Parity Blue and Yellow

Geometry Level 3

The figure shows a yellow equilateral triangle D E F DEF is inscribed inside a larger equilateral triangle A B C ABC with A B = 1 AB=1 .

If the blue area is equal to the yellow area, find A D = x AD =x . There are two values of x x . The smaller value of x x is equal A − A B \dfrac{A-\sqrt{A}}{B} , where A A and B B are integers and A A is square-free. Submit A + B A+B .

Inspiration - Blue or Yellow


The answer is 9.

Yuriy Kazakov by Yuriy Kazakov , 61, Russia

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3 solutions

Chew-Seong Cheong
May 24, 2021

When the blue area is equal to the yellow area, it means that the area of equilateral △ D E F \triangle DEF is half that of equilateral t r i a n g l e A B C triangle ABC . This means that the side length of △ D E F \triangle DEF , D E = 1 2 DE = \dfrac 1{\sqrt 2} . By cosine rule :

E B 2 + D B 2 − 2 E B ⋅ D B ⋅ cos ⁡ B = D E 2 Note that E B = A D = x x 2 + ( 1 − x ) 2 − 2 x ( 1 − x ) cos ⁡ 6 0 ∘ = 1 2 3 x 2 − 3 x + 1 2 = 0 x = 3 ± 9 − 6 6 = 3 − 3 6 Choosing the smaller value \begin{aligned} \blue{EB}^2 + DB^2 - 2\blue{EB} \cdot DB \cdot \cos B & = DE^2 & \small \blue{\text{Note that }EB = AD = x} \\ \blue x^2 + (1-x)^2 - 2 x(1-x) \cos 60^\circ & = \frac 12 \\ 3x^2 - 3x + \frac 12 & = 0 \\ x & = \frac {3\pm \sqrt{9-6}}6 \\ & = \frac {3-\sqrt 3}6 & \small \blue{\text{Choosing the smaller value}} \end{aligned}

The required answer A + B = 3 + 6 = 9 A+B = 3+6 = \boxed 9 .

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Your latex math is broken. Hard to read :)

Richard Costen - 2 weeks, 2 days ago
Sathvik Acharya
May 23, 2021

Let x x denote the length of segment A D AD . So, we have, A D = B E = C F = x and A F = B D = C E = 1 − x AD=BE=CF=x \;\; \text{and}\;\; AF=BD=CE=1-x Using the Cosine Rule , in △ A F D , \triangle AFD, D F 2 = A D 2 + A F 2 − 2 ⋅ A D ⋅ A F ⋅ cos ⁡ ∠ D A F = x 2 + ( 1 − x ) 2 − 2 ⋅ x ⋅ ( 1 − x ) ⋅ cos ⁡ 6 0 ∘ = 3 x 2 − 3 x + 1 ∴ D F = F E = E D = 3 x 2 − 3 x + 1 \begin{aligned} DF^2&=AD^2+AF^2-2\cdot AD\cdot AF\cdot \cos \angle DAF\\ &=x^2+(1-x)^2-2\cdot x\cdot (1-x)\cdot \cos 60^{\circ} \\ &=3x^2-3x+1\\ \therefore \; DF&=FE=ED=\sqrt{3x^2-3x+1} \end{aligned} Yellow Area = area ( △ D E F ) = D F 2 ⋅ 3 4 = ( 3 x 2 − 3 x + 1 ) 3 4 Yellow Area + Blue Area = area ( △ A B C ) 2 ⋅ Yellow Area = A B 2 ⋅ 3 4 ( 3 x 2 − 3 x + 1 ) 3 2 = 3 4 6 x 2 − 6 x + 1 = 0 ∴ x = 3 ± 3 6 \begin{aligned} {\color{#CEBB00}\text{Yellow Area}}&=\text{area}(\triangle DEF) \\ &=DF^2\cdot \frac{\sqrt{3}}{4} \\ &=\frac{(3x^2-3x+1)\sqrt{3}}{4} \\ \\ {\color{#CEBB00}\text{Yellow Area}}+{\color{#0C6AC7}\text{Blue Area}}&=\text{area}(\triangle ABC) \\ 2\cdot {\color{#CEBB00}\text{Yellow Area}}&=AB^2\cdot \frac{\sqrt{3}}{4} \\ \frac{(3x^2-3x+1)\sqrt{3}}{2}&=\frac{\sqrt{3}}{4} \\ 6x^2-6x+1&=0 \\ \therefore \; x&=\frac{3\pm\sqrt{3}}{6} \\ \end{aligned} Therefore, x min ⁡ = 3 − 3 6 ⟹ A = 3 , B = 6 ⟹ A + B = 9 x_{\min}=\dfrac{3-\sqrt{3}}{6}\implies A=3,\;B=6\implies A+B=\boxed{9}

Alternative: Blue Area = 3 ⋅ area ( △ A D F ) = 3 ⋅ 1 2 ⋅ A D ⋅ A F ⋅ sin ⁡ ∠ A F D = 3 2 ⋅ x ⋅ ( 1 − x ) ⋅ sin ⁡ 6 0 ∘ = ( 3 x − 3 x 2 ) 3 4 \begin{aligned}{\color{#0C6AC7}\text{Blue Area}}&=3\cdot\text{area}(\triangle ADF) \\ &=3\cdot \frac{1}{2}\cdot AD\cdot AF\cdot \sin \angle AFD \\ &=\frac{3}{2}\cdot x\cdot (1-x)\cdot \sin 60^{\circ} \\ &=\frac{(3x-3x^2)\sqrt{3}}{4} \end{aligned} Since Yellow Area = Blue Area , {\color{#CEBB00}\text{Yellow Area}}={\color{#0C6AC7}\text{Blue Area}}, 3 x 2 − 3 x + 1 = 3 x − 3 x 2 ⟺ 6 x 2 − 6 x + 1 = 0 ⟺ x = 3 ± 3 6 3x^2-3x+1=3x-3x^2\;\Longleftrightarrow\; 6x^2-6x+1=0 \;\Longleftrightarrow\; x=\frac{3\pm\sqrt{3}}{6}

2 Helpful 1 Interesting 2 Brilliant 0 Confused

Note: A simple angle chase leads to ∠ A F D = ∠ B D E = ∠ C E F \angle AFD=\angle BDE=\angle CEF , which combined with ∠ A = ∠ B = ∠ C = 6 0 ∘ \angle A=\angle B=\angle C=60^{\circ} and D F = F E = E D DF=FE=ED implies that △ A D F ≅ △ C F E ≅ △ B D E \triangle ADF\cong \triangle CFE\cong \triangle BDE .

Sathvik Acharya - 2 weeks, 5 days ago
Saya Suka
May 24, 2021

AD = x = BE = CF so AF = 1 – FC = 1 – x.

1 blue area
= (1/2) × x × (1 – x) × sin 60°
= (Equal division of 3 blue area) × (Equal division of yellow-blue area) × (Full area)
= (1/3) × (1/2) × (1/2) × 1² × sin 60°


x (1 – x) = 1/6
x (x – 1) = –1/6
(x – 1/2)² = (1/2)² – (1/6) = 1/12
x = 1/2 ± 1/√12
= (√3 ± 1) / √12
= (3 ± √3) / 6

Answer = 3 + 6 = 9

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