Dynamic Geometry: P70

Geometry Level pending

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 p q q \dfrac{p}{q}-\sqrt{q} , where p p and q q are coprime positive integers and q q is square-free. Find p q p-q .


The answer is 1.

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1 solution

Chew-Seong Cheong
Mar 12, 2021

Label the big equilateral triangle A B C ABC and let its centroid be O O and median be A M AM . Label the orange equilateral triangle D E F DEF and let its side length be a a . Due to symmetry, O O is also the centroid of the orange triangle. Since the centroid is 2 3 \dfrac 23 of a median length from the vertex, A O = 1 3 AO=\dfrac 1{\sqrt 3} and O N = a 2 3 ON = \dfrac a{2\sqrt 3} . The area of a white triangle:

A w = 1 2 E F A N = 1 2 E F ( A O O N ) = a 2 ( 1 3 a 2 3 ) = a ( 2 a ) 4 3 \begin{aligned} A_w & = \frac 12 \cdot EF \cdot AN = \frac 12 \cdot EF (AO-ON) \\ & = \frac a2 \left(\frac 1{\sqrt 3} - \frac a{2\sqrt 3} \right) \\ & = \frac {a(2-a)}{4\sqrt 3} \end{aligned}

Now the area of the big equilateral triangle A = 1 2 1 2 sin 6 0 = 3 4 A_\triangle = \dfrac 12 \cdot 1^2 \cdot \sin 60^\circ = \dfrac {\sqrt 3}4 . And

3 A w = A 2 3 a ( 2 a ) 4 3 = 3 8 4 a 2 a 2 = 1 2 a 2 4 a + 1 = 0 a = 1 1 2 Since a < 1 \begin{aligned} 3A_w & = \frac {A_\triangle}2 \\ \frac {3a(2-a)}{4\sqrt 3} & = \frac {\sqrt 3}8 \\ 4a - 2a^2 & = 1 \\ 2a^2 - 4a + 1 & = 0 \\ a & = 1 - \frac 1{\sqrt 2} & \small \blue{\text{Since }a < 1} \end{aligned}

The ratio of the orange triangle to that of the big triangle is a 2 1 1 = ( 1 1 2 ) 2 = 3 2 2 \dfrac {a^2}{1^1} = \left(1-\dfrac 1{\sqrt 2} \right)^2 = \dfrac 32 - \sqrt 2 . Therefore p q = 3 2 = 1 p-q = 3-2 = \boxed 1 .

Thank you for posting.

Valentin Duringer - 3 months ago

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