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China Mathematical Competition

Geometry Difficulty 4.8 AIME Find the answer China

Given $\triangle ABC$ and $\triangle AEF$ such that $B$ is the midpoint of $EF$ . Also, $AB = EF = 1$ , $BC = 6$ , $CA = \sqrt{33}$ , and $\overrightarrow{AB} \cdot \overrightarrow{AE} + \overrightarrow{AC} \cdot \overrightarrow{AF} = 2$ . The cosine of the angle between $\overrightarrow{EF}$ and $\overrightarrow{BC}$ is ______.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have
$\begin{aligned} 2 &= \overrightarrow{AB} \cdot \overrightarrow{AE} + \overrightarrow{AC} \cdot \overrightarrow{AF} \\ &= \overrightarrow{AB} \cdot (\overrightarrow{AB} + \overrightarrow{BE}) + \overrightarrow{AC} \cdot (\overrightarrow{AB} + \overrightarrow{BF}), \end{aligned}$
i.e.
$\overrightarrow{AB}^2 + \overrightarrow{AB} \cdot \overrightarrow{BE} + \overrightarrow{AC} \cdot \overrightarrow{AB} + \overrightarrow{AC} \cdot \overrightarrow{BF} = 2.$
As $\overrightarrow{AB}^2 = 1$ ,
$\overrightarrow{AC} \cdot \overrightarrow{AB} = \sqrt{33} \times 1 \times \frac{33+1-36}{2 \times \sqrt{33} \times 1} = -1$
and $\overrightarrow{BE} = -\overrightarrow{BF}$ , we get
$1 + \overrightarrow{BF} \cdot (\overrightarrow{AC} - \overrightarrow{AB}) - 1 = 2,$
i.e. $\overrightarrow{BF} \cdot \overrightarrow{BC} = 2$ . Defining $\theta$ as the angle between $\overrightarrow{EF}$ and $\overrightarrow{BC}$ , we get $|\overrightarrow{BF}| \cdot |\overrightarrow{BC}| \cdot \cos \theta = 2$ or $3\cos \theta = 2$ . So $\cos \theta = \frac{2}{3}$ .

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