面積最大的圓內(nèi)接四邊形（2022全國乙，9）

2022-12-27 22:43 作者:數(shù)學老頑童 0人讀過 | 我要投稿

（2022全國乙，9）已知球 $O$ 的半徑為 $1$ ，四棱錐的頂點為 $O$ ，底面的四個頂點均在球 $O$ 的球面上，則當該四棱錐的體積最大時，其高為（? ? ）

A. $%5Cfrac%7B1%7D%7B3%7D$

B. $%5Cfrac%7B1%7D%7B2%7D$

C. $%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D$

D. $%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D$

解：

設(shè)底面四邊形為 $ABCD$ ，

其所在圓的半徑為 $r$ ，

設(shè) $%5Cangle%20AOB%3D%5Calpha%20$ ， $%5Cangle%20BOC%3D%5Cbeta%20$ ，

$%5Cangle%20COD%3D%5Cgamma%20$ ， $%5Cangle%20DOA%3D%5Cdelta%20$ ，則

$%5Cbegin%7Baligned%7D%0A%09S_%7B%5Ctext%7B%E5%9B%9B%E8%BE%B9%E5%BD%A2%7DABCD%7D%26%3D%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Csin%20%5Calpha%20%2B%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Csin%20%5Cbeta%20%2B%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Csin%20%5Cgamma%20%2B%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Csin%20%5Cdelta%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Cleft(%20%5Csin%20%5Calpha%20%2B%5Csin%20%5Cbeta%20%2B%5Csin%20%5Cgamma%20%2B%5Csin%20%5Cdelta%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Cleft(%202%5Csin%20%5Cfrac%7B%5Calpha%20%2B%5Cbeta%7D%7B2%7D%5Ccos%20%5Cfrac%7B%5Calpha%20-%5Cbeta%7D%7B2%7D%2B2%5Csin%20%5Cfrac%7B%5Cgamma%20%2B%5Cdelta%7D%7B2%7D%5Ccos%20%5Cfrac%7B%5Cgamma%20-%5Cdelta%7D%7B2%7D%20%5Cright)%5C%5C%0A%09%26%5Cleqslant%20%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Cleft(%202%5Csin%20%5Cfrac%7B%5Calpha%20%2B%5Cbeta%7D%7B2%7D%5Ccdot%201%2B2%5Csin%20%5Cfrac%7B%5Cgamma%20%2B%5Cdelta%7D%7B2%7D%5Ccdot%201%20%5Cright)%5C%5C%0A%09%26%3Dr%5E2%5Cleft(%20%5Csin%20%5Calpha%20%2B%5Csin%20%5Cgamma%20%5Cright)%5C%5C%0A%09%26%3Dr%5E2%5Ccdot%202%5Csin%20%5Cfrac%7B%5Calpha%20%2B%5Cgamma%7D%7B2%7D%5Ccos%20%5Cfrac%7B%5Calpha%20-%5Cgamma%7D%7B2%7D%5C%5C%0A%09%26%5Cleqslant%20r%5E2%5Ccdot%202%5Csin%20%5Cfrac%7B%5Calpha%20%2B%5Cgamma%7D%7B2%7D%5Ccdot%201%5C%5C%0A%09%26%3Dr%5E2%5Ccdot%202%5Csin%20%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B2%7D%3D2r%5E2%5C%5C%0A%5Cend%7Baligned%7D$

當且僅當 $%5Calpha%20%3D%5Cbeta%20%3D%5Cgamma%20%3D%5Cdelta%20%3D%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20$ 時，取得最大值.

因為 $r%5E2%2Bh%5E2%3D1$ ，所以

$%5Cbegin%7Baligned%7D%0A%09V_%7B%5Ctext%7B%E5%9B%9B%E6%A3%B1%E9%94%A5%7D%7D%26%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot%20h%5Ccdot%202r%5E2%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot%20h%5Ccdot%202%5Cleft(%201-h%5E2%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B2%7D%7B3%7Dh%5Cleft(%201-h%5E2%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

令 $f%5Cleft(%20h%20%5Cright)%20%3D%5Cfrac%7B2%7D%7B3%7Dh%5Cleft(%201-h%5E2%20%5Cright)%20$ ， $h%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20$ ，

$%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20h%20%5Cright)%20%26%3D%5Cfrac%7B2%7D%7B3%7D%5Cleft(%201-h%5E2%20%5Cright)%20%2B%5Cfrac%7B2%7D%7B3%7Dh%5Ccdot%20%5Cleft(%20-2h%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B2%7D%7B3%7D%5Cleft(%201%2B%5Csqrt%7B3%7Dh%20%5Cright)%20%5Cleft(%201-%5Csqrt%7B3%7Dh%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

令 $f'%5Cleft(%20h%20%5Cright)%20%3D0$ ，得 $h%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D$ ，

當 $h%5Cin%20%5Cleft(%200%2C%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%20%5Cright)%20$ , $f'%5Cleft(%20h%20%5Cright)%20%3E0$ , $f%5Cleft(%20h%20%5Cright)%20%5Cnearrow%20$

當 $h%5Cin%20%5Cleft(%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%2C1%20%5Cright)%20$ , $f'%5Cleft(%20h%20%5Cright)%20%3C0$ , $f%5Cleft(%20h%20%5Cright)%20%5Csearrow%20$