Track / Stage 5 / 341 of 400 #941 of 2000
Problem 941 AIME late Difficulty 5.8 Prove it
[Median of a pyramid (tetrahedron).]
Given three vectors a ⃗ , b ⃗ \vec{a}, \vec{b} a , b c ⃗ \vec{c} c c ⃗ \vec{c} c ( b ⃗ ⋅ c ⃗ ) a ⃗ − ( a ⃗ ⋅ c ⃗ ) b ⃗ (\vec{b} \cdot \vec{c}) \vec{a}-(\vec{a} \cdot \vec{c}) \vec{b} ( b ⋅ c ) a − ( a ⋅ c ) b
This one wants a proof. Work it on paper, read the official solution, then mark
yourself honestly — the ladder only means something if the record is true.
I solved it I didn't Skip
Official solution By the properties of the scalar product
c ⃗ ⋅ ( ( b ⃗ ⋅ c ⃗ ) a ⃗ − ( a ⃗ ⋅ c ⃗ ) b ⃗ ) = = ( b ⃗ ⋅ c ⃗ ) ( a ⃗ ⋅ c ⃗ ) − ( a ⃗ ⋅ c ⃗ ) ( b ⃗ ⋅ c ⃗ ) = = ( b ⃗ ⋅ c ⃗ ) ( a ⃗ ⋅ c ⃗ ) − ( b ⃗ ⋅ c ⃗ ) ( a ⃗ ⋅ c ⃗ ) = 0
\begin{aligned}
& \vec{c} \cdot((\vec{b} \cdot \vec{c}) \vec{a}-(\vec{a} \cdot \vec{c}) \vec{b})= \\
= & (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c})-(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c})= \\
= & (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c})-(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c})=0
\end{aligned}
= = c ⋅ (( b ⋅ c ) a − ( a ⋅ c ) b ) = ( b ⋅ c ) ( a ⋅ c ) − ( a ⋅ c ) ( b ⋅ c ) = ( b ⋅ c ) ( a ⋅ c ) − ( b ⋅ c ) ( a ⋅ c ) = 0
Therefore,
c ⃗ ⊥ ( ( b ⃗ ⋅ c ⃗ ) a ⃗ − ( a ⃗ ⋅ c ⃗ ) b ⃗ )
\vec{c} \perp((\vec{b} \cdot \vec{c}) \vec{a}-(\vec{a} \cdot \vec{c}) \vec{b})
c ⊥ (( b ⋅ c ) a − ( a ⋅ c ) b )
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Source: NuminaMath-1.5 ,
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