+3점 · 피타고라스 정리 1-1. (10점)
먼저, α \displaystyle \alpha α ∣ a → ∣ \displaystyle | {\overrightarrow{a}} | ∣ a ∣ β = ∣ b → ∣ \displaystyle \beta = | {\overrightarrow{b}} | β = ∣ b ∣ γ = ∣ a → − b → ∣ \displaystyle \gamma = | {\overrightarrow{a}} - {\overrightarrow{b}} | γ = ∣ a − b ∣ \triangleOAB \displaystyle \triangleOAB \triangleOAB A \displaystyle A A q = ∣ a → ∣ cos θ \displaystyle q = | {\overrightarrow{a}} | \cos \theta q = ∣ a ∣ cos θ α 2 = q 2 + d 2 = q 2 + γ 2 − ( β − q ) 2 = q 2 + γ 2 − β 2 + 2 β q − q 2 = γ 2 − β 2 + 2 β q \displaystyle \alpha^{2} = q^{2} + d^{2} = q^{2} + \gamma^{2} - ( \beta - q )^{2} = q^{2} + \gamma^{2} - \beta^{2} + 2 \beta q - q^{2} = \gamma^{2} - \beta^{2} + 2 \beta q α 2 = q 2 + d 2 = q 2 + γ 2 − ( β − q ) 2 = q 2 + γ 2 − β 2 + 2 β q − q 2 = γ 2 − β 2 + 2 β q
+3점 · 벡터의 크기 이고, 벡터의 크기에 의하여 α 2 = a 1 2 + a 2 2 , β 2 = b 1 2 + b 2 2 \displaystyle \alpha^{2} = a_{1}^{2} + a_{2}^{2} , \beta^{2} = b_{1}^{2} + b_{2}^{2} α 2 = a 1 2 + a 2 2 , β 2 = b 1 2 + b 2 2 γ 2 = ( a 1 − b 1 ) 2 + ( a 2 − b 2 ) 2 = a 1 2 + a 2 2 + b 1 2 + b 2 2 − 2 ( a 1 b 1 + a 2 b 2 ) = α 2 + β 2 − 2 ( a 1 b 1 + a 2 b 2 ) . \displaystyle \gamma^{2} = ( a_{1} - b_{1} )^{2} + ( a_{2} - b_{2} )^{2} = a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} - 2 ( a_{1} b_{1} + a_{2} b_{2} ) = \alpha^{2} + \beta^{2} - 2 ( a_{1} b_{1} + a_{2} b_{2} ) . γ 2 = ( a 1 − b 1 ) 2 + ( a 2 − b 2 ) 2 = a 1 2 + a 2 2 + b 1 2 + b 2 2 − 2 ( a 1 b 1 + a 2 b 2 ) = α 2 + β 2 − 2 ( a 1 b 1 + a 2 b 2 ) .
+4점 · 내적 위의 식들에 의하여, ∣ a → ∣ ∣ b → ∣ cos θ = β q \displaystyle | {\overrightarrow{a}} | | {\overrightarrow{b}} | \cos \theta = \beta q ∣ a ∣∣ b ∣ cos θ = β q = a 1 b 1 + a 2 b 2 \displaystyle = a_{1} b_{1} + a_{2} b_{2} = a 1 b 1 + a 2 b 2
+2점 · 1-1에 의한 코사인 1-2. (10점) 1-1에 의하여 cos θ = a → ∙ b → ∣ a → ∣ ∣ b → ∣ \displaystyle \cos \theta = \frac{{{\overrightarrow{a}} \bullet {\overrightarrow{b}}}}{| {\overrightarrow{a}} | | {\overrightarrow{b}} |} cos θ = ∣ a ∣∣ b ∣ a ∙ b
+3점 · 삼각형의 넓이 또 삼각형 \triangleOAB \displaystyle \triangleOAB \triangleOAB S = 1 2 ∣ a → ∣ ∣ b → ∣ sin θ \displaystyle S = \frac{{1}}{2} | {\overrightarrow{a}} | | {\overrightarrow{b}} | \sin \theta S = 2 1 ∣ a ∣∣ b ∣ sin θ
+5점 · 넓이 식 정리 그러므로 S = 1 2 ∣ a → ∣ ∣ b → ∣ 1 − ( a → ∙ b → ∣ a → ∣ ∣ b → ∣ ) 2 = 1 2 ( ∣ a → ∣ ∣ b → ∣ ) 2 − ( a → ∙ b → ) 2 = 1 2 ( a 1 2 + a 2 2 ) ( b 1 2 + b 2 2 ) − ( a 1 b 1 + a 2 b 2 ) 2 = 1 2 a 1 2 b 2 2 + a 2 2 b 1 2 − 2 a 1 b 1 a 2 b 2 = 1 2 ( a 1 b 2 − a 2 b 1 ) 2 = 1 2 ∣ a 1 b 2 − a 2 b 1 ∣ \displaystyle \begin{aligned}S &= \frac{{1}}{2} | {\overrightarrow{a}} | | {\overrightarrow{b}} | \sqrt{1 - \left( \frac{{{\overrightarrow{a}} \bullet {\overrightarrow{b}}}}{| {\overrightarrow{a}} | | {\overrightarrow{b}} |} \right)^{2}} = \frac{{1}}{2} \sqrt{( | {\overrightarrow{a}} | | {\overrightarrow{b}} | )^{2} - ( {\overrightarrow{a}} \bullet {\overrightarrow{b}} )^{2}} = \frac{{1}}{2} \sqrt{( a_{1}^{2} + a_{2}^{2} ) ( b_{1}^{2} + b_{2}^{2} ) - ( a_{1} b_{1} + a_{2} b_{2} )^{2}} \\ &= \frac{{1}}{2} \sqrt{a_{1}^{2} b_{2}^{2} + a_{2}^{2} b_{1}^{2} - 2 a_{1} b_{1} a_{2} b_{2}} = \frac{{1}}{2} \sqrt{( a_{1} b_{2} - a_{2} b_{1} )^{2}} = \frac{{1}}{2} | a_{1} b_{2} - a_{2} b_{1} |\end{aligned} S = 2 1 ∣ a ∣∣ b ∣ 1 − ( ∣ a ∣∣ b ∣ a ∙ b ) 2 = 2 1 ( ∣ a ∣∣ b ∣ ) 2 − ( a ∙ b ) 2 = 2 1 ( a 1 2 + a 2 2 ) ( b 1 2 + b 2 2 ) − ( a 1 b 1 + a 2 b 2 ) 2 = 2 1 a 1 2 b 2 2 + a 2 2 b 1 2 − 2 a 1 b 1 a 2 b 2 = 2 1 ( a 1 b 2 − a 2 b 1 ) 2 = 2 1 ∣ a 1 b 2 − a 2 b 1 ∣ 이다.
+3점 · 정다각형의 꼭지점 1-3. (10점) 중심이 원점O \displaystyle O O r \displaystyle r r n \displaystyle n n A k = ( r cos 2 k π n , r sin 2 k π n ) \displaystyle A_{k} = ( r \cos \frac{{2 k \pi}}{n} , r \sin \frac{{2 k \pi}}{n} ) A k = ( r cos n 2 k π , r sin n 2 k π ) ( k = 0 , 1 , . . . , n − 1 ) \displaystyle ( k = 0 , 1 , . . . , n - 1 ) ( k = 0 , 1 , ... , n − 1 ) n \displaystyle n n A 0 = ( r , 0 ) , A 1 = ( r cos 2 π n , r sin 2 π n ) \displaystyle A_{0} = ( r , 0 ) , A_{1} = ( r \cos \frac{{2 \pi}}{n} , r \sin \frac{{2 \pi}}{n} ) A 0 = ( r , 0 ) , A 1 = ( r cos n 2 π , r sin n 2 π )
+4점 · 정다각형의 넓이 S n \displaystyle S_{n} S n \triangleOA 0 A 1 \displaystyle \triangleOA_{0} A_{1} \triangleOA 0 A 1 n \displaystyle n n S n = n ⋅ △ O A 0 A 1 = n ⋅ 1 2 ∣ r ⋅ r sin 2 π n + 0 ⋅ r cos 2 π n ∣ = n r 2 2 sin 2 π n \displaystyle S_{n} = n \cdot \triangle OA_{0} A_{1} = n \cdot \frac{{1}}{2} | r \cdot r \sin \frac{{2 \pi}}{n} + 0 \cdot r \cos \frac{{2 \pi}}{n} | = \frac{{n r^{2}}}{2} \sin \frac{{2 \pi}}{n} S n = n ⋅ △ O A 0 A 1 = n ⋅ 2 1 ∣ r ⋅ r sin n 2 π + 0 ⋅ r cos n 2 π ∣ = 2 n r 2 sin n 2 π
+3점 · 넓이의 극한 이다. 그러므로 lim n → ∞ S n = lim n → ∞ n r 2 2 sin 2 π n = π r 2 ⋅ lim n → ∞ sin 2 π n 2 π n = π r 2 ⋅ 1 = π r 2 \displaystyle \lim_{n \to \infty} {S_{n} = \lim_{n \to \infty} {\frac{{n r^{2}}}{2} \sin \frac{{2 \pi}}{n} = \pi r^{2} \cdot \lim_{n \to \infty} {\frac{{\sin \frac{{2 \pi}}{n}}}{\frac{{2 \pi}}{n}} = \pi r^{2} \cdot 1 = \pi r^{2}}}} n → ∞ lim S n = n → ∞ lim 2 n r 2 sin n 2 π = π r 2 ⋅ n → ∞ lim n 2 π sin n 2 π = π r 2 ⋅ 1 = π r 2
(모범답안과 다른 방법으로 논술한 합당한 답안도 모두 인정한다)
수리논술, 배움에서 논증의 완성까지.