+2점 · A S → = 1 4 b ⃗ + d ⃗ \displaystyle \overrightarrow{AS}=\frac14\vec b+\vec d A S = 4 1 b + d A P → = b ⃗ + t d ⃗ ( 0 ≤ t ≤ 1 ) \displaystyle \overrightarrow{AP}=\vec b+t\vec d\;(0\le t\le1) A P = b + t d ( 0 ≤ t ≤ 1 ) (1-1) 제시문 (가)에 의해 A S → = 3 A D → + A C → 3 + 1 = 3 4 A D → + 1 4 A C → = 3 4 d ⃗ + 1 4 ( b ⃗ + d ⃗ ) = 1 4 b ⃗ + d ⃗ \displaystyle \overrightarrow{AS}=\frac{3\overrightarrow{AD}+\overrightarrow{AC}}{3+1}=\frac34\overrightarrow{AD}+\frac14\overrightarrow{AC}=\frac34\vec d+\frac14(\vec b+\vec d)=\frac14\vec b+\vec d A S = 3 + 1 3 A D + A C = 4 3 A D + 4 1 A C = 4 3 d + 4 1 ( b + d ) = 4 1 b + d A P → = b ⃗ + t d ⃗ ( 0 ≤ t ≤ 1 ) \displaystyle \overrightarrow{AP}=\vec b+t\vec d\;(0\le t\le1) A P = b + t d ( 0 ≤ t ≤ 1 )
+2점 · A Q → = m 4 + n m + n b ⃗ + m + t n m + n d ⃗ \displaystyle \overrightarrow{AQ}=\frac{\frac m4+n}{m+n}\vec b+\frac{m+tn}{m+n}\vec d A Q = m + n 4 m + n b + m + n m + t n d A Q → = m A S → + n A P → m + n = m m + n ( 1 4 b ⃗ + d ⃗ ) + n m + n ( b ⃗ + t d ⃗ ) = m 4 + n m + n b ⃗ + m + t n m + n d ⃗ \displaystyle \overrightarrow{AQ}=\frac{m\overrightarrow{AS}+n\overrightarrow{AP}}{m+n}=\frac m{m+n}(\frac14\vec b+\vec d)+\frac n{m+n}(\vec b+t\vec d)=\frac{\frac m4+n}{m+n}\vec b+\frac{m+tn}{m+n}\vec d A Q = m + n m A S + n A P = m + n m ( 4 1 b + d ) + m + n n ( b + t d ) = m + n 4 m + n b + m + n m + t n d
+4점 · t = 1 − 3 m 4 n \displaystyle t=1-\frac{3m}{4n} t = 1 − 4 n 3 m A P → = b ⃗ + ( 1 − 3 m 4 n ) d ⃗ \displaystyle \overrightarrow{AP}=\vec b+(1-\frac{3m}{4n})\vec d A P = b + ( 1 − 4 n 3 m ) d A Q → \displaystyle \overrightarrow{AQ} A Q A C → = b ⃗ + d ⃗ \displaystyle \overrightarrow{AC}=\vec b+\vec d A C = b + d t = 1 − 3 m 4 n \displaystyle t=1-\frac{3m}{4n} t = 1 − 4 n 3 m A P → = b ⃗ + ( 1 − 3 m 4 n ) d ⃗ \displaystyle \overrightarrow{AP}=\vec b+(1-\frac{3m}{4n})\vec d A P = b + ( 1 − 4 n 3 m ) d
+3점 · S P → = 3 4 b ⃗ − t d ⃗ ( 0 ≤ t ≤ 1 ) \displaystyle \overrightarrow{SP}=\frac34\vec b-t\vec d\;(0\le t\le1) S P = 4 3 b − t d ( 0 ≤ t ≤ 1 ) (1-2) (a) S P → = 3 4 b ⃗ − t d ⃗ \displaystyle \overrightarrow{SP}=\frac34\vec b-t\vec d S P = 4 3 b − t d ( 0 ≤ t ≤ 1 ) \displaystyle (0\le t\le1) ( 0 ≤ t ≤ 1 )
+5점 · 0 = A S → ⋅ S P → = 3 16 ∣ b ⃗ ∣ 2 − 1 4 t b ⃗ ⋅ d ⃗ + 3 4 b ⃗ ⋅ d ⃗ − t ∣ d ⃗ ∣ 2 \displaystyle 0=\overrightarrow{AS}\cdot\overrightarrow{SP}=\frac3{16}|\vec b|^2-\frac14t\vec b\cdot\vec d+\frac34\vec b\cdot\vec d-t|\vec d|^2 0 = A S ⋅ S P = 16 3 ∣ b ∣ 2 − 4 1 t b ⋅ d + 4 3 b ⋅ d − t ∣ d ∣ 2 t = 1 2 \displaystyle t=\frac12 t = 2 1 +2점 · A P → = b ⃗ + 1 2 d ⃗ \displaystyle \overrightarrow{AP}=\vec b+\frac12\vec d A P = b + 2 1 d 0 = A S → ⋅ S P → = ( 1 4 b ⃗ + d ⃗ ) ⋅ ( 3 4 b ⃗ − t d ⃗ ) = 3 16 ∣ b ⃗ ∣ 2 − 1 4 t b ⃗ ⋅ d ⃗ + 3 4 b ⃗ ⋅ d ⃗ − t ∣ d ⃗ ∣ 2 \displaystyle 0=\overrightarrow{AS}\cdot\overrightarrow{SP}=(\frac14\vec b+\vec d)\cdot(\frac34\vec b-t\vec d)=\frac3{16}|\vec b|^2-\frac14t\vec b\cdot\vec d+\frac34\vec b\cdot\vec d-t|\vec d|^2 0 = A S ⋅ S P = ( 4 1 b + d ) ⋅ ( 4 3 b − t d ) = 16 3 ∣ b ∣ 2 − 4 1 t b ⋅ d + 4 3 b ⋅ d − t ∣ d ∣ 2 를 만족해야 하므로 t = 1 2 \displaystyle t=\frac12 t = 2 1 A P → = b ⃗ + 1 2 d ⃗ \displaystyle \overrightarrow{AP}=\vec b+\frac12\vec d A P = b + 2 1 d
+6점 · Q \displaystyle Q Q P S \displaystyle PS P S m : n = 2 : 3 \displaystyle m:n=2:3 m : n = 2 : 3 2 R A → + 3 R P → + 2 R S → = 0 ⃗ \displaystyle 2\overrightarrow{RA}+3\overrightarrow{RP}+2\overrightarrow{RS}=\vec0 2 R A + 3 R P + 2 R S = 0 A , R , Q \displaystyle A,R,Q A , R , Q A R ‾ : R Q ‾ = 5 : 2 \displaystyle \overline{AR}:\overline{RQ}=5:2 A R : R Q = 5 : 2 A S \displaystyle AS A S M \displaystyle M M 2 R A → + 3 R P → + 2 R S → = 0 ⃗ \displaystyle 2\overrightarrow{RA}+3\overrightarrow{RP}+2\overrightarrow{RS}=\vec0 2 R A + 3 R P + 2 R S = 0 R \displaystyle R R M P \displaystyle MP M P 3 : 4 \displaystyle 3:4 3 : 4 2 R A → + 3 R P → + 2 R S → = 0 ⃗ \displaystyle 2\overrightarrow{RA}+3\overrightarrow{RP}+2\overrightarrow{RS}=\vec0 2 R A + 3 R P + 2 R S = 0 R \displaystyle R R △ R P S = 2 7 △ A P S \displaystyle \triangle RPS=\frac27\triangle APS △ R P S = 7 2 △ A P S △ R P S = 4 7 △ M P S \displaystyle \triangle RPS=\frac47\triangle MPS △ R P S = 7 4 △ M P S △ R A P : △ R A S : △ R P S = 2 : 3 : 2 \displaystyle \triangle RAP:\triangle RAS:\triangle RPS=2:3:2 △ R A P : △ R A S : △ R P S = 2 : 3 : 2 △ R P S = 2 7 △ A P S \displaystyle \triangle RPS=\frac27\triangle APS △ R P S = 7 2 △ A P S (b) A P → = b ⃗ + 1 2 d ⃗ \displaystyle \overrightarrow{AP}=\vec b+\frac12\vec d A P = b + 2 1 d Q \displaystyle Q Q P S \displaystyle PS P S m : n = 2 : 3 \displaystyle m:n=2:3 m : n = 2 : 3 제시문 (가)에 의해 R Q → = 2 R S → + 3 R P → 2 + 3 = 2 5 R S → + 3 5 R P → \displaystyle \overrightarrow{RQ}=\frac{2\overrightarrow{RS}+3\overrightarrow{RP}}{2+3}=\frac25\overrightarrow{RS}+\frac35\overrightarrow{RP} R Q = 2 + 3 2 R S + 3 R P = 5 2 R S + 5 3 R P 2 R A → + 3 R P → + 2 R S → = 0 ⃗ \displaystyle 2\overrightarrow{RA}+3\overrightarrow{RP}+2\overrightarrow{RS}=\vec0 2 R A + 3 R P + 2 R S = 0 5 R Q → = 2 A R → \displaystyle 5\overrightarrow{RQ}=2\overrightarrow{AR} 5 R Q = 2 A R A , R , Q \displaystyle A,R,Q A , R , Q A R ‾ : R Q ‾ = 5 : 2 \displaystyle \overline{AR}:\overline{RQ}=5:2 A R : R Q = 5 : 2 △ R A P : △ R A S : △ R P S = 2 : 3 : 2 \displaystyle \triangle RAP:\triangle RAS:\triangle RPS=2:3:2 △ R A P : △ R A S : △ R P S = 2 : 3 : 2 △ A P S = 1 2 ∣ A S → ∣ ∣ S P → ∣ \displaystyle \triangle APS=\frac12|\overrightarrow{AS}||\overrightarrow{SP}| △ A P S = 2 1 ∣ A S ∣∣ S P ∣
+6점 · △ R P S = 3 6 20 \displaystyle \triangle RPS=\frac{3\sqrt6}{20} △ R P S = 20 3 6 A S → ⋅ A S → = ( 1 4 b ⃗ + d ⃗ ) ⋅ ( 1 4 b ⃗ + d ⃗ ) = 1 16 ∣ b ⃗ ∣ 2 + 1 2 b ⃗ ⋅ d ⃗ + ∣ d ⃗ ∣ 2 = 14 5 \displaystyle \overrightarrow{AS}\cdot\overrightarrow{AS}=(\frac14\vec b+\vec d)\cdot(\frac14\vec b+\vec d)=\frac1{16}|\vec b|^2+\frac12\vec b\cdot\vec d+|\vec d|^2=\frac{14}5 A S ⋅ A S = ( 4 1 b + d ) ⋅ ( 4 1 b + d ) = 16 1 ∣ b ∣ 2 + 2 1 b ⋅ d + ∣ d ∣ 2 = 5 14 S P → ⋅ S P → = ( 3 4 b ⃗ − 1 2 d ⃗ ) ⋅ ( 3 4 b ⃗ − 1 2 d ⃗ ) = 9 16 ∣ b ⃗ ∣ 2 − 3 4 b ⃗ ⋅ d ⃗ + 1 4 ∣ d ⃗ ∣ 2 = 189 80 \displaystyle \overrightarrow{SP}\cdot\overrightarrow{SP}=(\frac34\vec b-\frac12\vec d)\cdot(\frac34\vec b-\frac12\vec d)=\frac9{16}|\vec b|^2-\frac34\vec b\cdot\vec d+\frac14|\vec d|^2=\frac{189}{80} S P ⋅ S P = ( 4 3 b − 2 1 d ) ⋅ ( 4 3 b − 2 1 d ) = 16 9 ∣ b ∣ 2 − 4 3 b ⋅ d + 4 1 ∣ d ∣ 2 = 80 189 로부터 △ A P S = 21 6 40 \displaystyle \triangle APS=\frac{21\sqrt6}{40} △ A P S = 40 21 6 R P S \displaystyle RPS R P S 2 7 △ A P S = 3 6 20 \displaystyle \frac27\triangle APS=\frac{3\sqrt6}{20} 7 2 △ A P S = 20 3 6
수리논술, 배움에서 논증의 완성까지.