Track / Stage 6 / 273 of 400 #1273 of 1964
Problem 1273 National Olympiad, first round Difficulty 6.2 Prove it
As shown in Figure 15-8, in the spatial quadrilateral A B C D ABCD A B C D P , Q P, Q P , Q A C , B D AC, BD A C , B D A B = C D , A D = B C AB=CD, AD=BC A B = C D , A D = B C P Q ⊥ A C PQ \perp AC P Q ⊥ A C P Q ⊥ B D PQ \perp BD P Q ⊥ B D P Q ⊥ A C , P Q ⊥ B D PQ \perp AC, PQ \perp BD P Q ⊥ A C , P Q ⊥ B D A B = C D AB=CD A B = C D A D = B C AD=BC A D = B C
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Official solution Prove (1) P Q → = P A → + A Q → = − 1 2 A C → + 1 2 ( A B → + A D → ) = 1 2 ( A B → + A D → − \overrightarrow{P Q}=\overrightarrow{P A}+\overrightarrow{A Q}=-\frac{1}{2} \overrightarrow{A C}+\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A D})=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A D}- P Q = P A + A Q = − 2 1 A C + 2 1 ( A B + A D ) = 2 1 ( A B + A D − A C → ) , A B = C D ⇔ A B 2 → = C D 2 → = ( A D → − A C → ) 2 = A D 2 → + A C 2 → − 2 A D → ⋅ A C → \overrightarrow{A C}), A B=C D \Leftrightarrow \overrightarrow{A B^{2}}=\overrightarrow{C D^{2}}=(\overrightarrow{A D}-\overrightarrow{A C})^{2}=\overrightarrow{A D^{2}}+\overrightarrow{A C^{2}}-2 \overrightarrow{A D} \cdot \overrightarrow{A C} A C ) , A B = C D ⇔ A B 2 = C D 2 = ( A D − A C ) 2 = A D 2 + A C 2 − 2 A D ⋅ A C A D = B C ⇔ A D 2 → = B C 2 → = ( A C → − A B → ) 2 = A C 2 → + A B 2 → − 2 A C → ⋅ A B → A D=B C \Leftrightarrow \overrightarrow{A D^{2}}=\overrightarrow{B C^{2}}=(\overrightarrow{A C}-\overrightarrow{A B})^{2}=\overrightarrow{A C^{2}}+\overrightarrow{A B^{2}}-2 \overrightarrow{A C} \cdot \overrightarrow{A B} A D = B C ⇔ A D 2 = B C 2 = ( A C − A B ) 2 = A C 2 + A B 2 − 2 A C ⋅ A B A B 2 → + A D 2 → = A D 2 → + A C 2 → − 2 A D → ⋅ A C → + A C 2 → + A B 2 → − 2 A C → ⋅ A B → \overrightarrow{A B^{2}}+\overrightarrow{A D^{2}}=\overrightarrow{A D^{2}}+\overrightarrow{A C^{2}}-2 \overrightarrow{A D} \cdot \overrightarrow{A C}+\overrightarrow{A C^{2}}+\overrightarrow{A B^{2}}-2 \overrightarrow{A C} \cdot \overrightarrow{A B} A B 2 + A D 2 = A D 2 + A C 2 − 2 A D ⋅ A C + A C 2 + A B 2 − 2 A C ⋅ A B A D → ⋅ A C → + A B → ⋅ A C → = A C 2 → \overrightarrow{A D} \cdot \overrightarrow{A C}+\overrightarrow{A B} \cdot \overrightarrow{A C}=\overrightarrow{A C^{2}} A D ⋅ A C + A B ⋅ A C = A C 2 P Q → ⋅ A C → = 1 2 ( A B → + A D → − A C → ) ⋅ A C → = 1 2 ( A B → ⋅ A C → + A D → ⋅ A C → − A C 2 → ) = 0 ⇒ P Q → ⊥ A C → \overrightarrow{P Q} \cdot \overrightarrow{A C}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A D}-\overrightarrow{A C}) \cdot \overrightarrow{A C}=\frac{1}{2}(\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{A D} \cdot \overrightarrow{A C}-\overrightarrow{A C^{2}})=0 \Rightarrow \overrightarrow{P Q} \perp \overrightarrow{A C} P Q ⋅ A C = 2 1 ( A B + A D − A C ) ⋅ A C = 2 1 ( A B ⋅ A C + A D ⋅ A C − A C 2 ) = 0 ⇒ P Q ⊥ A C P Q → ⊥ B D → \overrightarrow{P Q} \perp \overrightarrow{B D} P Q ⊥ B D P Q ⊥ A C ⇒ P Q → ⋅ A C → = 0 ⇒ 1 2 ( A B → + A D → − A C → ) ⋅ A C → = 0 ⇒ A B → ⋅ A C → + A D → ⋅ A C → = A C 2 → , P Q ⊥ B D ⇒ P Q → ⋅ B D → = 0 ⇒ 1 2 ( A B → + A D → − A C → ) ⋅ ( A D → − A B → ) = 0 ⇒ A D 2 → = A B 2 → + A C → ⋅ A D → − A C → ⋅ A B → , C D 2 → = ( A D → − A C → ) 2 = P Q \perp A C \Rightarrow \overrightarrow{P Q} \cdot \overrightarrow{A C}=0 \Rightarrow \frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A D}-\overrightarrow{A C}) \cdot \overrightarrow{A C}=0 \Rightarrow \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{A D} \cdot \overrightarrow{A C}=\overrightarrow{A C^{2}}, P Q \perp B D \Rightarrow \overrightarrow{P Q} \cdot \overrightarrow{B D}=0 \Rightarrow \frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A D}-\overrightarrow{A C}) \cdot (\overrightarrow{A D}-\overrightarrow{A B})=0 \Rightarrow \overrightarrow{A D^{2}}=\overrightarrow{A B^{2}}+\overrightarrow{A C} \cdot \overrightarrow{A D}-\overrightarrow{A C} \cdot \overrightarrow{A B}, \overrightarrow{C D^{2}}=(\overrightarrow{A D}-\overrightarrow{A C})^{2}= P Q ⊥ A C ⇒ P Q ⋅ A C = 0 ⇒ 2 1 ( A B + A D − A C ) ⋅ A C = 0 ⇒ A B ⋅ A C + A D ⋅ A C = A C 2 , P Q ⊥ B D ⇒ P Q ⋅ B D = 0 ⇒ 2 1 ( A B + A D − A C ) ⋅ ( A D − A B ) = 0 ⇒ A D 2 = A B 2 + A C ⋅ A D − A C ⋅ A B , C D 2 = ( A D − A C ) 2 = A D 2 → + A C 2 → − 2 A D → ⋅ A C → = ( A B → ⋅ A C → + A D → ⋅ A C → ) + ( A B 2 → + A C → ⋅ A D → − A C → ⋅ A B → ) − 2 A D → ⋅ A C → = A B 2 → \overrightarrow{A D^{2}}+\overrightarrow{A C^{2}}-2 \overrightarrow{A D} \cdot \overrightarrow{A C}=(\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{A D} \cdot \overrightarrow{A C})+(\overrightarrow{A B^{2}}+\overrightarrow{A C} \cdot \overrightarrow{A D}-\overrightarrow{A C} \cdot \overrightarrow{A B})-2 \overrightarrow{A D} \cdot \overrightarrow{A C}=\overrightarrow{A B^{2}} A D 2 + A C 2 − 2 A D ⋅ A C = ( A B ⋅ A C + A D ⋅ A C ) + ( A B 2 + A C ⋅ A D − A C ⋅ A B ) − 2 A D ⋅ A C = A B 2 C D = A B C D=A B C D = A B A D = B C A D=B C A D = B C
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Source: NuminaMath-1.5 ,
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