题目
定义：多面体 M M M P P P
Φ P = 1 − 1 2 π ( ∠ Q 1 P Q 2 + ∠ Q 2 P Q 3 + ⋯ + ∠ Q k − 1 P Q k + ∠ Q k P Q 1 ) \Phi_P=1-\frac{1}{2\pi}\bigl(\angle Q_1PQ_2+\angle Q_2PQ_3+\cdots+\angle Q_{k-1}PQ_k+\angle Q_kPQ_1\bigr)
Φ P = 1 − 2 π 1 ( ∠ Q 1 P Q 2 + ∠ Q 2 P Q 3 + ⋯ + ∠ Q k − 1 P Q k + ∠ Q k P Q 1 )
其中 P P P M M M Q i Q_i Q i i = 1 , 2 , ⋯ , k i=1,2,\cdots,k i = 1 , 2 , ⋯ , k k ≥ 3 k\ge3 k ≥ 3 k ∈ N ∗ k\in\mathbf N^* k ∈ N ∗ M M M P P P Q 1 P Q 2 Q_1PQ_2 Q 1 P Q 2 Q 2 P Q 3 Q_2PQ_3 Q 2 P Q 3 ⋯ \cdots ⋯ Q k − 1 P Q k Q_{k-1}PQ_k Q k − 1 P Q k Q k P Q 1 Q_kPQ_1 Q k P Q 1 M M M P P P P − A B C D P-ABCD P − A B C D P D ⊥ PD\perp P D ⊥ A B C D ABCD A B C D A B C D ABCD A B C D C D = 2 CD=2 C D = 2 D P = 2 3 DP=2\sqrt3 D P = 2 3
(1) 求四棱锥 P − A B C D P-ABCD P − A B C D C C C
(2) 求四棱锥 P − A B C D P-ABCD P − A B C D
(3) 若 Q Q Q P B PB P B C Q CQ C Q A B C D ABCD A B C D
参考解答
解析：
（1）顶点 C C C
因为 P D ⊥ PD\perp P D ⊥ A B C D ABCD A B C D C D ⊂ CD\subset C D ⊂ A B C D ABCD A B C D P D ⊥ C D PD\perp CD P D ⊥ C D
tan ∠ D C P = P D C D = 2 3 2 = 3 ⟹ ∠ D C P = π 3 \tan\angle DCP=\frac{PD}{CD}=\frac{2\sqrt3}{2}=\sqrt3\qquad\Longrightarrow\qquad\angle DCP=\frac{\pi}{3}
tan ∠ D C P = C D P D = 2 2 3 = 3 ⟹ ∠ D C P = 3 π
因为 P D ⊥ PD\perp P D ⊥ A B C D ABCD A B C D B C ⊂ BC\subset B C ⊂ A B C D ABCD A B C D P D ⊥ B C PD\perp BC P D ⊥ B C
又 B C ⊥ C D BC\perp CD B C ⊥ C D P D ∩ C D = D PD\cap CD=D P D ∩ C D = D P D , C D ⊂ PD,\,CD\subset P D , C D ⊂ P C D PCD P C D B C ⊥ BC\perp B C ⊥ P C D PCD P C D
又 P C ⊂ PC\subset P C ⊂ P C D PCD P C D B C ⊥ P C BC\perp PC B C ⊥ P C ∠ P C B = π 2 \angle PCB=\dfrac{\pi}{2} ∠ P C B = 2 π
由离散曲率的定义得
Φ C = 1 − 1 2 π ( ∠ D C B + ∠ P C B + ∠ D C P ) = 1 − 1 2 π ( π 2 + π 2 + π 3 ) = 1 3 \begin{aligned}
\Phi_C &= 1-\frac{1}{2\pi}\bigl(\angle DCB+\angle PCB+\angle DCP\bigr) \\
&= 1-\frac{1}{2\pi}\left(\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{3}\right) = \frac{1}{3}
\end{aligned}
Φ C = 1 − 2 π 1 ( ∠ D C B + ∠ P C B + ∠ D C P ) = 1 − 2 π 1 ( 2 π + 2 π + 3 π ) = 3 1
（2）内切球的表面积
因为四边形 A B C D ABCD A B C D A B ⊥ A D AB\perp AD A B ⊥ A D
因为 P D ⊥ PD\perp P D ⊥ A B C D ABCD A B C D A B ⊂ AB\subset A B ⊂ A B C D ABCD A B C D A B ⊥ P D AB\perp PD A B ⊥ P D
因为 A D ∩ P D = D AD\cap PD=D A D ∩ P D = D A D , P D ⊂ AD,\,PD\subset A D , P D ⊂ P A D PAD P A D A B ⊥ AB\perp A B ⊥ P A D PAD P A D
因为 P A ⊂ PA\subset P A ⊂ P A D PAD P A D A B ⊥ P A AB\perp PA A B ⊥ P A
设四棱锥 P − A B C D P-ABCD P − A B C D S 1 S_1 S 1
S 1 = S △ P A D + S △ P C D + S △ P A B + S △ P B C + S 正方形 A B C D = 2 S △ P C D + 2 S △ P B C + S 正方形 A B C D = 2 × 1 2 × 2 × 2 3 + 2 × 1 2 × 4 × 2 + 4 = 4 3 + 12 \begin{aligned}
S_1 &= S_{\triangle PAD}+S_{\triangle PCD}+S_{\triangle PAB}+S_{\triangle PBC}+S_{\text{正方形}ABCD} \\
&= 2S_{\triangle PCD}+2S_{\triangle PBC}+S_{\text{正方形}ABCD} \\
&= 2\times\frac12\times2\times2\sqrt3+2\times\frac12\times4\times2+4 \\
&= 4\sqrt3+12
\end{aligned}
S 1 = S △ P A D + S △ P C D + S △ P A B + S △ P B C + S 正方形 A B C D = 2 S △ P C D + 2 S △ P B C + S 正方形 A B C D = 2 × 2 1 × 2 × 2 3 + 2 × 2 1 × 4 × 2 + 4 = 4 3 + 1 2
设四棱锥 P − A B C D P-ABCD P − A B C D r r r
V P − A B C D = 1 3 S 正方形 A B C D ⋅ P D = 1 3 S 1 ⋅ r V_{P-ABCD}=\frac13 S_{\text{正方形}ABCD}\cdot PD=\frac13 S_1\cdot r
V P − A B C D = 3 1 S 正方形 A B C D ⋅ P D = 3 1 S 1 ⋅ r
所以
r = S 正方形 A B C D ⋅ P D S 1 = 2 2 × 2 3 4 3 + 12 = 3 − 1 r = \frac{S_{\text{正方形}ABCD}\cdot PD}{S_1}
= \frac{2^2\times2\sqrt3}{4\sqrt3+12}
= \sqrt3-1
r = S 1 S 正方形 A B C D ⋅ P D = 4 3 + 1 2 2 2 × 2 3 = 3 − 1
内切球的表面积
S 2 = 4 π r 2 = 4 π ( 3 − 1 ) 2 = ( 16 − 8 3 ) π S_2 = 4\pi r^2 = 4\pi(\sqrt3-1)^2 = (16-8\sqrt3)\pi
S 2 = 4 π r 2 = 4 π ( 3 − 1 ) 2 = ( 1 6 − 8 3 ) π
（3）直线 C Q CQ C Q A B C D ABCD A B C D
过 Q Q Q Q G ∥ P D QG\parallel PD Q G ∥ P D B D BD B D G G G C G CG C G
因为 P D ⊥ PD\perp P D ⊥ A B C D ABCD A B C D Q G ⊥ QG\perp Q G ⊥ A B C D ABCD A B C D
则 ∠ G C Q \angle GCQ ∠ G C Q C Q CQ C Q A B C D ABCD A B C D
当 Q Q Q B B B ∠ G C Q = 0 \angle GCQ=0 ∠ G C Q = 0
当 Q Q Q B B B B G = x ( 0 < x ≤ 2 2 ) BG=x\ (0<x\le 2\sqrt2) B G = x ( 0 < x ≤ 2 2 )
在 △ B C G \triangle BCG △ B C G
C G 2 = B C 2 + B G 2 − 2 ⋅ B C ⋅ B G ⋅ cos ∠ D B C = 4 + x 2 − 2 ⋅ 2 ⋅ x ⋅ 2 2 = x 2 − 2 2 x + 4 \begin{aligned}
CG^2 &= BC^2+BG^2-2\cdot BC\cdot BG\cdot\cos\angle DBC \\
&= 4+x^2-2\cdot2\cdot x\cdot\frac{\sqrt2}{2} \\
&= x^2-2\sqrt2 x+4
\end{aligned}
C G 2 = B C 2 + B G 2 − 2 ⋅ B C ⋅ B G ⋅ cos ∠ D B C = 4 + x 2 − 2 ⋅ 2 ⋅ x ⋅ 2 2 = x 2 − 2 2 x + 4
因为 Q G ∥ P D QG\parallel PD Q G ∥ P D △ B G Q ∼ △ B D P \triangle BGQ\sim\triangle BDP △ B G Q ∼ △ B D P
Q G P D = B G B D ⟹ Q G = B G ⋅ P D B D = x ⋅ 2 3 2 2 = 6 2 x \frac{QG}{PD}=\frac{BG}{BD}\quad\Longrightarrow\quad
QG = \frac{BG\cdot PD}{BD} = \frac{x\cdot2\sqrt3}{2\sqrt2} = \frac{\sqrt6}{2}x
P D Q G = B D B G ⟹ Q G = B D B G ⋅ P D = 2 2 x ⋅ 2 3 = 2 6 x
所以
tan 2 ∠ G C Q = Q G 2 C G 2 = 3 2 x 2 x 2 − 2 2 x + 4 = 3 2 1 − 2 2 x + 4 x 2 = 3 2 ( 2 x − 2 2 ) 2 + 1 2 \begin{aligned}
\tan^2\angle GCQ &= \frac{QG^2}{CG^2}
= \frac{\dfrac32x^2}{x^2-2\sqrt2x+4} \\
&= \frac{\dfrac32}{1-\dfrac{2\sqrt2}{x}+\dfrac4{x^2}}
= \frac{\dfrac32}{\left(\dfrac2x-\dfrac{\sqrt2}{2}\right)^2+\dfrac12}
\end{aligned}
tan 2 ∠ G C Q = C G 2 Q G 2 = x 2 − 2 2 x + 4 2 3 x 2 = 1 − x 2 2 + x 2 4 2 3 = ( x 2 − 2 2 ) 2 + 2 1 2 3
当分母 ( 2 x − 2 2 ) 2 + 1 2 \left(\dfrac2x-\dfrac{\sqrt2}{2}\right)^2+\dfrac12 ( x 2 − 2 2 ) 2 + 2 1 tan 2 ∠ G C Q \tan^2\angle GCQ tan 2 ∠ G C Q
此时 x = 2 2 x=2\sqrt2 x = 2 2 Q Q Q P P P
由 tan 2 ∠ G C Q ≤ 3 \tan^2\angle GCQ\le3 tan 2 ∠ G C Q ≤ 3 tan ∠ G C Q ≤ 3 \tan\angle GCQ\le\sqrt3 tan ∠ G C Q ≤ 3 ∠ G C Q ≤ π 3 \angle GCQ\le\dfrac{\pi}{3} ∠ G C Q ≤ 3 π
所以 ∠ G C Q \angle GCQ ∠ G C Q π 3 \dfrac{\pi}{3} 3 π
故直线 C Q CQ C Q A B C D ABCD A B C D [ 0 , π 3 ] \displaystyle\left[0,\frac{\pi}{3}\right] [ 0 , 3 π ]
答案：
(1) Φ C = 1 3 \displaystyle\Phi_C=\frac13 Φ C = 3 1
(2) 内切球表面积 ( 16 − 8 3 ) π \displaystyle(16-8\sqrt3)\pi ( 1 6 − 8 3 ) π
(3) [ 0 , π 3 ] \displaystyle\left[0,\frac{\pi}{3}\right] [ 0 , 3 π ]
