[문제 1]
30점 · 2 r 1 ( t ) r ′ ( t ) = ( t − 1 3 ) r 1 ( t ) \displaystyle 2r_1(t)r\prime(t)=\left(t-\frac13\right)r_1(t) 2 r 1 ( t ) r ′ ( t ) = ( t − 3 1 ) r 1 ( t ) r ′ ( t ) = t 2 − 1 6 \displaystyle r\prime(t)=\frac t2-\frac16 r ′ ( t ) = 2 t − 6 1 (1-1) 준식 r 1 ( t ) 2 = ∫ 0 t ( x − 1 3 ) r 1 ( x ) d x + ∫ 0 1 r 1 ( x ) d x \displaystyle r_1(t)^2=\int_0^t\left(x-\frac13\right)r_1(x)\,dx+\int_0^1r_1(x)\,dx r 1 ( t ) 2 = ∫ 0 t ( x − 3 1 ) r 1 ( x ) d x + ∫ 0 1 r 1 ( x ) d x 2 r 1 ( t ) r 1 ′ ( t ) = ( t − 1 3 ) r 1 ( t ) \displaystyle 2r_1(t)r_1\prime(t)=\left(t-\frac13\right)r_1(t) 2 r 1 ( t ) r 1 ′ ( t ) = ( t − 3 1 ) r 1 ( t ) r 1 ′ ( t ) = 1 2 t − 1 6 \displaystyle r_1\prime(t)=\frac12t-\frac16 r 1 ′ ( t ) = 2 1 t − 6 1
60점 따라서 t = 2 \displaystyle t=2 t = 2 C 1 \displaystyle C_1 C 1 ∣ r 1 ′ ( 2 ) ∣ = r 1 ′ ( 2 ) = 5 6 \displaystyle |r_1\prime(2)|=r_1\prime(2)=\frac56 ∣ r 1 ′ ( 2 ) ∣ = r 1 ′ ( 2 ) = 6 5
10점 · r 1 ( t ) = 1 4 t 2 − 1 6 t + k \displaystyle r_1(t)=\frac14t^2-\frac16t+k r 1 ( t ) = 4 1 t 2 − 6 1 t + k (1-2) r 1 ′ ( t ) = 1 2 t − 1 6 \displaystyle r_1\prime(t)=\frac12t-\frac16 r 1 ′ ( t ) = 2 1 t − 6 1 r 1 ( t ) = 1 4 t 2 − 1 6 t + k \displaystyle r_1(t)=\frac14t^2-\frac16t+k r 1 ( t ) = 4 1 t 2 − 6 1 t + k t ≥ 0 \displaystyle t\ge0 t ≥ 0 r 1 ( t ) > 0 \displaystyle r_1(t)>0 r 1 ( t ) > 0 k = r 1 ( 0 ) > 0 \displaystyle k=r_1(0)>0 k = r 1 ( 0 ) > 0
준식 r 1 ( t ) 2 = ∫ 0 t ( x − 1 3 ) r 1 ( x ) d x + ∫ 0 1 r 1 ( x ) d x \displaystyle r_1(t)^2=\int_0^t\left(x-\frac13\right)r_1(x)\,dx+\int_0^1r_1(x)\,dx r 1 ( t ) 2 = ∫ 0 t ( x − 3 1 ) r 1 ( x ) d x + ∫ 0 1 r 1 ( x ) d x r 1 ( 0 ) 2 = ∫ 0 1 r 1 ( x ) d x \displaystyle r_1(0)^2=\int_0^1r_1(x)\,dx r 1 ( 0 ) 2 = ∫ 0 1 r 1 ( x ) d x
50점 · k 2 = k \displaystyle k^2=k k 2 = k r 1 ( 0 ) = k \displaystyle r_1(0)=k r 1 ( 0 ) = k ∫ 0 1 r 1 ( x ) d x = ∫ 0 1 ( 1 4 x 2 − 1 6 x + k ) d x = k \displaystyle \int_0^1r_1(x)\,dx=\int_0^1\left(\frac14x^2-\frac16x+k\right)\,dx=k ∫ 0 1 r 1 ( x ) d x = ∫ 0 1 ( 4 1 x 2 − 6 1 x + k ) d x = k k 2 = k \displaystyle k^2=k k 2 = k
60점 k > 0 \displaystyle k>0 k > 0 k = 1 \displaystyle k=1 k = 1 따라서 r 1 ( t ) = 1 4 t 2 − 1 6 t + 1 \displaystyle r_1(t)=\frac14t^2-\frac16t+1 r 1 ( t ) = 4 1 t 2 − 6 1 t + 1 r 1 ( 2 ) = 5 3 \displaystyle r_1(2)=\frac53 r 1 ( 2 ) = 3 5
(1-3) 시각 t \displaystyle t t C 2 \displaystyle C_2 C 2 O 2 \displaystyle \mathrm O_2 O 2
O 2 = ( ( r 1 ( t ) + r 2 ( t ) ) 2 − ( r 1 ( t ) − r 2 ( t ) ) 2 , r 2 ( t ) ) = ( 2 r 1 ( t ) r 2 ( t ) , r 2 ( t ) ) \displaystyle \mathrm O_2=\left(\sqrt{(r_1(t)+r_2(t))^2-(r_1(t)-r_2(t))^2},r_2(t)\right)=\left(2\sqrt{r_1(t)r_2(t)},r_2(t)\right) O 2 = ( ( r 1 ( t ) + r 2 ( t ) ) 2 − ( r 1 ( t ) − r 2 ( t ) ) 2 , r 2 ( t ) ) = ( 2 r 1 ( t ) r 2 ( t ) , r 2 ( t ) )
접점 P는 선분 O 1 O 2 \displaystyle \mathrm O_1\mathrm O_2 O 1 O 2 r 1 ( t ) : r 2 ( t ) \displaystyle r_1(t):r_2(t) r 1 ( t ) : r 2 ( t )
( r 1 ⋅ 2 r 1 r 2 + r 2 ⋅ 0 r 1 + r 2 , r 1 r 2 + r 2 r 1 r 1 + r 2 ) = ( 2 r 1 r 1 r 2 r 1 + r 2 , 2 r 1 r 2 r 1 + r 2 ) \displaystyle \left(\frac{r_1\cdot2\sqrt{r_1r_2}+r_2\cdot0}{r_1+r_2},\frac{r_1r_2+r_2r_1}{r_1+r_2}\right)=\left(\frac{2r_1\sqrt{r_1r_2}}{r_1+r_2},\frac{2r_1r_2}{r_1+r_2}\right) ( r 1 + r 2 r 1 ⋅ 2 r 1 r 2 + r 2 ⋅ 0 , r 1 + r 2 r 1 r 2 + r 2 r 1 ) = ( r 1 + r 2 2 r 1 r 1 r 2 , r 1 + r 2 2 r 1 r 2 )
+20점 · 점 P ( 25 3 42 , 5 14 ) \displaystyle \mathrm P\left(\frac{25\sqrt3}{42},\frac5{14}\right) P ( 42 25 3 , 14 5 ) r 1 ( 2 ) = 5 3 \displaystyle r_1(2)=\frac53 r 1 ( 2 ) = 3 5 r 2 ( 2 ) = 1 5 \displaystyle r_2(2)=\frac15 r 2 ( 2 ) = 5 1 t = 2 \displaystyle t=2 t = 2 ( 25 3 42 , 5 14 ) \displaystyle \left(\frac{25\sqrt3}{42},\frac5{14}\right) ( 42 25 3 , 14 5 )
직선 ℓ \displaystyle \ell ℓ C 1 \displaystyle C_1 C 1 C 1 : x 2 + ( y − r 1 ) 2 = r 1 2 \displaystyle C_1:x^2+(y-r_1)^2=r_1^2 C 1 : x 2 + ( y − r 1 ) 2 = r 1 2
P ( 2 r 1 r 1 r 2 r 1 + r 2 , 2 r 1 r 2 r 1 + r 2 ) = P ( a , b ) \displaystyle \mathrm P\left(\frac{2r_1\sqrt{r_1r_2}}{r_1+r_2},\frac{2r_1r_2}{r_1+r_2}\right)=\mathrm P(a,b) P ( r 1 + r 2 2 r 1 r 1 r 2 , r 1 + r 2 2 r 1 r 2 ) = P ( a , b ) C 1 \displaystyle C_1 C 1
ℓ : a x + ( b − r 1 ) ( y − r 1 ) = r 1 2 \displaystyle \ell:ax+(b-r_1)(y-r_1)=r_1^2 ℓ : a x + ( b − r 1 ) ( y − r 1 ) = r 1 2
(원 x 2 + y 2 = r 2 \displaystyle x^2+y^2=r^2 x 2 + y 2 = r 2 ( x 1 , y 1 ) \displaystyle (x_1,y_1) ( x 1 , y 1 ) x 1 x + y 1 y = r 2 \displaystyle x_1x+y_1y=r^2 x 1 x + y 1 y = r 2
Q의 좌표를 ( q , 0 ) \displaystyle (q,0) ( q , 0 ) a q + ( b − r 1 ) ( 0 − r 1 ) = r 1 2 \displaystyle aq+(b-r_1)(0-r_1)=r_1^2 a q + ( b − r 1 ) ( 0 − r 1 ) = r 1 2
+20점 · 점 Q ( 3 3 , 0 ) \displaystyle \mathrm Q\left(\frac{\sqrt3}3,0\right) Q ( 3 3 , 0 ) q ( t ) = b a r 1 ( t ) = r 1 ( t ) r 2 ( t ) \displaystyle q(t)=\frac ba r_1(t)=\sqrt{r_1(t)r_2(t)} q ( t ) = a b r 1 ( t ) = r 1 ( t ) r 2 ( t ) t = 2 \displaystyle t=2 t = 2 ( 3 3 , 0 ) \displaystyle \left(\frac{\sqrt3}3,0\right) ( 3 3 , 0 )
q ′ ( t ) = r 1 ′ ( t ) r 2 ( t ) + r 1 ( t ) r 2 ′ ( t ) 2 r 1 ( t ) r 2 ( t ) \displaystyle q\prime(t)=\frac{r_1\prime(t)r_2(t)+r_1(t)r_2\prime(t)}{2\sqrt{r_1(t)r_2(t)}} q ′ ( t ) = 2 r 1 ( t ) r 2 ( t ) r 1 ′ ( t ) r 2 ( t ) + r 1 ( t ) r 2 ′ ( t ) r 1 ( 2 ) = 5 3 \displaystyle r_1(2)=\frac53 r 1 ( 2 ) = 3 5 r 2 ( 2 ) = 1 5 \displaystyle r_2(2)=\frac15 r 2 ( 2 ) = 5 1 r 1 ′ ( 2 ) = 5 6 \displaystyle r_1\prime(2)=\frac56 r 1 ′ ( 2 ) = 6 5 r 2 ′ ( 2 ) = − 4 25 \displaystyle r_2\prime(2)=-\frac4{25} r 2 ′ ( 2 ) = − 25 4
+20점 · ∣ q ′ ( 2 ) ∣ = 3 20 \displaystyle |q\prime(2)|=\frac{\sqrt3}{20} ∣ q ′ ( 2 ) ∣ = 20 3 q ′ ( 2 ) = − 3 20 \displaystyle q\prime(2)=-\frac{\sqrt3}{20} q ′ ( 2 ) = − 20 3 t = 2 \displaystyle t=2 t = 2 ∣ q ′ ( 2 ) ∣ = − q ′ ( 2 ) = 3 20 \displaystyle |q\prime(2)|=-q\prime(2)=\frac{\sqrt3}{20} ∣ q ′ ( 2 ) ∣ = − q ′ ( 2 ) = 20 3
수리논술, 이게 다다.