di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Solid Geometry


Question: $如图，在棱长为a的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别是棱DD_{1}，AB的中点．$![Image](https://cdntk.53zaixian.com/tq/1703664421493010434/img/compression/M_d2b97d086c949171046ea814b56a1e7a.jpg)
$(1)若平面PQC与直线AA_{1}交于点R，求\frac{AR}{A_1R}的值；$

Solution:
$如图所示,$![Image](https://cdntk.53zaixian.com/tq/1703664421857914882/img/compression/M_7ad62414085dc578c26bae57326fe4a3.jpg)$因为平面ABB_{1}A_{1}\parallel 平面CDD_{1}C_{1}，平面ABB_{1}A_{1}\cap 平面PQC=RQ，平面CDD_{1}C_{1}\cap 平面PQC=PC，所以RQ\parallel PC，所以\triangle ARQ∽\triangle DPC，则\frac{AR}{DP}=\frac{AQ}{DC}，又DC=a，DP=\frac{1}{2}a，AQ=\frac{1}{2}a，所以\frac{AR}{\frac{1}{2}a}=\frac{\frac{1}{2}a}{a}，所以AR=\frac{1}{4}a，则A_{1}R=\frac{3}{4}a，所以\frac{AR}{A_1R}=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，在棱长为a的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别是棱DD_{1}，AB的中点．$<img_2775>
$(1)若平面PQC与直线AA_{1}交于点R，求\frac{AR}{A_1R}的值；$

Solution:
$如图所示,$<img_2429>$因为平面ABB_{1}A_{1}\parallel 平面CDD_{1}C_{1}，平面ABB_{1}A_{1}\cap 平面PQC=RQ，平面CDD_{1}C_{1}\cap 平面PQC=PC，所以RQ\parallel PC，所以\triangle ARQ∽\triangle DPC，则\frac{AR}{DP}=\frac{AQ}{DC}，又DC=a，DP=\frac{1}{2}a，AQ=\frac{1}{2}a，所以\frac{AR}{\frac{1}{2}a}=\frac{\frac{1}{2}a}{a}，所以AR=\frac{1}{4}a，则A_{1}R=\frac{3}{4}a，所以\frac{AR}{A_1R}=\frac{1}{3}.$

Final answer: $\frac{1}{3}$
Subfield: Solid Geometry


Question: $如图，在棱长为a的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别是棱DD_{1}，AB的中点．$![Image](https://cdntk.53zaixian.com/tq/1703664421493010434/img/compression/M_d2b97d086c949171046ea814b56a1e7a.jpg)
$(2)设M为棱CC_{1}上一点且CM＝\lambda CC_{1}(\lambda \in R)，若BM\parallel 平面PQC，求\lambda 的值．$

Solution:
$取AA_{1}的中点E，则R为AE的中点，连接EM,BE，则BE\parallel RQ，因为RQ\subset 平面PCQ，BE\not\subset 平面PCQ，所以BE\parallel 平面PCQ．又BM\parallel 平面PCQ，BM\cap BE=B，BM,BE\subset 平面BME，所以平面BME\parallel 平面PCQ.设DD_{1}\cap 平面BME=F，连接EF,FM，因为平面BME\parallel 平面PCQ，平面BME\cap 平面CDD_{1}C_{1}＝FM，平面PCQ\cap 平面CDD_{1}C_{1}＝PC，所以FM\parallel PC，又CM\parallel PF，所以四边形CPFM为平行四边形，同理，四边形PREF也是平行四边形，所以CM=PF=ER=\frac{1}{4}a，所以\lambda =\frac{CM}{CC_1}=\frac{\frac{1}{4}a}{a}=\frac{1}{4}．$

Final answer: $\frac{1}{4}$
Subfield: Solid Geometry


Question: $如图，在棱长为a的正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别是棱DD_{1}，AB的中点．$<img_2775>
$(2)设M为棱CC_{1}上一点且CM＝\lambda CC_{1}(\lambda \in R)，若BM\parallel 平面PQC，求\lambda 的值．$

Solution:
$取AA_{1}的中点E，则R为AE的中点，连接EM,BE，则BE\parallel RQ，因为RQ\subset 平面PCQ，BE\not\subset 平面PCQ，所以BE\parallel 平面PCQ．又BM\parallel 平面PCQ，BM\cap BE=B，BM,BE\subset 平面BME，所以平面BME\parallel 平面PCQ.设DD_{1}\cap 平面BME=F，连接EF,FM，因为平面BME\parallel 平面PCQ，平面BME\cap 平面CDD_{1}C_{1}＝FM，平面PCQ\cap 平面CDD_{1}C_{1}＝PC，所以FM\parallel PC，又CM\parallel PF，所以四边形CPFM为平行四边形，同理，四边形PREF也是平行四边形，所以CM=PF=ER=\frac{1}{4}a，所以\lambda =\frac{CM}{CC_1}=\frac{\frac{1}{4}a}{a}=\frac{1}{4}．$

Final answer: $\frac{1}{4}$
Subfield: Solid Geometry


Question: $已知圆柱O_{1}O_{2}的体积为2\pi ，侧面积为4\pi ．$<img_2370>
$(1)求圆柱O_{1}O_{2}的底面半径和母线长；$

Solution:
$设圆柱O_{1}O_{2}的底面半径为r，母线长为l，则由圆柱O_{1}O_{2}的体积为2\pi ，侧面积为4\pi 得\left\{\begin{matrix}\pi r^2l=2\pi ,\\ 2\pi rl=4\pi ,\end{matrix}\right.解得\left\{\begin{matrix}l=2,\\ r=1,\end{matrix}\right.所以圆柱O_{1}O_{2}的底面半径为1，母线长为2.$

Final answer: 1,2
Subfield: Solid Geometry


Question: $已知圆柱O_{1}O_{2}的体积为2\pi ，侧面积为4\pi ．$![Image](https://cdntk.53zaixian.com/tq/1703664354698719234/img/compression/M_5d5d75456bac650f18bbc92706e6de59.jpg)
$(1)求圆柱O_{1}O_{2}的底面半径和母线长；$

Solution:
$设圆柱O_{1}O_{2}的底面半径为r，母线长为l，则由圆柱O_{1}O_{2}的体积为2\pi ，侧面积为4\pi 得\left\{\begin{matrix}\pi r^2l=2\pi ,\\ 2\pi rl=4\pi ,\end{matrix}\right.解得\left\{\begin{matrix}l=2,\\ r=1,\end{matrix}\right.所以圆柱O_{1}O_{2}的底面半径为1，母线长为2.$

Final answer: 1,2
Subfield: Solid Geometry


Question: $已知圆柱O_{1}O_{2}的体积为2\pi ，侧面积为4\pi ．$![Image](https://cdntk.53zaixian.com/tq/1703664354698719234/img/compression/M_5d5d75456bac650f18bbc92706e6de59.jpg)
$(2)以上底面圆的圆心O_{1}和下底面圆构成圆锥O_{1}O_{2}，求此圆锥的表面积．$

Solution:
$由(1)知，圆锥O_{1}O_{2}的底面半径为1，高为2，所以圆锥O_{1}O_{2}的母线长为\sqrt{1^2+2^2}=\sqrt{5}，所以圆锥O_{1}O_{2}的表面积为\pi \times 1^{2}+\pi \times 1\times \sqrt{5}=(1+\sqrt{5})\pi ．$

Final answer: $(1+\sqrt 5)\pi $
Subfield: Solid Geometry


Question: $已知圆柱O_{1}O_{2}的体积为2\pi ，侧面积为4\pi ．$<img_2370>
$(2)以上底面圆的圆心O_{1}和下底面圆构成圆锥O_{1}O_{2}，求此圆锥的表面积．$

Solution:
$由(1)知，圆锥O_{1}O_{2}的底面半径为1，高为2，所以圆锥O_{1}O_{2}的母线长为\sqrt{1^2+2^2}=\sqrt{5}，所以圆锥O_{1}O_{2}的表面积为\pi \times 1^{2}+\pi \times 1\times \sqrt{5}=(1+\sqrt{5})\pi ．$

Final answer: $(1+\sqrt 5)\pi $
Subfield: Solid Geometry


Question: $如图所示，已知母线长为4\sqrt{3}的圆锥AO，其侧面展开图为半圆.$![Image](https://cdntk.53zaixian.com/tq/1703664358385512449/img/compression/M_4a437b6bcdbbf3c3a4e73aeb5d29150d.jpg)
$(1)求圆锥的底面积；$

Solution:
$由题意得圆锥的侧面展开图是以4\sqrt{3}为半径的半圆，设OB=R，因为半圆的弧长为4\sqrt{3}\pi ，所以2\pi R=4\sqrt{3}\pi ，所以R=2\sqrt{3}，故圆锥的底面积为\pi R^{2}=12\pi .$

Final answer: $12\pi $
Subfield: Solid Geometry


Question: $如图所示，已知母线长为4\sqrt{3}的圆锥AO，其侧面展开图为半圆.$<img_2693>
$(1)求圆锥的底面积；$

Solution:
$由题意得圆锥的侧面展开图是以4\sqrt{3}为半径的半圆，设OB=R，因为半圆的弧长为4\sqrt{3}\pi ，所以2\pi R=4\sqrt{3}\pi ，所以R=2\sqrt{3}，故圆锥的底面积为\pi R^{2}=12\pi .$

Final answer: $12\pi $
Subfield: Solid Geometry


Question: $如图所示，已知母线长为4\sqrt{3}的圆锥AO，其侧面展开图为半圆.$![Image](https://cdntk.53zaixian.com/tq/1703664358385512449/img/compression/M_4a437b6bcdbbf3c3a4e73aeb5d29150d.jpg)
$(2)在该圆锥内按如图所示的方式放置一个圆柱，当圆柱的侧面积最大时，求圆柱的体积.$

Solution:
$设圆柱的高OO_{1}=h，OD=r，在Rt\triangle AOB中，AO=\sqrt{AB^2-OB^2}=6，易知\triangle AO_{1}D_{1}∽\triangle AOB，所以\frac{AO_{1}}{AO}=\frac{O_1D_1}{OB}，即\frac{6-h}{6}=\frac{r}{2\sqrt{3}}，所以h=6-\sqrt{3}r，所以圆柱的侧面积S=2\pi rh=2\pi r(6-\sqrt{3}r)=-2\sqrt{3}\pi \times (r^{2}-2\sqrt{3}r)=-2\sqrt{3}\pi (r-\sqrt{3})^{2}+6\sqrt{3}\pi ，易得0所以当r=\sqrt{3}时，圆柱的侧面积最大，此时h=3,所以V_{圆柱}=\pi r^{2}h=9\pi .$

Final answer: $9\pi $
Subfield: Solid Geometry


Question: $如图所示，已知母线长为4\sqrt{3}的圆锥AO，其侧面展开图为半圆.$<img_2693>
$(2)在该圆锥内按如图所示的方式放置一个圆柱，当圆柱的侧面积最大时，求圆柱的体积.$

Solution:
$设圆柱的高OO_{1}=h，OD=r，在Rt\triangle AOB中，AO=\sqrt{AB^2-OB^2}=6，易知\triangle AO_{1}D_{1}∽\triangle AOB，所以\frac{AO_{1}}{AO}=\frac{O_1D_1}{OB}，即\frac{6-h}{6}=\frac{r}{2\sqrt{3}}，所以h=6-\sqrt{3}r，所以圆柱的侧面积S=2\pi rh=2\pi r(6-\sqrt{3}r)=-2\sqrt{3}\pi \times (r^{2}-2\sqrt{3}r)=-2\sqrt{3}\pi (r-\sqrt{3})^{2}+6\sqrt{3}\pi ，易得0所以当r=\sqrt{3}时，圆柱的侧面积最大，此时h=3,所以V_{圆柱}=\pi r^{2}h=9\pi .$

Final answer: $9\pi $
Subfield: Solid Geometry


Question: $如图所示，以线段AB为直径的半圆上有一点C，且BC=1，AC=\sqrt{3}，将图中阴影部分绕直线AB旋转180^\circ 得到一个几何体．$![Image](https://cdntk.53zaixian.com/tq/1703664360985980931/img/compression/M_d6024bd31930eefeaad5b3fa178641c0.jpg)
$(1)求形成的几何体的体积；$

Solution:
$所得几何体为半球挖掉两个半圆锥,过点C作CO_{1}\perp AB，垂足为O_{1}，如图.$![Image](https://cdntk.53zaixian.com/tq/1703664361275387907/img/compression/M_72bacd73330c33c3d7ba48cadddc7e46.jpg)$因为AC=\sqrt{3}，BC=1，所以AB=2，故CO_{1}=\frac{\sqrt{3}}{2}.以直线AB为轴将Rt\triangle AO_{1}C旋转180^\circ 得到一个半圆锥，其体积V_{1}=\frac{1}{2}\times \frac{1}{3}\pi \times CO^2_1\times AO_{1}，以直线AB为轴将Rt\triangle BO_{1}C旋转180^\circ 得到一个半圆锥，其体积V_{2}=\frac{1}{2}\times \frac{1}{3}\pi \times CO^2_1\times BO_{1}，则V_{1}+V_{2}=\frac{1}{6}\pi \times CO^2_1\times AO_{1}+\frac{1}{6}\pi \times CO^2_1\times BO_{1}=\frac{1}{6}\pi \times CO^2_1\times BA=\frac{1}{6}\pi \times \frac{3}{4}\times 2=\frac{\pi }{4}，以直线AB为轴将半圆旋转180^\circ 得到一个半球，其体积V_{3}=\frac{1}{2}\times \frac{4}{3}\pi \times 1^{3}=\frac{2}{3}\pi ，故V_{几何体}=V_{3}-V_{1}-V_{2}=\frac{2}{3}\pi -\frac{\pi }{4}=\frac{5}{12}\pi .$

Final answer: $\frac{5}{12}\pi$
Subfield: Solid Geometry


Question: $如图所示，以线段AB为直径的半圆上有一点C，且BC=1，AC=\sqrt{3}，将图中阴影部分绕直线AB旋转180^\circ 得到一个几何体．$<img_2244>
$(1)求形成的几何体的体积；$

Solution:
$所得几何体为半球挖掉两个半圆锥,过点C作CO_{1}\perp AB，垂足为O_{1}，如图.$<img_2568>$因为AC=\sqrt{3}，BC=1，所以AB=2，故CO_{1}=\frac{\sqrt{3}}{2}.以直线AB为轴将Rt\triangle AO_{1}C旋转180^\circ 得到一个半圆锥，其体积V_{1}=\frac{1}{2}\times \frac{1}{3}\pi \times CO^2_1\times AO_{1}，以直线AB为轴将Rt\triangle BO_{1}C旋转180^\circ 得到一个半圆锥，其体积V_{2}=\frac{1}{2}\times \frac{1}{3}\pi \times CO^2_1\times BO_{1}，则V_{1}+V_{2}=\frac{1}{6}\pi \times CO^2_1\times AO_{1}+\frac{1}{6}\pi \times CO^2_1\times BO_{1}=\frac{1}{6}\pi \times CO^2_1\times BA=\frac{1}{6}\pi \times \frac{3}{4}\times 2=\frac{\pi }{4}，以直线AB为轴将半圆旋转180^\circ 得到一个半球，其体积V_{3}=\frac{1}{2}\times \frac{4}{3}\pi \times 1^{3}=\frac{2}{3}\pi ，故V_{几何体}=V_{3}-V_{1}-V_{2}=\frac{2}{3}\pi -\frac{\pi }{4}=\frac{5}{12}\pi .$

Final answer: $\frac{5}{12}\pi$
Subfield: Solid Geometry


Question: $如图所示，以线段AB为直径的半圆上有一点C，且BC=1，AC=\sqrt{3}，将图中阴影部分绕直线AB旋转180^\circ 得到一个几何体．$![Image](https://cdntk.53zaixian.com/tq/1703664360985980931/img/compression/M_d6024bd31930eefeaad5b3fa178641c0.jpg)
$(2)求形成的几何体的表面积．$

Solution:
$以直线AB为轴将Rt\triangle AO_{1}C旋转180^\circ 得到一个半圆锥，其侧面积S_{1}=\frac{1}{2}\times \pi \times \frac{\sqrt{3}}{2}\times \sqrt{3}=\frac{3}{4}\pi ，以直线AB为轴将Rt\triangle BO_{1}C旋转180^\circ 得到一个半圆锥，其侧面积S_{2}=\frac{1}{2}\times \pi \times \frac{\sqrt{3}}{2}\times 1=\frac{\sqrt{3}}{4}\pi ，以直线AB为轴将半圆旋转180^\circ 得到一个半球，其表面积S_{3}=\frac{1}{2}\times 4\pi \times 1^{2}=2\pi ，正面为一个圆去掉两个三角形，其面积S_{4}=\pi \times 1^{2}-2\times \frac{1}{2}\times 1\times \sqrt{3}=\pi -\sqrt{3}，故S_{几何体}=S_{1}+S_{2}+S_{3}+S_{4}=\frac{3}{4}\pi +\frac{\sqrt{3}}{4}\pi +2\pi +\pi -\sqrt{3}=\frac{15+\sqrt{3}}{4}\pi -\sqrt{3}.$

Final answer: $\frac{15+\sqrt{3}}{4}\pi -\sqrt{3}$
Subfield: Solid Geometry


Question: $如图所示，以线段AB为直径的半圆上有一点C，且BC=1，AC=\sqrt{3}，将图中阴影部分绕直线AB旋转180^\circ 得到一个几何体．$<img_2244>
$(2)求形成的几何体的表面积．$

Solution:
$以直线AB为轴将Rt\triangle AO_{1}C旋转180^\circ 得到一个半圆锥，其侧面积S_{1}=\frac{1}{2}\times \pi \times \frac{\sqrt{3}}{2}\times \sqrt{3}=\frac{3}{4}\pi ，以直线AB为轴将Rt\triangle BO_{1}C旋转180^\circ 得到一个半圆锥，其侧面积S_{2}=\frac{1}{2}\times \pi \times \frac{\sqrt{3}}{2}\times 1=\frac{\sqrt{3}}{4}\pi ，以直线AB为轴将半圆旋转180^\circ 得到一个半球，其表面积S_{3}=\frac{1}{2}\times 4\pi \times 1^{2}=2\pi ，正面为一个圆去掉两个三角形，其面积S_{4}=\pi \times 1^{2}-2\times \frac{1}{2}\times 1\times \sqrt{3}=\pi -\sqrt{3}，故S_{几何体}=S_{1}+S_{2}+S_{3}+S_{4}=\frac{3}{4}\pi +\frac{\sqrt{3}}{4}\pi +2\pi +\pi -\sqrt{3}=\frac{15+\sqrt{3}}{4}\pi -\sqrt{3}.$

Final answer: $\frac{15+\sqrt{3}}{4}\pi -\sqrt{3}$
Subfield: Solid Geometry


Question: $如图，在正四面体S-ABC中，AB=4，E，F，R分别是SB，SC，SA的中点，分别取SE，SF的中点M，N，Q为平面SBC内一点.$<img_2248>
$(1)求证：平面MNR\parallel 平面AEF；$

Solution:
$证明：\because M，N分别为SE，SF的中点，\therefore MN\parallel EF，又\because MN\not\subset 平面AEF，EF\subset 平面AEF，\therefore MN\parallel 平面AEF.\because R，M分别为SA，SE的中点，\therefore RM\parallel AE，又\because RM\not\subset 平面AEF，AE\subset 平面AEF，\therefore RM\parallel 平面AEF.\because MN\cap RM=M，MN，RM\subset 平面MNR，\therefore 平面MNR\parallel 平面AEF.$
Subfield: Solid Geometry


Question: $如图，在正四面体S-ABC中，AB=4，E，F，R分别是SB，SC，SA的中点，分别取SE，SF的中点M，N，Q为平面SBC内一点.$![Image](https://cdntk.53zaixian.com/tq/1703664412679761922/img/compression/M_7a03c9d29a9d292556cf7786be0e4e29.jpg)
$(1)求证：平面MNR\parallel 平面AEF；$

Solution:
$证明：\because M，N分别为SE，SF的中点，\therefore MN\parallel EF，又\because MN\not\subset 平面AEF，EF\subset 平面AEF，\therefore MN\parallel 平面AEF.\because R，M分别为SA，SE的中点，\therefore RM\parallel AE，又\because RM\not\subset 平面AEF，AE\subset 平面AEF，\therefore RM\parallel 平面AEF.\because MN\cap RM=M，MN，RM\subset 平面MNR，\therefore 平面MNR\parallel 平面AEF.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在正四面体S-ABC中，AB=4，E，F，R分别是SB，SC，SA的中点，分别取SE，SF的中点M，N，Q为平面SBC内一点.$![Image](https://cdntk.53zaixian.com/tq/1703664412679761922/img/compression/M_7a03c9d29a9d292556cf7786be0e4e29.jpg)
$(2)若RQ\parallel 平面AEF，求线段RQ长度的最小值.$

Solution:
$由(1)知，平面MNR\parallel 平面AEF，若平面SBC内存在一点Q，使RQ\parallel 平面AEF，则Q在线段MN上，当RQ\perp MN时，线段RQ最短.\because E，F分别为SB，SC的中点，M，N分别为SE，SF的中点，\therefore MN=\frac{1}{2}EF=\frac{1}{4}BC=1.\because AS=AB=SB=4，\therefore AE\perp SB，AE=\sqrt{AB^2-BE^2}=2\sqrt{3}，\because R，M分别为SA，SE的中点，\therefore RM=\frac{1}{2}AE=\sqrt{3}，同理，RN=\sqrt{3}，\therefore 当Q为MN的中点时，RQ\perp MN，|RQ|取最小值，\therefore |RQ|_{min}=\sqrt{RM^2-MQ^2}=\sqrt{(\sqrt{3})^2-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{11}}{2}.$![Image](https://cdntk.53zaixian.com/tq/1703664413149523969/img/compression/M_8fa9fddbc95baa69722bebbad91c880a.jpg)

Final answer: $\frac{\sqrt{11}}{2}$
Subfield: Solid Geometry


Question: $如图，在正四面体S-ABC中，AB=4，E，F，R分别是SB，SC，SA的中点，分别取SE，SF的中点M，N，Q为平面SBC内一点.$<img_2248>
$(2)若RQ\parallel 平面AEF，求线段RQ长度的最小值.$

Solution:
$由(1)知，平面MNR\parallel 平面AEF，若平面SBC内存在一点Q，使RQ\parallel 平面AEF，则Q在线段MN上，当RQ\perp MN时，线段RQ最短.\because E，F分别为SB，SC的中点，M，N分别为SE，SF的中点，\therefore MN=\frac{1}{2}EF=\frac{1}{4}BC=1.\because AS=AB=SB=4，\therefore AE\perp SB，AE=\sqrt{AB^2-BE^2}=2\sqrt{3}，\because R，M分别为SA，SE的中点，\therefore RM=\frac{1}{2}AE=\sqrt{3}，同理，RN=\sqrt{3}，\therefore 当Q为MN的中点时，RQ\perp MN，|RQ|取最小值，\therefore |RQ|_{min}=\sqrt{RM^2-MQ^2}=\sqrt{(\sqrt{3})^2-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{11}}{2}.$<img_1782>

Final answer: $\frac{\sqrt{11}}{2}$
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别为面对角线BD，CD_{1}上的点，且\frac{CQ}{QD_1}=\frac{BP}{PD}=\frac{2}{3}．$![Image](https://cdntk.53zaixian.com/tq/1703664413896110082/img/compression/M_715fe99988010d7f8956f2494bb020aa.jpg)
$(1)求证：PQ\parallel 平面A_{1}D_{1}DA；$

Solution:
$证明：如图，连接CP并延长，与DA的延长线交于点M，连接D_{1}M.$![Image](https://cdntk.53zaixian.com/tq/1703664413514428419/img/compression/M_6771e809d2d7cf23f9f2529faaac5bdf.jpg)$因为四边形ABCD为正方形，所以BC\parallel AD，所以\triangle PBC∽\triangle PDM，所以\frac{CP}{PM}=\frac{BP}{PD}.又因为\frac{CQ}{QD_1}=\frac{BP}{PD}，所以\frac{CQ}{QD_1}=\frac{CP}{PM}，所以PQ\parallel MD_{1}．又MD_{1}\subset 平面A_{1}D_{1}DA，PQ\not\subset 平面A_{1}D_{1}DA，所以PQ\parallel 平面A_{1}D_{1}DA.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别为面对角线BD，CD_{1}上的点，且\frac{CQ}{QD_1}=\frac{BP}{PD}=\frac{2}{3}．$<img_3129>
$(1)求证：PQ\parallel 平面A_{1}D_{1}DA；$

Solution:
$证明：如图，连接CP并延长，与DA的延长线交于点M，连接D_{1}M.$<img_2525>$因为四边形ABCD为正方形，所以BC\parallel AD，所以\triangle PBC∽\triangle PDM，所以\frac{CP}{PM}=\frac{BP}{PD}.又因为\frac{CQ}{QD_1}=\frac{BP}{PD}，所以\frac{CQ}{QD_1}=\frac{CP}{PM}，所以PQ\parallel MD_{1}．又MD_{1}\subset 平面A_{1}D_{1}DA，PQ\not\subset 平面A_{1}D_{1}DA，所以PQ\parallel 平面A_{1}D_{1}DA.$
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别为面对角线BD，CD_{1}上的点，且\frac{CQ}{QD_1}=\frac{BP}{PD}=\frac{2}{3}．$![Image](https://cdntk.53zaixian.com/tq/1703664413896110082/img/compression/M_715fe99988010d7f8956f2494bb020aa.jpg)
$(2)若R是AB上的点，\frac{AR}{AB}的值为多少时，能使平面PQR\parallel 平面A_{1}D_{1}DA？请给出证明．$

Solution:
$当\frac{AR}{AB}=\frac{3}{5}时，能使平面PQR\parallel 平面A_{1}D_{1}DA.证明如下：因为\frac{AR}{AB}=\frac{3}{5}，所以\frac{BR}{RA}=\frac{2}{3}，故\frac{BR}{RA}=\frac{BP}{PD}，所以PR\parallel DA.因为DA\subset 平面A_{1}D_{1}DA，PR\not\subset 平面A_{1}D_{1}DA，所以PR\parallel 平面A_{1}D_{1}DA.由(1)知PQ\parallel 平面A_{1}D_{1}DA,因为PQ\cap PR=P，PQ,PR\subset 平面PQR,所以平面PQR\parallel 平面A_{1}D_{1}DA.$![Image](https://cdntk.53zaixian.com/tq/1703664413896110082/img/compression/M_920f2d4de609498c6be598f4f7115906.jpg)

Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $已知正方体ABCD-A_{1}B_{1}C_{1}D_{1}中，P，Q分别为面对角线BD，CD_{1}上的点，且\frac{CQ}{QD_1}=\frac{BP}{PD}=\frac{2}{3}．$<img_3129>
$(2)若R是AB上的点，\frac{AR}{AB}的值为多少时，能使平面PQR\parallel 平面A_{1}D_{1}DA？请给出证明．$

Solution:
$当\frac{AR}{AB}=\frac{3}{5}时，能使平面PQR\parallel 平面A_{1}D_{1}DA.证明如下：因为\frac{AR}{AB}=\frac{3}{5}，所以\frac{BR}{RA}=\frac{2}{3}，故\frac{BR}{RA}=\frac{BP}{PD}，所以PR\parallel DA.因为DA\subset 平面A_{1}D_{1}DA，PR\not\subset 平面A_{1}D_{1}DA，所以PR\parallel 平面A_{1}D_{1}DA.由(1)知PQ\parallel 平面A_{1}D_{1}DA,因为PQ\cap PR=P，PQ,PR\subset 平面PQR,所以平面PQR\parallel 平面A_{1}D_{1}DA.$<img_1572>

Final answer: $\frac{3}{5}$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}的所有棱长都是2，AA_{1}\perp 平面ABC，D是AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664484742098946/img/compression/M_948e51f8098890f407f8845822bcc9cc.jpg)
$(1)证明:BD\perp 平面A_{1}ACC_{1};$

Solution:
$证明：因为AA_{1}\perp 平面ABC，BD\subset 平面ABC，所以AA_{1}\perp BD.因为AB=AC=BC，D是AC中点，所以BD\perp AC.因为AC\cap AA_{1}=A，AC,AA_{1}\subset 平面A_{1}ACC_{1}，所以BD\perp 平面A_{1}ACC_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}的所有棱长都是2，AA_{1}\perp 平面ABC，D是AC的中点.$<img_2036>
$(1)证明:BD\perp 平面A_{1}ACC_{1};$

Solution:
$证明：因为AA_{1}\perp 平面ABC，BD\subset 平面ABC，所以AA_{1}\perp BD.因为AB=AC=BC，D是AC中点，所以BD\perp AC.因为AC\cap AA_{1}=A，AC,AA_{1}\subset 平面A_{1}ACC_{1}，所以BD\perp 平面A_{1}ACC_{1}.$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}的所有棱长都是2，AA_{1}\perp 平面ABC，D是AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664484742098946/img/compression/M_948e51f8098890f407f8845822bcc9cc.jpg)
$(2)求异面直线B_{1}C_{1}与A_{1}D所成角的余弦值;$

Solution:
$取AB的中点F，连接DF，A_{1}F，则DF\parallel BC，又BC\parallel B_{1}C_{1}，所以DF\parallel B_{1}C_{1}，所以异面直线B_{1}C_{1}与A_{1}D所成的角为\angle A_{1}DF(或其补角).在\triangle A_{1}DF中，DF=\frac{1}{2}BC=1，A_{1}D=\sqrt{2^2+1^2}=\sqrt{5}，A_{1}F=\sqrt{2^2+1^2}=\sqrt{5}，所以由余弦定理的推论得\\cos  \angle A_{1}DF=\frac{5+1-5}{2\times 1\times \sqrt{5}}=\frac{\sqrt{5}}{10}，故异面直线B_{1}C_{1}与A_{1}D所成角的余弦值为\frac{\sqrt{5}}{10}.$![Image](https://cdntk.53zaixian.com/tq/1703664485107003394/img/compression/M_c97412babf646bacd01a1f53ed6f612d.jpg)

Final answer: $\frac{\sqrt{5}}{10}$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}的所有棱长都是2，AA_{1}\perp 平面ABC，D是AC的中点.$<img_2036>
$(2)求异面直线B_{1}C_{1}与A_{1}D所成角的余弦值;$

Solution:
$取AB的中点F，连接DF，A_{1}F，则DF\parallel BC，又BC\parallel B_{1}C_{1}，所以DF\parallel B_{1}C_{1}，所以异面直线B_{1}C_{1}与A_{1}D所成的角为\angle A_{1}DF(或其补角).在\triangle A_{1}DF中，DF=\frac{1}{2}BC=1，A_{1}D=\sqrt{2^2+1^2}=\sqrt{5}，A_{1}F=\sqrt{2^2+1^2}=\sqrt{5}，所以由余弦定理的推论得\\cos  \angle A_{1}DF=\frac{5+1-5}{2\times 1\times \sqrt{5}}=\frac{\sqrt{5}}{10}，故异面直线B_{1}C_{1}与A_{1}D所成角的余弦值为\frac{\sqrt{5}}{10}.$<img_2218>

Final answer: $\frac{\sqrt{5}}{10}$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}的所有棱长都是2，AA_{1}\perp 平面ABC，D是AC的中点.$<img_2036>
$(3)求点B_{1}到平面A_{1}BD的距离.$

Solution:
$连接B_{1}D，易知V_{B_{1}-A_{1}BD}=V_{D-A_{1}B_{1}B}.过点D作DH\perp AB于H，因为AA_{1}\perp 平面ABC，HD\subset 平面ABC，所以HD\perp AA_{1}，因为AA_{1}\cap AB=A，AA_{1},AB\subset 平面ABB_{1}A_{1}，所以HD\perp 平面ABB_{1}A_{1}.易得HD=AD\\sin   60^\circ =\frac{\sqrt{3}}{2},故V_{D-A_{1}B_{1}B}=\frac{1}{3}\times S_{\triangle A_{1}B_{1}B}\times HD=\frac{1}{3}\times 2\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}，由(1)知BD\perp 平面A_{1}ACC_{1}，所以BD\perp A_{1}D，故S_{\triangle A_{1}BD}=\frac{1}{2}\times \sqrt{5}\times \sqrt{3}=\frac{\sqrt{15}}{2}，设B_{1}到平面A_{1}BD的距离为d，则V_{B_{1}-A_{1}BD}=\frac{1}{3}\times S_{\triangle A_{1}BD}\times d=\frac{1}{3}\times \frac{\sqrt{15}}{2}d=\frac{\sqrt{3}}{3}，解得d=\frac{2\sqrt{5}}{5}.故B_{1}到平面A_{1}BD的距离为\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Solid Geometry


Question: $如图，三棱柱ABC-A_{1}B_{1}C_{1}的所有棱长都是2，AA_{1}\perp 平面ABC，D是AC的中点.$![Image](https://cdntk.53zaixian.com/tq/1703664484742098946/img/compression/M_948e51f8098890f407f8845822bcc9cc.jpg)
$(3)求点B_{1}到平面A_{1}BD的距离.$

Solution:
$连接B_{1}D，易知V_{B_{1}-A_{1}BD}=V_{D-A_{1}B_{1}B}.过点D作DH\perp AB于H，因为AA_{1}\perp 平面ABC，HD\subset 平面ABC，所以HD\perp AA_{1}，因为AA_{1}\cap AB=A，AA_{1},AB\subset 平面ABB_{1}A_{1}，所以HD\perp 平面ABB_{1}A_{1}.易得HD=AD\\sin   60^\circ =\frac{\sqrt{3}}{2},故V_{D-A_{1}B_{1}B}=\frac{1}{3}\times S_{\triangle A_{1}B_{1}B}\times HD=\frac{1}{3}\times 2\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}，由(1)知BD\perp 平面A_{1}ACC_{1}，所以BD\perp A_{1}D，故S_{\triangle A_{1}BD}=\frac{1}{2}\times \sqrt{5}\times \sqrt{3}=\frac{\sqrt{15}}{2}，设B_{1}到平面A_{1}BD的距离为d，则V_{B_{1}-A_{1}BD}=\frac{1}{3}\times S_{\triangle A_{1}BD}\times d=\frac{1}{3}\times \frac{\sqrt{15}}{2}d=\frac{\sqrt{3}}{3}，解得d=\frac{2\sqrt{5}}{5}.故B_{1}到平面A_{1}BD的距离为\frac{2\sqrt{5}}{5}.$

Final answer: $\frac{2\sqrt{5}}{5}$
Subfield: Solid Geometry


Question: $如图，在梯形ABCD中，AB\parallel DC，AB=2AD=2DC=4，E，F分别是AD，BC的中点，且EF交AC于点O，现将梯形ABCD沿对角线AC翻折成直二面角P-A_{1}C_{1}-B_{1}.$<img_2626>
$(1)证明：A_{1}B_{1}\parallel 平面O_{1}E_{1}F_{1}；$

Solution:
$证明：易得O是AC的中点，即O_{1}是A_{1}C_{1}的中点，\therefore O_{1}F_{1}是\triangle A_{1}B_{1}C_{1}的中位线，\therefore A_{1}B_{1}\parallel O_{1}F_{1}，又O_{1}F_{1}\subset 平面O_{1}E_{1}F_{1}，A_{1}B_{1}\not\subset 平面O_{1}E_{1}F_{1}，\therefore A_{1}B_{1}\parallel 平面O_{1}E_{1}F_{1}.$
Subfield: Solid Geometry


Question: $如图，在梯形ABCD中，AB\parallel DC，AB=2AD=2DC=4，E，F分别是AD，BC的中点，且EF交AC于点O，现将梯形ABCD沿对角线AC翻折成直二面角P-A_{1}C_{1}-B_{1}.$![Image](https://cdntk.53zaixian.com/tq/1703664491771752451/img/compression/M_7af0263f6480cfc89587f298142bb643.jpg)
$(1)证明：A_{1}B_{1}\parallel 平面O_{1}E_{1}F_{1}；$

Solution:
$证明：易得O是AC的中点，即O_{1}是A_{1}C_{1}的中点，\therefore O_{1}F_{1}是\triangle A_{1}B_{1}C_{1}的中位线，\therefore A_{1}B_{1}\parallel O_{1}F_{1}，又O_{1}F_{1}\subset 平面O_{1}E_{1}F_{1}，A_{1}B_{1}\not\subset 平面O_{1}E_{1}F_{1}，\therefore A_{1}B_{1}\parallel 平面O_{1}E_{1}F_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在梯形ABCD中，AB\parallel DC，AB=2AD=2DC=4，E，F分别是AD，BC的中点，且EF交AC于点O，现将梯形ABCD沿对角线AC翻折成直二面角P-A_{1}C_{1}-B_{1}.$<img_2626>
$(2)证明：C_{1}B_{1}\perp PA_{1}；$

Solution:
$证明：取AB的中点G，连接CG，如图，$<img_1681>$\because AB=2DC=4，AB\parallel DC,\therefore AG\xlongequal{//}DC,\therefore 四边形ADCG为平行四边形，\therefore CG=AD=2，\therefore CG=AG=GB，\therefore \angle ACB=90^\circ ，即C_{1}B_{1}\perp C_{1}A_{1}，由题意知平面PA_{1}C_{1}\perp 平面A_{1}C_{1}B_{1}，又平面PA_{1}C_{1}\cap 平面A_{1}C_{1}B_{1}=A_{1}C_{1}，\therefore C_{1}B_{1}\perp 平面PA_{1}C_{1}，又PA_{1}\subset 平面PA_{1}C_{1},\therefore C_{1}B_{1}\perp PA_{1}.$
Subfield: Solid Geometry


Question: $如图，在梯形ABCD中，AB\parallel DC，AB=2AD=2DC=4，E，F分别是AD，BC的中点，且EF交AC于点O，现将梯形ABCD沿对角线AC翻折成直二面角P-A_{1}C_{1}-B_{1}.$![Image](https://cdntk.53zaixian.com/tq/1703664491771752451/img/compression/M_7af0263f6480cfc89587f298142bb643.jpg)
$(2)证明：C_{1}B_{1}\perp PA_{1}；$

Solution:
$证明：取AB的中点G，连接CG，如图，$![Image](https://cdntk.53zaixian.com/tq/1703664492207960066/img/compression/M_cb57d2145f4c1dcc15cae23e6b0db722.jpg)$\because AB=2DC=4，AB\parallel DC,\therefore AG\xlongequal{//}DC,\therefore 四边形ADCG为平行四边形，\therefore CG=AD=2，\therefore CG=AG=GB，\therefore \angle ACB=90^\circ ，即C_{1}B_{1}\perp C_{1}A_{1}，由题意知平面PA_{1}C_{1}\perp 平面A_{1}C_{1}B_{1}，又平面PA_{1}C_{1}\cap 平面A_{1}C_{1}B_{1}=A_{1}C_{1}，\therefore C_{1}B_{1}\perp 平面PA_{1}C_{1}，又PA_{1}\subset 平面PA_{1}C_{1},\therefore C_{1}B_{1}\perp PA_{1}.$

Final answer: 证明题
Subfield: Solid Geometry


Question: $如图，在梯形ABCD中，AB\parallel DC，AB=2AD=2DC=4，E，F分别是AD，BC的中点，且EF交AC于点O，现将梯形ABCD沿对角线AC翻折成直二面角P-A_{1}C_{1}-B_{1}.$<img_2626>
$(3)若\angle ADC=60^\circ ，试问在线段PB_{1}上是否存在点M，使得三棱锥P-A_{1}C_{1}M的体积为\frac{4}{3}？若存在，求出\frac{PM}{PB_1}的值；若不存在，请说明理由.$

Solution:
$假设存在满足条件的点M.\because \angle ADC=60^\circ ，AD=DC，\therefore \triangle ACD是等边三角形，\therefore AC=2，\angle BAC=60^\circ ，在\triangle ABC中，由余弦定理可得BC^{2}=2^{2}+4^{2}-2\times 2\times 4\times \\cos   60^\circ =12，故BC=2\sqrt{3}.由(2)知，C_{1}B_{1}\perp 平面PA_{1}C_{1}.\therefore V_{P-A_{1}B_{1}C_{1}}=V_{B_{1}-PA_{1}C_{1}}=\frac{1}{3}\times \frac{1}{2}\times 2\times \sqrt{3}\times 2\sqrt{3}=2.又\because V_{P-A_{1}C_{1}M}=\frac{4}{3}，\therefore V_{M-A_{1}B_{1}C_{1}}=2-\frac{4}{3}=\frac{2}{3},\because V_{M-A_{1}B_{1}C_{1}}=\frac{1}{3}V_{P-A_{1}B_{1}C_{1}}，\therefore \frac{MB_1}{PB_1}=\frac{1}{3},\therefore \frac{PM}{PB_1}=\frac{2}{3}.故在线段PB_{1}上存在点M，使得三棱锥P-A_{1}C_{1}M的体积为\frac{4}{3}，此时\frac{PM}{PB_1}=\frac{2}{3}.$<img_2838>

Final answer: $\frac{2}{3}$
Subfield: Solid Geometry


Question: $如图，在梯形ABCD中，AB\parallel DC，AB=2AD=2DC=4，E，F分别是AD，BC的中点，且EF交AC于点O，现将梯形ABCD沿对角线AC翻折成直二面角P-A_{1}C_{1}-B_{1}.$![Image](https://cdntk.53zaixian.com/tq/1703664491771752451/img/compression/M_7af0263f6480cfc89587f298142bb643.jpg)
$(3)若\angle ADC=60^\circ ，试问在线段PB_{1}上是否存在点M，使得三棱锥P-A_{1}C_{1}M的体积为\frac{4}{3}？若存在，求出\frac{PM}{PB_1}的值；若不存在，请说明理由.$

Solution:
$假设存在满足条件的点M.\because \angle ADC=60^\circ ，AD=DC，\therefore \triangle ACD是等边三角形，\therefore AC=2，\angle BAC=60^\circ ，在\triangle ABC中，由余弦定理可得BC^{2}=2^{2}+4^{2}-2\times 2\times 4\times \\cos   60^\circ =12，故BC=2\sqrt{3}.由(2)知，C_{1}B_{1}\perp 平面PA_{1}C_{1}.\therefore V_{P-A_{1}B_{1}C_{1}}=V_{B_{1}-PA_{1}C_{1}}=\frac{1}{3}\times \frac{1}{2}\times 2\times \sqrt{3}\times 2\sqrt{3}=2.又\because V_{P-A_{1}C_{1}M}=\frac{4}{3}，\therefore V_{M-A_{1}B_{1}C_{1}}=2-\frac{4}{3}=\frac{2}{3},\because V_{M-A_{1}B_{1}C_{1}}=\frac{1}{3}V_{P-A_{1}B_{1}C_{1}}，\therefore \frac{MB_1}{PB_1}=\frac{1}{3},\therefore \frac{PM}{PB_1}=\frac{2}{3}.故在线段PB_{1}上存在点M，使得三棱锥P-A_{1}C_{1}M的体积为\frac{4}{3}，此时\frac{PM}{PB_1}=\frac{2}{3}.$![Image](https://cdntk.53zaixian.com/tq/1703664492497367042/img/compression/M_a1997a7d290af8887a9e970818d80f00.jpg)

Final answer: $\frac{2}{3}$