변의 길이 (A B ‾ = 5 t \overline{AB}=5t A B = 5 t
A D ‾ : D B ‾ = 3 : 2 \overline{AD}:\overline{DB}=3:2 A D : D B = 3 : 2 A D ‾ = 3 t \overline{AD}=3t A D = 3 t D B ‾ = 2 t \overline{DB}=2t D B = 2 t A B ‾ = 5 t \overline{AB}=5t A B = 5 t t > 0 t>0 t > 0
외접원 반지름이 7 7 7
B C ‾ sin A = A B ‾ sin C = 2 × 7 = 14
\dfrac{\overline{BC}}{\sin A}=\dfrac{\overline{AB}}{\sin C}=2\times 7=14
sin A B C = sin C A B = 2 × 7 = 14 sin A : sin C = 8 : 5 \sin A:\sin C=8:5 sin A : sin C = 8 : 5 B C ‾ : A B ‾ = 8 : 5 \overline{BC}:\overline{AB}=8:5 B C : A B = 8 : 5 B C ‾ = 8 t \overline{BC}=8t B C = 8 t
A E ‾ \overline{AE} A E A C ‾ \overline{AC} A C 두 삼각형은 ∠ A \angle A ∠ A
△ A D E △ A B C = A D ‾ ⋅ A E ‾ A B ‾ ⋅ A C ‾ = 9 35
\dfrac{\triangle ADE}{\triangle ABC}
=\dfrac{\overline{AD}\cdot\overline{AE}}{\overline{AB}\cdot\overline{AC}}
=\dfrac{9}{35}
△ A B C △ A D E = A B ⋅ A C A D ⋅ A E = 35 9 3 t ⋅ A E ‾ 5 t ⋅ A C ‾ = 9 35 ⇒ 3 A E ‾ : 5 A C ‾ = 9 : 35
\dfrac{3t\cdot\overline{AE}}{5t\cdot\overline{AC}}=\dfrac{9}{35}
\quad\Rightarrow\quad
3\overline{AE}:5\overline{AC}=9:35
5 t ⋅ A C 3 t ⋅ A E = 35 9 ⇒ 3 A E : 5 A C = 9 : 35 A E ‾ = 3 7 A C ‾
\overline{AE}=\dfrac{3}{7}\overline{AC}
A E = 7 3 A C 원 O O O A A A A D ‾ \overline{AD} A D E E E A D ‾ = A E ‾ = 3 t \overline{AD}=\overline{AE}=3t A D = A E = 3 t
따라서 A C ‾ = 7 t \overline{AC}=7t A C = 7 t
∠ B \angle B ∠ B t t t 코사인법칙:
cos ( ∠ A B C ) = ( 5 t ) 2 + ( 8 t ) 2 − ( 7 t ) 2 2 ⋅ 5 t ⋅ 8 t = 1 2
\cos(\angle ABC)
=\dfrac{(5t)^{2}+(8t)^{2}-(7t)^{2}}{2\cdot 5t\cdot 8t}
=\dfrac{1}{2}
cos ( ∠ A B C ) = 2 ⋅ 5 t ⋅ 8 t ( 5 t ) 2 + ( 8 t ) 2 − ( 7 t ) 2 = 2 1 ∠ A B C = 60 ∘ \angle ABC=60^{\circ} ∠ A B C = 6 0 ∘
A C ‾ sin ( ∠ A B C ) = 14 ⇒ 7 t = 14 sin 60 ∘ = 7 3
\dfrac{\overline{AC}}{\sin(\angle ABC)}=14
\quad\Rightarrow\quad
7t=14\sin 60^{\circ}=7\sqrt{3}
sin ( ∠ A B C ) A C = 14 ⇒ 7 t = 14 sin 6 0 ∘ = 7 3 t = 3 , B C ‾ = 8 3 , A D ‾ = A E ‾ = 3 3
t=\sqrt{3},\quad
\overline{BC}=8\sqrt{3},\quad
\overline{AD}=\overline{AE}=3\sqrt{3}
t = 3 , B C = 8 3 , A D = A E = 3 3 △ P B C \triangle PBC △ P B C 점 A A A B C BC B C H H H
A H ‾ = A B ‾ sin 60 ∘ = 5 3 ⋅ 3 2 = 15 2
\overline{AH}=\overline{AB}\sin 60^{\circ}
=5\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}
=\dfrac{15}{2}
A H = A B sin 6 0 ∘ = 5 3 ⋅ 2 3 = 2 15 △ P B C \triangle PBC △ P B C
1 2 × B C ‾ × ( 점 P 와 직선 B C 사이 거리 )
\dfrac{1}{2}\times\overline{BC}\times(\text{점 }P\text{와 직선 }BC\text{ 사이 거리})
2 1 × B C × ( 점 P 와 직선 B C 사이 거리 ) 점 P P P O O O A A A 3 3 3\sqrt{3} 3 3 B C BC B C A A A B C BC B C A H ‾ + A P ‾ \overline{AH}+\overline{AP} A H + A P A P ‾ ≤ 3 3 \overline{AP}\le 3\sqrt{3} A P ≤ 3 3
△ P B C 의 넓이 ≤ 1 2 × ( A H ‾ + 3 3 ) × 8 3 = 1 2 × ( 15 2 + 3 3 ) × 8 3 = 36 + 30 3
\begin{aligned}
\triangle PBC\text{의 넓이}
&\le \dfrac{1}{2}\times(\overline{AH}+3\sqrt{3})\times 8\sqrt{3} \\
&=\dfrac{1}{2}\times\left(\dfrac{15}{2}+3\sqrt{3}\right)\times 8\sqrt{3} \\
&=36+30\sqrt{3}
\end{aligned}
△ P B C 의 넓이 ≤ 2 1 × ( A H + 3 3 ) × 8 3 = 2 1 × ( 2 15 + 3 3 ) × 8 3 = 36 + 30 3 (등호는 P P P A H AH A H
따라서 정답은 ④ 예요.
