Quick Problem!

Geometry Level 3

In △ A B C \triangle{ABC} , the altitude C D ‾ = a \overline{CD} = a and A B ‾ = a + 1 \overline{AB} = a + 1 . Let E E and F F be the midpoints of A D ‾ \overline{AD} and C B ‾ \overline{CB} respectively and form segment E F ‾ \overline{EF} .

If the value of a a for which E F ‾ = 5 + 2 5 2 \overline{EF} = \dfrac{\sqrt{5 + 2\sqrt{5}}}{2} can be expressed as a = α + β λ a = \dfrac{\alpha + \sqrt{\beta}}{\lambda} , where α , β \alpha, \beta and λ \lambda are coprime positive integers, find α + β + λ \alpha + \beta + \lambda .


The answer is 8.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Apr 26, 2021

Using the above diagram △ C D B ∼ △ F G B ⟹ y h = 2 y a ⟹ h = a 2 \triangle{CDB} \sim \triangle{FGB} \implies \dfrac{y}{h} = \dfrac{2y}{a} \implies h = \dfrac{a}{2}

and G B ‾ x = h a = 1 2 ⟹ G B ‾ = x 2 ⟹ E G ‾ = a + 1 − x 2 − ( a + 1 − x 2 ) = a + 1 2 \dfrac{\overline{GB}}{x} = \dfrac{h}{a} = \dfrac{1}{2} \implies \overline{GB} = \dfrac{x}{2} \implies \overline{EG} = a + 1 - \dfrac{x}{2} - (\dfrac{a + 1 - x}{2}) = \dfrac{a + 1}{2}

⟹ E F ‾ = 2 a 2 + 2 a + 1 2 = 5 + 2 5 2 ⟹ \implies \overline{EF} = \dfrac{\sqrt{2a^2 + 2a + 1}}{2} = \dfrac{\sqrt{5 + 2\sqrt{5}}}{2} \implies

2 a 2 + 2 a + 1 = 5 + 2 5 ⟹ 2 a 2 + 2 a − 2 ( 2 + 5 ) = 0 ⟹ 2a^2 + 2a + 1 = 5 + 2\sqrt{5} \implies 2a^2 + 2a - 2(2 + \sqrt{5}) = 0 \implies

a 2 + a − ( 2 + 5 ) = 0 a^2 + a - (2 + \sqrt{5}) = 0 and dropping the negative root

⟹ a = − 1 + 9 + 4 5 2 = − 1 + ( 5 + 2 ) 2 = 1 + 5 2 = α + β λ \implies a = \dfrac{-1 + \sqrt{9 + 4\sqrt{5}}}{2} = \dfrac{-1 + (\sqrt{5} + 2)}{2} = \dfrac{1 + \sqrt{5}}{2} = \dfrac{\alpha + \sqrt{\beta}}{\lambda}

⟹ α + β + λ = 8 \implies \alpha + \beta + \lambda = \boxed{8} .

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