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Geometry Difficulty 4.8 AIME Find the answer China

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

A number or a short expression. Spacing and $ signs are ignored.

Solution

We have
2=AB→⋅AE→+AC→⋅AF→=AB→⋅(AB→+BE→)+AC→⋅(AB→+BF→), \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.
AB→2+AB→⋅BE→+AC→⋅AB→+AC→⋅BF→=2. \overrightarrow{AB}^2 + \overrightarrow{AB} \cdot \overrightarrow{BE} + \overrightarrow{AC} \cdot \overrightarrow{AB} + \overrightarrow{AC} \cdot \overrightarrow{BF} = 2.
As AB→2=1\overrightarrow{AB}^2 = 1,
AC→⋅AB→=33×1×33+1−362×33×1=−1 \overrightarrow{AC} \cdot \overrightarrow{AB} = \sqrt{33} \times 1 \times \frac{33+1-36}{2 \times \sqrt{33} \times 1} = -1
and BE→=−BF→\overrightarrow{BE} = -\overrightarrow{BF}, we get
1+BF→⋅(AC→−AB→)−1=2, 1 + \overrightarrow{BF} \cdot (\overrightarrow{AC} - \overrightarrow{AB}) - 1 = 2,
i.e. BF→⋅BC→=2\overrightarrow{BF} \cdot \overrightarrow{BC} = 2. Defining θ\theta as the angle between EF→\overrightarrow{EF} and BC→\overrightarrow{BC}, we get ∣BF→∣⋅∣BC→∣⋅cos⁡θ=2|\overrightarrow{BF}| \cdot |\overrightarrow{BC}| \cdot \cos \theta = 2 or 3cos⁡θ=23\cos \theta = 2. So cos⁡θ=23\cos \theta = \frac{2}{3}.

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