한 변의 길이가 3 3 3 A B C \mathrm{ABC} ABC 0 < k < 1 0<k<1 0 < k < 1 k k k A B \mathrm{AB} AB B C \mathrm{BC} BC ( 1 − k ) : k (1-k):k ( 1 − k ) : k P \mathrm P P Q \mathrm Q Q A B \mathrm{AB} AB B C \mathrm{BC} BC k : ( k + 1 ) k:(k+1) k : ( k + 1 ) P ′ \mathrm P^{\prime} P ′ Q ′ \mathrm Q^{\prime} Q ′ P B Q \mathrm{PBQ} PBQ S 1 S_1 S 1 P ′ Q ′ B \mathrm{P^{\prime}Q^{\prime}B} P ′ Q ′ B S 2 S_2 S 2 S 1 : S 2 = 1 : 4 S_1:S_2=1:4 S 1 : S 2 = 1 : 4 k k k
두 선분 A B \mathrm{AB} AB B C \mathrm{BC} BC 3 3 3
A P ‾ = B Q ‾ = (가) \overline{\mathrm{AP}}=\overline{\mathrm{BQ}}=\boxed{\text{(가)}} AP = BQ = ( 가 ) A P ′ ‾ = B Q ′ ‾ = 3 k \overline{\mathrm{AP^{\prime}}}=\overline{\mathrm{BQ^{\prime}}}=3k A P ′ = B Q ′ = 3 k
이다. 두 점 P \mathrm P P P ′ \mathrm P^{\prime} P ′ B C \mathrm{BC} BC H \mathrm H H H ′ \mathrm H^{\prime} H ′ P B H \mathrm{PBH} PBH P ′ B H ′ \mathrm{P^{\prime}BH^{\prime}} P ′ B H ′
P H ‾ : P ′ H ′ ‾ = P B ‾ : P ′ B ‾ = { 3 − ( (가) ) } : ( (나) ) \overline{\mathrm{PH}}:\overline{\mathrm{P^{\prime}H^{\prime}}}=\overline{\mathrm{PB}}:\overline{\mathrm{P^{\prime}B}}=\{3-(\boxed{\text{(가)}})\}:(\boxed{\text{(나)}}) PH : P ′ H ′ = PB : P ′ B = { 3 − ( ( 가 ) )} : ( ( 나 ) )
이므로 S 1 : S 2 = ( 1 2 × B Q ‾ × P H ‾ ) : ( 1 2 × B Q ′ ‾ × P ′ H ′ ‾ ) = ( B Q ‾ × P B ‾ ) : ( B Q ′ ‾ × P ′ B ‾ ) S_1:S_2=(\frac12\times\overline{\mathrm{BQ}}\times\overline{\mathrm{PH}}):(\frac12\times\overline{\mathrm{BQ^{\prime}}}\times\overline{\mathrm{P^{\prime}H^{\prime}}})=(\overline{\mathrm{BQ}}\times\overline{\mathrm{PB}}):(\overline{\mathrm{BQ^{\prime}}}\times\overline{\mathrm{P^{\prime}B}}) S 1 : S 2 = ( 2 1 × BQ × PH ) : ( 2 1 × B Q ′ × P ′ H ′ ) = ( BQ × PB ) : ( B Q ′ × P ′ B )
이다. 따라서 k = (다) k=\boxed{\text{(다)}} k = ( 다 )
위의 (가), (나)에 알맞은 식을 각각 f ( k ) f(k) f ( k ) g ( k ) g(k) g ( k ) p p p f ( p ) × g ( p ) f(p)\times g(p) f ( p ) × g ( p ) [4점]
