SAT1000 - P965

Geometry Level 5

In an acute △ A B C \triangle ABC , tan ⁡ A = 1 2 \tan A = \dfrac{1}{2} , D D is a point on B C BC , [ A B D ] = 2 , [ A C D ] = 4 [ABD]=2, [ACD]=4 .

If E , F E,F are points on A B , A C AB, AC respectively, and D E ⊥ A B , D F ⊥ A C DE \perp AB, DF \perp AC , then find the value of D E → ⋅ D F → \overrightarrow{DE} \cdot \overrightarrow{DF} .

Let A A denote the answer. Submit ⌊ 1000 A ⌋ \lfloor 1000A \rfloor .

Have a look at my problem set: SAT 1000 problems


The answer is -1067.

Alice Smith by Alice Smith , 20, China

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

tan ⁡ A = 1 2 ⟹ sin ⁡ A = 1 5 \tan A=\dfrac 12\implies \sin A=\dfrac {1}{\sqrt 5} ,

cos ⁡ A = 2 5 \cos A=\dfrac {2}{\sqrt 5}

∣ A B ‾ ∣ × ∣ D E ‾ ∣ = 4 , ∣ A C ‾ ∣ × ∣ D F ‾ ∣ = 8 |\overline {AB}|\times |\overline {DE}|=4,|\overline {AC}|\times |\overline {DF}|=8

∣ A B ‾ ∣ × ∣ A C ‾ ∣ × sin ⁡ A = 12 |\overline {AB}|\times |\overline {AC}|\times \sin A=12

⟹ ∣ A B ‾ ∣ × ∣ A C ‾ ∣ = 12 5 \implies |\overline {AB}|\times |\overline {AC}|=12\sqrt 5

So, ∣ D E ‾ ∣ × ∣ D F ‾ ∣ = 4 × 8 12 5 = 8 3 5 |\overline {DE}|\times |\overline {DF}|=\dfrac {4\times 8}{12\sqrt 5}=\dfrac {8}{3\sqrt 5}

So, D E ⃗ . D F ⃗ = ∣ D E ‾ ∣ × ∣ D F ‾ ∣ × cos ⁡ ( π − A ) = − 8 3 5 × 2 5 = − 16 15 \vec {DE}.\vec {DF}=|\overline {DE}|\times |\overline {DF}|\times \cos (π-A) =-\dfrac {8}{3\sqrt 5}\times \dfrac {2}{\sqrt 5}=-\dfrac {16}{15} .

Hence the required answer is

⌊ − 1000 × 16 15 ⌋ = − 1067 \lfloor -1000\times \dfrac {16}{15}\rfloor =\boxed {-1067} .

1 Helpful 0 Interesting 0 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...