, where and are coprime positive integers and is square-free. Find .
The diagram shows a blue equilateral triangle. Each purple triangle is an isosceles triangle and its base is a side of the blue equilateral triangle. They grow, shrink and meet at the triangle's center. The orange triangle is drawn using the meeting vertices of each purple circle. When the white area is half the blue equilateral triangle's area, the ratio of the orange triangle area to the area of the blue triangle can be expressed as
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Label the big equilateral triangle A B C and let its centroid be O and median be A M . Label the orange equilateral triangle D E F and let its side length be a . Due to symmetry, O is also the centroid of the orange triangle. Since the centroid is 3 2 of a median length from the vertex, A O = 3 1 and O N = 2 3 a . The area of a white triangle:
A w = 2 1 ⋅ E F ⋅ A N = 2 1 ⋅ E F ( A O − O N ) = 2 a ( 3 1 − 2 3 a ) = 4 3 a ( 2 − a )
Now the area of the big equilateral triangle A △ = 2 1 ⋅ 1 2 ⋅ sin 6 0 ∘ = 4 3 . And
3 A w 4 3 3 a ( 2 − a ) 4 a − 2 a 2 2 a 2 − 4 a + 1 a = 2 A △ = 8 3 = 1 = 0 = 1 − 2 1 Since a < 1
The ratio of the orange triangle to that of the big triangle is 1 1 a 2 = ( 1 − 2 1 ) 2 = 2 3 − 2 . Therefore p − q = 3 − 2 = 1 .