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[几何] 一道立体几何球心位置的确定

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hjfmhh posted 2025-6-5 16:48 |Read mode

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original poster hjfmhh posted 2025-6-5 23:01
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original poster hjfmhh posted 2025-6-5 23:17
Last edited by hbghlyj 2025-6-17 01:32过C作底面的垂线,垂足为H
由 $C A=C B=C P$ 知 $H A=H B=H P$ 即H是$\triangle P A B$的外心。
又 $\angle A P B=90^{\circ}, \therefore \triangle P A B$ 的外心是 $A B$ 的中点 $\therefore$ 点C在平面PAB上的射影是 $A B$ 中点M
$\therefore \triangle A B C$ 的外心即是三棱锥外接球的球心
即外接球半径即 $\triangle A B C$ 外接圆半径。
$C M=\frac{3}{2} . \quad \sin\angle A B C=\frac{1}{2}, \quad 2 R=\frac{A C}{\sin \angle A B C}=\frac{3}{1/2}=6$
$R=3$.
$4 \pi R^2=36 \pi$

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押中题了。I卷立几大题就有球心问题。  posted 2025-6-8 23:03

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