[Guest]

1. construct an equilateral triangle. then construct three circles such that the center of each circle is on a midpoint of each side and the radius is half the side length.

find the area where all three circles intersect, proportional to the triangle.

2. construct an equilateral triangle. then construct a circle such that the center is at the center of the triangle, and the circle goes through the six 1/3 marks on the sides. remove the portions of the circle outside the triangle, then remove the portions of the triangle outside the remaining circle. what is this area, proportional to triangle area?

[Guest]

Triangles!!.bmp

1. Imagine an equilateral triangle DEF with side length of 2 (and area 3^0.5) with side midpoints A,B and C (See fig 1 attached). The grey area is the area we want to find. The triangles ABD, ACF and BCE are all equilateral with side length 1 and so triangle ABC is as well (and has area [3^0.5]/4). Consider circle C. Because CB and CA are of length 1, they are radii of the circle. The area of the 60 degree wedge ACB (Dark grey) has area pi/6. Triple this area for each of the three circle arcs and subtact the area of the triangle ABC twice to get the area we want: (pi-[3^0.5])/2. Or proportional to the triangle DEF: [pi-(3^0.5):2(3^0.5)]. About 0.407:1.

2. Imagine an equilateral triangle HIJ with side length of 3 (and area [9{3^0.5}]/4) with points at 1 unit intervals along each side A,B,C,D,E and F (See fig 2 attached). The grey area isthe area we want to find. All the small triangles formed in the figure are equilateral triangles with side length 1. This is apparent if we look at the smaller triangles CFH, BEI and ADJ, which are all the same as the triangle from the previous solution. To get the area we want we add triangle DEG to Wedge ABG (The dark grey areas) and multiply by 3. We get: (2pi + 3[3^0.5])/4. Proportional to Triangle HIJ: (2pi + 3[3^0.5])/(9[3^0.5]). About 0.736:1

I may have made some computational errors here somewhere but I think the method is correct.

-Tuck

[Guest]

i only did #1

[x(.5x*√3).5]-3[.5x(.25x*√3)-(pi*x^2*.167-(.5x*√3*.5x))]

x=side length

take the triangle area

break up the bigger triangle into 4 triangles

subtract the area of one of the smaller triangles

add the circle arc to the subtraction

substract a triangle from the arc

multiply step 2 to now by 3 and subtract it all from the regualr triangle area

i hope i did it right

Edited September 20, 2009 by houston9388

[Guest]

Let Area of overlap be O (DGEHFID)

Let C = area of each circle

T = area of triangle ABC

t = area of triangle DEF

DEF is also equilateral triangle with sides 1/2 the size of ABC

then, O = 1/6C - 2t

where, t = 1/4T

then O = 1/6C - 1/2T

C = pi.x²

T = root(3)/4x²

C = 4.pi.T/root(3)

C = 7.26T

then, O = 0.7092 T

O ~= t/ root(2)

Let P be the areas pf remaining pieces

T = area of triangle ABC

C = area of circle

c = area of arc GFJ

t = area of triangle OGF

then, P = T - C + 3c

OGF is an equilateral triangle with sides 1/3rd the size of triangle ABC

then, c = 1/6C - t

where, t = T/9

C = pi.x²

T = root(3)/4 . 9x²

C = pi.T.4 / 9 root(3)

C = 0.80 T

c = 0.80 T/6 - T/9

c = 0.02 T

P = T - 0,80 T + 0,06T

P = 0,2636 T