0. As c > 0; hence c = [sqrt(5)+1]/2. Yes c happens to be none other than the golden ratio arising from simple geometrical considerations, even without the pentagon having to be regular.
Putting this value of c in eqn.(4) we get the required value of w/x.