SAT1000 - P988

Geometry Level 3

Given that $\hat{e_1}, {\hat{e_2}}$ are unit vectors, and $\textbf b = x \hat{e_1} + y \hat{e_2}\ (x,y \in \mathbb R)$ is a nonzero vector.

If the angle between $\hat{e_1}$ and $\hat{e_2}$ is $\dfrac{\pi}{6}$ , then find the maximum value of $\dfrac{|x|}{|\textbf b|}$ .

Let $A$ deonte the answer. Submit $\lfloor 1000A \rfloor$ .

Have a look at my problem set: SAT 1000 problems

1 solution

A Former Brilliant Member
Jul 17, 2020

$b^2=x^2+y^2+2xy\hat e_1.\hat e_2=x^2+y^2+\sqrt 3xy$

$\implies y^2+\sqrt 3n|b|y+(n^2-1)b^2=0$ , where $n=\dfrac {|x|}{|b|}$

Since $y$ is real, $3n^2b^2\geq 4(n^2-1)b^2\implies n\leq 2$

Hence $A=2$ and the answer is $\boxed {2000}$ .

You are a foolish learner

Raji Nair - 9 months, 4 weeks ago