Tri-Blades (Posted on 2009-05-28)

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].

	Submitted by brianjn
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This was inspired by a very first 'hacking' experiment with "Inkscape", an SVG (Scalable Vector Graphics) program. It is freeware under GNU license and available from http://inkscape.org for download. Let P be the midpoint of AC and Q the midpoint of AB. OB/QB = AB/PB (By proportion of similar triangles). Radius of inner circle: OB = AB * QB/PB = 1 * 0.5/ (√3/2) = 1/√3 Perpendicular from side midpoint to centre of ABD: OP = PB - OB = √3/2 - 1/√3 = √3/6 Radius of outer circle: OH = √ (OP2 + PH2) = √{(√3/6) 2 + (5/2) 2} = √ (1/12 + 25/4) = √ (19/3) Area of ABC: Area = AC * BP/2 = 1 * √3/2 * 0.5 = √3/4 Area of a right scalene: AI = 3 AF = 3/2 (∠BAC = 60°, ½ side of equilateral) FI = 3√3/2 Area AIF = FI * AF /2 = (3√3/2 * 3/2)/2 = 9√3/8 Total Pink area: 3 * AIF - 2 * ABC = 27√3/8 - 2 * √3/4 = 27√3/8 - √3/2 = 27√3/8 - 4√3/8 = 23√3/8 Area of Circle: Π (OH) 2 = Π (√ (19/3)) 2 = 19 Π /3 Area of White: Area of Circle - Pink = 19 Π/3 - 23√3/8 Ratio White to Pink: 19 Π /3 - 23 √3/8):23 √3/8 = (8 * 19 Π)/(3 * 23√3) - 1: 1 = (8 * 19 Π √3)/(9 * 23) -1 : 1 = 152√3/261 - 1 : 1 ≅ 2.9956 : 1 If the Pink area equals the White area then Pink covers half the area of the circle. Let K be a point on AF such that KI is perpendicular to AF. Now KI, being the altitude of Δ AIF, is 3√3/2, that of a side of a right scalene. Area of AIF is: 3√3/2 * AF/2 = 3√3*AF/4 Pink area is: 3(3√3*AF/4) - 2 * √3/4 [3 triangles - 2 * ABC] = 9√3*AF/4 -2 √3/4 = √3/4(9*AF -2) 2 * Pink = Area of Circle so, 19 Π /3 = 2 * √3/4(9*AF - 2) 19 Π /3 = √3/2(9*AF - 2) (19 Π /3) * (2/√3)= (9*AF - 2) 38 Π√3/9 = (9*AF - 2) 38 Π√3/9 + 2 = 9 * AF AF = (38 Π√3/9)/9 + 2/9 AF = 38 Π√3/81 + 2/9 AF = 38 Π√3/81 + 18/81 AF = (38 Π√3 +18)/81 AF ≅ 2.7749

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-03-27 22:39:44
	Solution	Dej Mar	2009-05-29 18:07:31
	complete solution	Daniel	2009-05-28 16:19:04