LFM Pure and Mechanics

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gaokao 2015 Q17 13 marks Multi-Part 3D Geometry Problem View
17. (13 points) As shown in the figure, $AA _ { 1 } \perp$ plane $ABC$, $BB _ { 1 } \parallel AA _ { 1 }$, $AB = AC = 3$, $BC = 2 \sqrt { 5 }$, $AA _ { 1 } = \sqrt { 7 }$, $BB _ { 1 } = 2 \sqrt { 7 }$. Points $E$ and $F$ are the midpoints of $BC$ and $A _ { 1 } C$ respectively. (I) Prove that $EF \parallel$ plane $A _ { 1 } B _ { 1 } BA$; (II) Prove that plane $AEA _ { 1 } \perp$ plane $BCB _ { 1 }$. (III) Find the angle between line $A _ { 1 } B _ { 1 }$ and plane $BCB _ { 1 }$. [Figure]
gaokao 2015 Q17 13 marks Multi-Part 3D Geometry Problem View
As shown in the figure, in the quadrangular prism $\mathrm{ABCD} - A_1B_1C_1D_1$, the lateral edge $AA_1 \perp$ base $\mathrm{ABCD}$, $\mathrm{AB} \perp \mathrm{AC}$, $\mathrm{AB} = 1$, $\mathrm{AC} = AA_1 = 2$, $AD = CD = \sqrt{5}$, and points M and N are the midpoints of $B_1C$ and $D_1D$ respectively.
(I) Prove: $\mathrm{MN} \parallel$ plane ABCD
(II) Find the sine value of the dihedral angle $D_1 - AC - B_1$;
(III) Let E be a point on edge $A_1B_1$. If the sine value of the angle between line NE and plane ABCD is $\frac{1}{3}$, find the length of segment $A_1E$.
gaokao 2015 Q17 Multi-Part 3D Geometry Problem View
17. (This question is worth 15 points) As shown in the figure, in the triangular prism $ABC - A _ { 1 } B _ { 1 } C _ { 1 }$, $\angle BAC = 90 ^ { \circ }$ , $AB = AC = 2$ , $A _ { 1 } A = 4$ , the projection of $A _ { 1 }$ on the base plane $ABC$ is the midpoint of $BC$, and $D$ is the midpoint of $B _ { 1 } C _ { 1 }$ . (I) Prove that $A _ { 1 } D \perp$ plane $A _ { 1 } B C _ { 1 }$ ; (II) Find the cosine of the plane angle of the dihedral angle $A _ { 1 } - BD - B _ { 1 }$ . [Figure]
gaokao 2015 Q18 14 marks Multi-Part 3D Geometry Problem View
As shown in the figure, in the triangular pyramid $E - ABC$, plane $EAB \perp$ plane $ABC$, triangle $EAB$ is equilateral, $AC \perp BC$, and $AC = BC = \sqrt { 2 }$. $O$ and $M$ are the midpoints of $AB$ and $EA$ respectively.\n(1) Prove that $EB \parallel$ plane $MOC$.\n(2) Prove that plane $MOC \perp$ plane $EAB$.\n(3) Find the volume of the triangular pyramid $E - ABC$.
gaokao 2015 Q18 Multi-Part 3D Geometry Problem View
18. (This question is worth 12 points) As shown in Figure 4, the right triangular prism $\mathrm { ABC } - \mathrm { A } _ { 1 } \mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$ has an equilateral triangle base with side length 2. $\mathrm { E }$ and $\mathrm { F }$ are the midpoints of $\mathrm { BC }$ and $\mathrm { CC } _ { 1 }$ respectively. (I) Prove that: plane $\mathrm { AEF } \perp$ plane $\mathrm { B } _ { 1 } \mathrm { BCC } _ { 1 }$ (II) If the angle between line $A _ { 1 } C$ and plane $A _ { 1 } A B B _ { 1 }$ is $45 ^ { \circ }$, find the volume of the triangular pyramid F-AEC. [Figure]
gaokao 2015 Q18 12 marks Dihedral Angle Computation View
18. (12 points) As shown in Figure 1, in right trapezoid ABCD, $\mathrm { AD } / / \mathrm { BC } , ~ \angle \mathrm { BAD } = \frac { \pi } { 2 } , \mathrm { AB } = \mathrm { BC } = 1$, $\mathrm { AD } = 2$, E is the midpoint of AD, and O is the intersection of AC and BE. Fold $\triangle \mathrm { ABE }$ along BE to the position $\Delta \mathrm { A } _ { 1 } \mathrm { BE }$ as shown in Figure 2.
[Figure]
Figure 1
[Figure]
Figure 2
(I) Prove that $\mathrm { CD } \perp$ plane $\mathrm { A } _ { 1 } \mathrm { OC }$; (II) If plane $\mathrm { A } _ { 1 } \mathrm { BE } \perp$ plane BCDE, find the cosine of the dihedral angle between plane $\mathrm { A } _ { 1 } \mathrm { BC }$ and plane $\mathrm { A } _ { 1 } \mathrm { CD }$.
gaokao 2015 Q18 Multi-Part 3D Geometry Problem View
18. A net of a cube and a schematic diagram of the cube are shown in the figure. In the cube, let $M$ be the midpoint of $BC$ and $N$ be the midpoint of $GH$.
(1) Mark the letters $F$, $G$, $H$ at the corresponding vertices of the cube (no explanation needed).
(2) Prove: line $MN \parallel$ plane $BDH$.
(3) Find the cosine of the dihedral angle $A - E G - M$. [Figure]
gaokao 2015 Q18 15 marks Multi-Part 3D Geometry Problem View
18. (15 points) As shown in the figure, in the triangular prism $A B C - A _ { 1 } B _ { 1 } C _ { 1 }$ , $\angle \mathrm { ABC } = 90 ^ { \circ } , \mathrm { AB } = \mathrm { AC } = 2 , \mathrm { AA } _ { 1 } = 4$ , the projection of $A _ { 1 }$ on the base plane ABC is the midpoint of BC, and D is the midpoint of $B _ { 1 } C _ { 1 }$.
(1) Prove that $A _ { 1 } \mathrm { D } \perp$ plane $\mathrm { A } _ { 1 } \mathrm { BC }$ ;
(2) Find the sine of the angle between line $\mathrm { A } _ { 1 } \mathrm { B}$ and plane $\mathrm { BB } _ { 1 } \mathrm { C } C _ { 1 }$ . [Figure]
gaokao 2015 Q19 Multi-Part 3D Geometry Problem View
19. (This question is worth 12 points). In the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$, $A B = 16 , B C = 10 , A A _ { 1 } = 8$. Points $E$ and $F$ are on $A _ { 1 } B _ { 1 }$ and $D _ { 1 } C _ { 1 }$ respectively, with $A _ { 1 } E = D _ { 1 } F = 4$. A plane $\alpha$ passes through points $E$ and $F$ and intersects the faces of the rectangular prism, with the intersection lines forming a square. [Figure] (I) Draw this square in the figure (no need to explain the method or reasoning); (II) Find the ratio of the volumes of the two parts that plane $\alpha$ divides the rectangular prism into.
gaokao 2015 Q19 12 marks Multi-Part 3D Geometry Problem View
As shown in the figure, in rectangular prism $\mathrm { ABCD } - \mathrm { A } _ { 1 } \mathrm { B } _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$, we have $\mathrm { AB } = 16 , \mathrm { BC } = 10 , \mathrm { AA } _ { 1 } = 8$. Points $\mathrm { E }$ and $\mathrm { F }$ are on $\mathrm { A } _ { 1 } \mathrm { B } _ { 1 }$ and $\mathrm { D } _ { 1 } \mathrm { C } _ { 1 }$ respectively, with $\mathrm { A } _ { 1 } \mathrm { E } = \mathrm { D } _ { 1 } \mathrm { F }$. A plane $\alpha$ passes through points $E$ and $F$ and intersects the faces of the rectangular prism, with the intersection lines forming a square.
(I) Draw this square in the figure (no need to explain the method or reasoning)
(II) Find the sine of the angle between line
gaokao 2015 Q20 12 marks Multi-Part 3D Geometry Problem View
As shown in the figure for question (20), in the triangular pyramid $\mathrm { P } - \mathrm { ABC }$, plane $\mathrm { PAC } \perp$ plane $\mathrm { ABC }$, $\angle \mathrm { ABC } = \frac { \pi } { 2 }$. Points $D$ and $E$ lie on segment $AC$ with $\mathrm { AD } = \mathrm { DE } = \mathrm { EC } = 2$, $\mathrm { PD } = \mathrm { PC } = 4$. Point $F$ lies on segment $AB$ with $\mathrm { EF } \parallel \mathrm { BC }$ .
(I) Prove that $\mathrm { AB } \perp$ plane $PFE$ .
(II) If the volume of the quadrangular pyramid $\mathrm { P } - \mathrm { DFBC }$ is 7, find the length of segment $BC$.
gaokao 2015 Q21 14 marks Circle-Related Locus Problems View
21. (14 marks) (I) Let point $D ( t , 0 ) ( | t | \leq 2 ) , N \left( x _ { 0 } , y _ { 0 } \right) , M ( x , y )$. According to the problem,
$$\overrightarrow { M D } = 2 \overrightarrow { D N } \text {, and } | \overrightarrow { D N } | = | \overrightarrow { O N } | = 1,$$
[Figure]
so $( t - x , - y ) = 2 \left( x _ { 0 } - t , y _ { 0 } \right)$, and $\left\{ \begin{array} { l } \left( x _ { 0 } - t \right) ^ { 2 } + y _ { 0 } ^ { 2 } = 1 , \\ x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } = 1 . \end{array} \right.$ That is, $\left\{ \begin{array} { l } t - x = 2 x _ { 0 } - 2 t , \\ y = - 2 y _ { 0 } . \end{array} \right.$ and $t \left( t - 2 x _ { 0 } \right) = 0$. Since when point $D$ is fixed, point $N$ is also fixed, $t$ is not identically zero, thus $t = 2 x _ { 0 }$, so $x _ { 0 } = \frac { x } { 4 } , y _ { 0 } = - \frac { y } { 2 }$. Substituting into $x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } = 1$, we obtain $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$, that is, the equation of the required curve $C$ is $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$. (II) (1) When the slope of line $l$ does not exist, line $l$ is $x = 4$ or $x = - 4$, and we have $S _ { \triangle O P Q } = \frac { 1 } { 2 } \times 4 \times 4 = 8$.
(2) When the slope of line $l$ exists, let line $l : y = k x + m \left( k \neq \pm \frac { 1 } { 2 } \right)$. From $\left\{ \begin{array} { l } y = k x + m , \\ x ^ { 2 } + 4 y ^ { 2 } = 16 , \end{array} \right.$ eliminating $y$, we obtain $\left( 1 + 4 k ^ { 2 } \right) x ^ { 2 } + 8 k m x + 4 m ^ { 2 } - 16 = 0$. Since line $l$ always has exactly one common point with ellipse $C$, we have $\Delta = 64 k ^ { 2 } m ^ { 2 } - 4 \left( 1 + 4 k ^ { 2 } \right) \left( 4 m ^ { 2 } - 16 \right) = 0$, that is, $m ^ { 2 } = 16 k ^ { 2 } + 4$.
From $\left\{ \begin{array} { l } y = k x + m , \\ x - 2 y = 0 , \end{array} \right.$ we obtain $P \left( \frac { 2 m } { 1 - 2 k } , \frac { m } { 1 - 2 k } \right)$; similarly, we obtain $Q \left( \frac { - 2 m } { 1 + 2 k } , \frac { m } { 1 + 2 k } \right)$. From the distance from origin $O$ to line $P Q$ being $d = \frac { | m | } { \sqrt { 1 + k ^ { 2 } } }$ and $| P Q | = \sqrt { 1 + k ^ { 2 } } \left| x _ { P } - x _ { Q } \right|$, we obtain $S _ { \triangle O P Q } = \frac { 1 } { 2 } | P Q | \cdot d = \frac { 1 } { 2 } | m | \left| x _ { P } - x _ { Q } \right| = \frac { 1 } { 2 } \cdot | m | \left| \frac { 2 m } { 1 - 2 k } + \frac { 2 m } { 1 + 2 k } \right| = \left| \frac { 2 m ^ { 2 } } { 1 - 4 k ^ { 2 } } \right|$.
Substituting (1) into (2), we obtain $S _ { \triangle O P Q } = \left| \frac { 2 m ^ { 2 } } { 1 - 4 k ^ { 2 } } \right| = 8 \frac { \left| 4 k ^ { 2 } + 1 \right| } { \left| 4 k ^ { 2 } - 1 \right| }$. When $k ^ { 2 } > \frac { 1 } { 4 }$, $S _ { \triangle O P Q } = 8 \left( \frac { 4 k ^ { 2 } + 1 } { 4 k ^ { 2 } - 1 } \right) = 8 \left( 1 + \frac { 2 } { 4 k ^ { 2 } - 1 } \right) > 8$; When $0 \leq k ^ { 2 } < \frac { 1 } { 4 }$, $S _ { \triangle O P Q } = 8 \left( \frac { 4 k ^ { 2 } + 1 } { 1 - 4 k ^ { 2 } } \right) = 8 \left( - 1 + \frac
gaokao 2015 Q22 Multi-Part 3D Geometry Problem View
22. As shown in the figure, in the quadrangular pyramid $P - A B C D$, given that $P A \perp$ plane $A B C D$, and the quadrilateral $A B C D$ is a right trapezoid, $\angle A B C = \angle B A D = \frac { \pi } { 2 } , P A = A D = 2 , A B = B C = 1$
(1) Find the cosine of the dihedral angle between plane $P A B$ and plane $P C D$;
(2) Point Q is a moving point on segment BP. When the angle between line CQ and DP is minimized, find the length of segment BQ. [Figure]
gaokao 2017 Q10 MCQ: Angle Between Skew Lines View
10. In a right triangular prism $A B C - A _ { 1 } B _ { 1 } C _ { 1 }$, $\angle A B C = 120 ^ { \circ } , A B = 2 , B C = C C _ { 1 } = 1$. The cosine of the angle between skew lines $A B _ { 1 }$ and $B C _ { 1 }$ is
A. $\frac { \sqrt { 3 } } { 2 }$
B. $\frac { \sqrt { 15 } } { 5 }$
C. $\frac { \sqrt { 10 } } { 5 }$
D. $\frac { \sqrt { 3 } } { 3 }$ [Figure]
gaokao 2018 Q9 5 marks MCQ: Angle Between Skew Lines View
In the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$, $E$ is the midpoint of edge $C C _ { 1 }$. The tangent of the angle between skew lines $A E$ and $C D$ is
A. $\frac { \sqrt { 2 } } { 2 }$
B. $\frac { \sqrt { 3 } } { 2 }$
C. $\frac { \sqrt { 5 } } { 2 }$
D. $\frac { \sqrt { 7 } } { 2 }$
gaokao 2018 Q9 5 marks MCQ: Angle Between Skew Lines View
In rectangular prism $A B C D - A _ { 1 } B C _ { 1 } D _ { 1 }$, $A B = B C = 1 , A A _ { 1 } = \sqrt { 3 }$, the cosine of the angle between skew lines $A D _ { 1 }$ and $D B _ { 1 }$ is
A. $\frac { 1 } { 5 }$
B. $\frac { \sqrt { 5 } } { 6 }$
C. $\frac { \sqrt { 5 } } { 5 }$
D. $\frac { \sqrt { 2 } } { 2 }$
gaokao 2018 Q10 5 marks Volume of a 3D Solid View
In rectangular parallelepiped $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$, $A B = B C = 2$, and the angle between $A C _ { 1 }$ and plane $B B _ { 1 } C _ { 1 } C$ is $30 ^ { \circ }$. Then the volume of the rectangular parallelepiped is
A. 8
B. $6 \sqrt { 2 }$
C. $8 \sqrt { 2 }$
D. $8 \sqrt { 3 }$
As shown in the figure, in the triangular pyramid $P - A B C$, $A B = B C = 2 \sqrt { 2 }$, $P A = P B = P C = A C = 4$, and $O$ is the midpoint of $A C$.
(1) Prove: $P O \perp$ plane $A B C$;
(2) If point $M$ is on edge $B C$ such that $M C = 2 M B$, find the distance from point $C$ to plane $P O M$.
gaokao 2018 Q20 12 marks Dihedral Angle or Angle Between Planes/Lines View
(12 points)
As shown in the figure, in triangular pyramid $P - A B C$, $A B = B C = 2 \sqrt { 2 }$, $P A = P B = P C = A C = 4$, $O$ is the midpoint of $A C$.
(1) Prove that $P O \perp$ plane $A B C$;
(2) Point $M$ is on edge $B C$ such that the dihedral angle $M - P A - C$ is $30 ^ { \circ }$. Find the sine of the angle between $P C$ and plane $P A M$.
As shown in the figure, point $N$ is the center of square $ABCD$, $\triangle ECD$ is an equilateral triangle, plane $ECD \perp$ plane $ABCD$, and $M$ is the midpoint of segment $ED$. Then
A. $BM = EN$, and lines $BM$ and $EN$ are intersecting lines
B. $BM \neq EN$, and lines $BM$ and $EN$ are intersecting lines
C. $BM = EN$, and lines $BM$ and $EN$ are skew lines
D. $BM \neq EN$, and lines $BM$ and $EN$ are skew lines
8. As shown in the figure, point $N$ is the center of square $A B C D$ , $\triangle E C D$ is an equilateral triangle, plane $E C D \perp$ plane $A B C D$ , and $M$ is the midpoint of segment $E D$ . Then [Figure]
A. $B M = E N$ , and lines $B M$ and $E N$ are intersecting lines
B. $B M \neq E N$ , and lines $B M$ and $E N$ are intersecting lines
C. $B M = E N$ , and lines $B M$ and $E N$ are skew lines
D. $B M \neq E N$ , and lines $B M$ and $E N$ are skew lines
8. As shown in the figure, point $N$ is the center of square $A B C D$ , $\triangle E C D$ is an equilateral triangle, plane $E C D \perp$ plane $A B C D$ , and $M$ is the midpoint of segment $E D$ . Then [Figure]
A. $B M = E N$ , and lines $B M$ and $E N$ are intersecting lines
B. $B M \neq E N$ , and lines $B M$ and $E N$ are intersecting lines
C. $B M = E N$ , and lines $B M$ and $E N$ are skew lines
D. $B M \neq E N$ , and lines $B M$ and $E N$ are skew lines
gaokao 2019 Q16 14 marks Multi-Part 3D Geometry Problem View
As shown in the figure, in the quadrangular pyramid $P - A B C D$, $P A \perp$ plane $A B C D$, $A D \perp C D$, $A D \| B C$, $P A = A D = C D = 2$, $B C =$ 3. $E$ is the midpoint of $P D$, and point $F$ is on $P C$ such that $\frac { P F } { P C } = \frac { 1 } { 3 }$. (I) Prove that: $C D \perp$ plane $P A D$; (II) Find the cosine of the dihedral angle $F - A E - P$; (III) Let point $G$ be on $P B$ such that $\frac { P G } { P B } = \frac { 2 } { 3 }$. Determine whether line $A G$ lies in plane $A E F$, and explain the reason.
gaokao 2019 Q17 12 marks Multi-Part 3D Geometry Problem View
17. (12 points) As shown in the figure, the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ has a square base $A B C D$. Point $E$ is on edge $A A _ { 1 }$, and $B E \perp E C _ { 1 }$. [Figure]
(1) Prove: $B E \perp$ plane $E B _ { 1 } C _ { 1 }$;
(2) If $A E = A _ { 1 } E , A B = 3$, find the volume of the quadrangular pyramid $E - B B _ { 1 } C _ { 1 } C$.
gaokao 2019 Q17 12 marks Multi-Part 3D Geometry Problem View
17. (12 points) As shown in the figure, the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ has a square base $A B C D$. Point $E$ is on edge $A A _ { 1 }$, and $B E \perp E C _ { 1 }$. [Figure]
(1) Prove that $B E \perp$ plane $E B _ { 1 } C _ { 1 }$;
(2) If $A E = A _ { 1 } E$, find the sine of the dihedral angle $B - E C - C _ { 1 }$.