Each set of points {A,B,C}, {D,E,F} and {G,H,I} are equilaterally spaced and lie respectively upon the circumferences of 3 concentric circles centred at O.

The points D, A, B & I are collineal in that order as are F, C, A & H and E, B, C & G.

Scalene triangles GDB, HEC and IFA are formed by these points and contain Δ ABC in their overlapping.


Note: This is not scaled.
Area "X" is that which is bounded by the scalene triangles (pink) while Area "Y" is all other areas (white) within the largest circle.

If line segment AB has a length of 1 unit and segment AI is 3 units:
1. What is the value of segment AF when the Pink region fills exactly half of the largest circle?
2. What is the proportion, Pink:White (or X:Y), when / AFI is a right angle? [I have considered X, the pink region, to equal 1 in my solution].

Part 1:
/\ ABC is an equilateral triangle with side length 1. AO is length 1/SQRT(3). The area of /\ ABC = (1/4)*SQRT(3).

Let T equal the area of triangle AFI,
then T = X/3 + (1/6)*SQRT(3).

The length of OI, the radius of the large circle, is SQRT(19/3).
The area of the large circle is (19/3)*pi.

The area of X is, then, (19/6)*pi; and, then,
Area /\ AFI = (19/18)*pi + (1/6)*SQRT(3).

From the general formula: Area = (ab sin C)/2
Given a = 3; b = AF; C = 60 degrees
Area /\ AFI = AF*(3/4)*SQRT(3)

Thus ... (19/18)*pi + (1/6)*SQRT(3) = (3/4)*SQRT(3)*AF,
 and simplifying,

 AF = (38/81)*pi*SQRT(3) + 2/9 ~= 2.774976882996

Part 2:
In a 30°-60°-90° triangle the sides are in the ratio 1:SQRT(3):2. As we are given the hypotenuse AI = 3, AF = 3/2 and FI = 3*SQRT(3)/2, the area of triangle AFI is (9/8)*SQRT(3).
As the area for triangle ABC is (1/4)*SQRT(3), the pink area, X, and the white area, Y, can be equated as follows: 

X = (23/8)*SQRT(3).
Y = (19/3)*pi - X

The proportion, X:Y, after simplifying, is 

1:((152*pi*SQRT(3) - 207)/207) ~= 1:2.995615990773

Posted by Dej Mar on 2009-05-29 18:07:31