解:方法一：(1)证明：${\displaystyle \because O、D}$分别是${\displaystyle AC、PC}$的中点

，又${\displaystyle PA\subset }$平面${\displaystyle PAB}$     平面${\displaystyle PAB}$ 　     (2)${\displaystyle \because AB\perp BC}$，${\displaystyle OA=OC}$     ${\displaystyle \therefore OA=OB=OC}$     又${\displaystyle \because OP\perp }$平面${\displaystyle ABC}$     ${\displaystyle \therefore PA=PB=PC}$，取${\displaystyle BC}$的中点${\displaystyle E}$，连结${\displaystyle PE}$     则${\displaystyle BC\perp }$平面${\displaystyle POE}$，作${\displaystyle OF\perp PE}$于点${\displaystyle F}$，连结${\displaystyle DF}$     ${\displaystyle \therefore OF\perp }$平面${\displaystyle PBC}$     ${\displaystyle \therefore \angle ODF}$是${\displaystyle OD}$与平面${\displaystyle PBC}$所成的角     又     ${\displaystyle \therefore PA}$与平面${\displaystyle PBC}$的大小等于${\displaystyle \angle ODF}$     在${\displaystyle Rt△ODF}$中，${\displaystyle sin\angle ODF=\frac{OF}{OD}=\frac{\sqrt{210}}{30}}$     ${\displaystyle \therefore }$直线${\displaystyle PA}$与平面${\displaystyle PBC}$所成角为${\displaystyle arcsin\left(\frac{\sqrt{210}}{30}\right)}$

    (3)由(2)知，${\displaystyle OF\perp }$平面${\displaystyle PBC}$

    ${\displaystyle \therefore F}$是${\displaystyle O}$在平面${\displaystyle PBC}$内的射影

    ${\displaystyle \because D}$是${\displaystyle PC}$的中点

    若点${\displaystyle F}$是${\displaystyle △PBC}$的重心，则${\displaystyle B、F、D}$三点共线

    ${\displaystyle \therefore }$直线${\displaystyle OB}$在平面${\displaystyle PBC}$内的射影为直线${\displaystyle BD}$

    ${\displaystyle \because OB\perp PC}$

    ${\displaystyle \therefore PC\perp BD}$

    ${\displaystyle \therefore PB=BC}$，即${\displaystyle k=1}$ 

    反之，当${\displaystyle k=1}$时，三棱锥${\displaystyle O-PBC}$为正三棱锥

    ${\displaystyle \therefore O}$在平面${\displaystyle PBC}$内的射影恰好为${\displaystyle △PBC}$的重心

    方法二：${\displaystyle \because PO\perp }$平面${\displaystyle ABC}$，${\displaystyle OA=OC}$，${\displaystyle AB=BC}$

    ${\displaystyle \therefore OA\perp OB}$，${\displaystyle OA\perp OP}$，${\displaystyle OB\perp OP}$

    以${\displaystyle O}$为原点，射线${\displaystyle OP}$为非负${\displaystyle z}$轴，如图，建立空间直角坐标系${\displaystyle O-xyz}$，设${\displaystyle AB=a}$，于是${\displaystyle A(\frac{\sqrt{2}}{2}a，0，0)}$，${\displaystyle B(0，\frac{\sqrt{2}}{2}a，0)}$，${\displaystyle C(-\frac{\sqrt{2}}{2}a，0，0)}$

设${\displaystyle OP=h}$，则${\displaystyle P(0，0，h)}$     (1)证明：${\displaystyle \because }$点${\displaystyle D}$是${\displaystyle PC}$的中点，     ${\displaystyle \therefore \overrightarrow{OD}=(-\frac{\sqrt{2}}{4}a，0，\frac{1}{2}h)}$，${\displaystyle \overrightarrow{PA}=(\frac{\sqrt{2}}{2}a，0，-h)}$     ${\displaystyle \therefore \overrightarrow{OD}=-\frac{1}{2}\overrightarrow{PA}}$         平面${\displaystyle PAB}$ 　     (2)${\displaystyle \because k=\frac{1}{2}}$，即${\displaystyle PA=2a}$     ${\displaystyle \therefore h=\sqrt{\frac{7}{2}}a}$        ${\displaystyle \therefore \overrightarrow{PA}=(\frac{\sqrt{2}}{2}a，0，-\sqrt{\frac{7}{2}}a)}$            可求得平面${\displaystyle PBC}$的法向量${\displaystyle \overrightarrow{n}=(1，-1，-\sqrt{\frac{1}{7}})}$         设直线${\displaystyle PA}$与平面${\displaystyle PBC}$所成的角为${\displaystyle \theta }$     则     ${\displaystyle \therefore }$直线${\displaystyle PA}$与平面${\displaystyle PBC}$所成角为${\displaystyle arcsin\left(\frac{\sqrt{210}}{30}\right)}$

    (3)${\displaystyle △PBC}$的重心${\displaystyle G(-\frac{\sqrt{2}}{6}a，\frac{\sqrt{2}}{6}a，\frac{1}{3}h)}$

    ${\displaystyle \therefore \overrightarrow{OG}=(-\frac{\sqrt{2}}{6}a，\frac{\sqrt{2}}{6}a，\frac{1}{3}h)}$

    ${\displaystyle \because OG\perp }$平面${\displaystyle PBC}$

    ${\displaystyle \therefore \overrightarrow{OG}\perp \overrightarrow{PB}}$，又${\displaystyle \overrightarrow{PB}=(0，\frac{\sqrt{2}}{2}a，-h)}$

    ${\displaystyle \therefore \overrightarrow{OG}·\overrightarrow{PB}=\frac{1}{6}{a}^2-\frac{1}{3}{h}^2=0}$

    ${\displaystyle \therefore h=\frac{\sqrt{2}}{2}a}$

    ${\displaystyle \therefore PA=\sqrt{O{A}^2+h^2}=a}$，即${\displaystyle k=1}$

    反之，当${\displaystyle k=1}$时，三棱锥${\displaystyle O-PBC}$为正三棱锥

    ${\displaystyle \therefore O}$在平面${\displaystyle PBC}$内的射影恰好为${\displaystyle △PBC}$的重心

    评注:本题主要考察空间线面关系、空间向量的概念与运算等基础知识，同时考察空间想象能力和推理运算能力．

    解:${\displaystyle \because A、B}$纬度均为${\displaystyle 60°}$如图，设北纬${\displaystyle 45°}$圈小圆球心为${\displaystyle O_1}$，半径为${\displaystyle r}$，则${\displaystyle O_{1}A=O_{1}B=r}$

${\displaystyle \because \angle OA{O}_1=45°}$     ${\displaystyle \therefore r=O_{1}A=AO·cos}$45°=sqrt2/2R${\displaystyle }$     设${\displaystyle \angle A{O}_{1}B=\theta }$，则${\displaystyle r\theta =\frac{\sqrt{2}}{4}\pi R}$     ${\displaystyle \therefore \theta =90°}$     ${\displaystyle 在Rt△A{O}_{1}B中}$，${\displaystyle AB=\sqrt{O_1{A}^2+O_1{B}^2}=\sqrt{2}r=R}$     ${\displaystyle \therefore △AOB}$为正三角形     ${\displaystyle \therefore \angle AOB=\frac{\pi }{3}}$     ${\displaystyle \therefore A、B}$两点的球面距离等于${\displaystyle \frac{\pi }{3}R}$

    评注:求球面距离时，用好两个截面，即过球面两点的大圆截面和小圆截面．抓住它的边角的互相联系：弦${\displaystyle AB}$既是${\displaystyle {\odot }_1}$中的弦又是${\displaystyle \odot }$中的弦。具体做法如下：先计算线段${\displaystyle AB}$的长，再计算圆心角${\displaystyle \angle AOB}$，最后计算大圆上${\displaystyle A}$、${\displaystyle B}$两点间劣弧的长。    

    解:方法一：(1)不妨设正三角形${\displaystyle ABC}$的边长为${\displaystyle 3}$

    评注:1、

    解:在

    (提示内容)

    解:由．

    评注:解决此题

    一、选择题
    1、若正棱锥的底面边长与侧棱长相等，则该棱锥一定不是(   )
    A.三棱锥     B.四棱锥     C.五棱锥     D.六棱锥
    2、(06·浙江9)如图，${\displaystyle O}$是半径为${\displaystyle 1}$的球的球心，点${\displaystyle A、B、C}$在球面上，${\displaystyle OA、OB、OC}$两两垂直，${\displaystyle E、F}$分别是大圆 与 的中点，则点${\displaystyle E、F}$在该球面上的球面距离是(   )
    A.${\displaystyle \frac{\pi }{4}}$         B.${\displaystyle \frac{\pi }{3}}$          C.${\displaystyle \frac{\pi }{2}}$          D.${\displaystyle \frac{\sqrt{2}\pi }{4}}$

                                                               
    3、(06·安徽9)表面积为${\displaystyle 2\sqrt{3}}$的正八面体的各个顶点都在同一个球面上，则此球的体积为(   )
    A.${\displaystyle \frac{\sqrt{2}}{3}\pi }$       B.${\displaystyle \frac{1}{3}\pi }$         C.${\displaystyle \frac{2}{3}\pi }$         D.${\displaystyle \frac{2\sqrt{2}\pi }{3}}$
    4、如图，在正三棱锥${\displaystyle A-BCD}$中，${\displaystyle E、F}$分别是${\displaystyle AB、BC}$的中点，且${\displaystyle DE\perp EF}$，${\displaystyle AB=\sqrt{2}}$，则正三棱锥${\displaystyle A-BCD}$的体积是(   )
    A.${\displaystyle \sqrt{2}}$        B.${\displaystyle \frac{\sqrt{2}}{3}}$         C.${\displaystyle \frac{\sqrt{2}}{6}}$        D.${\displaystyle \frac{\sqrt{3}}{12}}$

                                                     
    5、在长方体${\displaystyle ABCD-A_1{B}_1{C}_1{D}_1}$中，侧面${\displaystyle A_{1}AD{D}_1}$是正方形，${\displaystyle M}$是棱${\displaystyle CD}$的中点，${\displaystyle AM}$与${\displaystyle C{D}_1}$成的角为${\displaystyle \theta }$，若${\displaystyle sin\theta =\frac{\sqrt{78}}{9}}$，则${\displaystyle \frac{A{A}_1}{AB}}$的值为(   )
    A.${\displaystyle \sqrt{2}}$        B.${\displaystyle \frac{2\sqrt{2}}{3}}$        C.${\displaystyle \frac{\sqrt{2}}{2}}$        D.${\displaystyle \frac{\sqrt{3}}{2}}$
    二、填空题
    6、${\displaystyle P、Q}$是半径为${\displaystyle R}$的球面上的两点，它们的球面距离是${\displaystyle \frac{\pi }{2}R}$，则过${\displaystyle P、Q}$的平面中，与球心最大的距离是______
    7、(06·浙江14)正四面体${\displaystyle ABCD}$的棱长为${\displaystyle 1}$，棱平面${\displaystyle \alpha }$，则正四面体上的所有点在平面${\displaystyle \alpha }$内的射影构成的图形面积的取值范围是______
    三、解答题
    8、(04·北京)如图，在正三棱柱${\displaystyle ABC-A_1{B}_1{C}_1}$中，${\displaystyle AB=2}$，${\displaystyle A{A}_1=2}$，由顶点${\displaystyle B}$沿棱柱侧面经过棱${\displaystyle A{A}_1}$到顶点${\displaystyle C_1}$的最短路线与${\displaystyle A{A}_1}$的交点记为${\displaystyle M}$，求：

(1)三棱柱的侧面展开图的对角线长     (2)该最短路线的长及${\displaystyle \frac{A_{1}M}{AM}}$的值     (3)平面${\displaystyle C_{1}MB}$与平面${\displaystyle ABC}$所成二面角(锐角)的大小

    9、(05·江西)如图，在长方体${\displaystyle ABCD-A_1{B}_1{C}_1{D}_1}$中，${\displaystyle AD=A{A}_1=1}$，${\displaystyle AB=2}$，点${\displaystyle E}$在棱${\displaystyle AB}$上移动

(1)求证：${\displaystyle D_{1}E\perp A_{1}D}$     (2)当${\displaystyle E}$为${\displaystyle AB}$的中点时，求点${\displaystyle E}$到平面${\displaystyle AC{D}_1}$的距离     (3)${\displaystyle AE}$为何值时，二面角${\displaystyle D_1-EC-D}$的大小为${\displaystyle \frac{\pi }{4}}$

    10、(探究创新题)正三棱柱${\displaystyle ABC-A_1{B}_1{C}_1}$的所有棱长为${\displaystyle 2}$，点${\displaystyle P}$是侧棱${\displaystyle A{A}_1}$上任意一点

(1)求证：${\displaystyle B_{1}P}$不可能与平面${\displaystyle AC{C}_1{A}_1}$垂直     (2)当${\displaystyle B{C}_1\perp B_{1}P}$时，求线段${\displaystyle AP}$的长     (3)在(2)的条件下，求二面角${\displaystyle C-B_{1}P-C-1}$的大小