사면체 O A B C \mathrm{OABC} OABC O \mathrm{O} O A B C \mathrm{ABC} ABC H \mathrm{H} H H \mathrm{H} H A C ‾ \overline{\mathrm{AC}} AC A B , B C \mathrm{AB},\ \mathrm{BC} AB , BC D , E \mathrm{D},\ \mathrm{E} D , E O A B C \mathrm{OABC} OABC S S S 1 1 S 2 = q p \vphantom{\begin{matrix}1\\1\end{matrix}}S^{\scriptstyle 2}=\dfrac{q}{p} 1 1 S 2 = p q 1 1 q p + 1 \vphantom{\begin{matrix}1\\1\end{matrix}}\dfrac{q}{p+1} 1 1 p + 1 q p , q p,\ q p , q
(가) C H ‾ ⊥ A B ‾ \overline{\mathrm{CH}}\perp\overline{\mathrm{AB}} CH ⊥ AB 1 1 ∠ A O C = π 2 \vphantom{\begin{matrix}1\\1\end{matrix}}\angle \mathrm{AOC}=\dfrac{\pi}{2} 1 1 ∠ AOC = 2 π ∠ B D H + ∠ B H C = π \angle \mathrm{BDH}+\angle \mathrm{BHC}=\pi ∠ BDH + ∠ BHC = π
(나) O A ‾ = 5 \overline{\mathrm{OA}}=\sqrt5 OA = 5 O B ‾ = 2 11 \overline{\mathrm{OB}}=2\sqrt{11} OB = 2 11 1 1 O H ‾ = 11 3 \vphantom{\begin{matrix}1\\1\end{matrix}}\overline{\mathrm{OH}}=\sqrt{\dfrac{11}{3}} 1 1 OH = 3 11
(다)
A C ‾ 2 + A B ‾ 2 + 2 B C ‾ 2 + 20 = 2 O A ‾ 2 + 3 O B ‾ 2 + 4 O C ‾ 2 \begin{aligned}&\overline{\mathrm{AC}}^{\scriptstyle 2}+\overline{\mathrm{AB}}^{\scriptstyle 2}+2\overline{\mathrm{BC}}^{\scriptstyle 2}+20\\[0.5em]&\qquad=2\overline{\mathrm{OA}}^{\scriptstyle 2}+3\overline{\mathrm{OB}}^{\scriptstyle 2}+4\overline{\mathrm{OC}}^{\scriptstyle 2}\end{aligned} AC 2 + AB 2 + 2 BC 2 + 20 = 2 OA 2 + 3 OB 2 + 4 OC 2 
