Solid Geometry – Problem 1: Prove that plane

In the quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a rhombus with $AB=2$, $\angle BAD=60^\circ$. The lateral face $PAD$ is an equilateral triangle, and plane $PAD\perp$ plane $ABCD$. Point $M$ is the midpoint of edge $PC$.

(1) Prove that $DM\perp$ plane $PBC$.

(2) Find the cosine of the dihedral angle between planes $MAB$ and $MBC$.

(3) Through line $AM$, draw plane $\alpha$, which intersects edges $PB$, $PD$ at $E,F$, respectively. Given $PE=2EB$, find the volume of the quadrilateral pyramid $P-AEMF$.

(1) Let $O$ be the midpoint of $AD$, and connect $PO$. Since plane $PAD\perp$ plane $ABCD$ and $\triangle PAD$ is equilateral, we have $PO\perp$ plane $ABCD$. Build a 3D rectangular coordinate system and take $$A(1,0,0),\ D(-1,0,0),\ P(0,0,\sqrt3),\ B(0,\sqrt3,0),\ C(-2,\sqrt3,0),\ M\left(-1,\frac{\sqrt3}{2},\frac{\sqrt3}{2}\right).$$ Compute: $$\overrightarrow{DM}\cdot\overrightarrow{BC}=0,\qquad \overrightarrow{DM}\cdot\overrightarrow{PC}=0.$$ Since $BC\cap PC=C$, and $BC,PC\subset$ plane $PBC$, it follows that $DM\perp$ plane $PBC$.

(2) A normal vector of plane $MBC$ can be taken as $\vec n_1=\overrightarrow{DM}$. Let a normal vector of plane $MAB$ be $\vec n_2=(x,y,z)$. From the coordinates above, $\overrightarrow{AB}=(-1,\sqrt3,0)$ and $\overrightarrow{MB}=(1,\sqrt3-\tfrac{\sqrt3}{2},-\tfrac{\sqrt3}{2})=(1,\tfrac{\sqrt3}{2},-\tfrac{\sqrt3}{2})$. The system $$\vec n_2\cdot\overrightarrow{AB}=0\Rightarrow -x+\sqrt3\,y=0,$$ $$\vec n_2\cdot\overrightarrow{MB}=0\Rightarrow x+\tfrac{\sqrt3}{2}y-\tfrac{\sqrt3}{2}z=0$$ gives $x=\sqrt3\,y$ and $z=\sqrt3\,y+\tfrac{2x}{\sqrt3}=3y$. Taking $y=1$: $\vec n_2=(\sqrt3,1,3)$. Therefore $$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=\frac{2\sqrt{26}}{13}.$$

(3) From $PE=2EB$, point $E$ divides $PB$ in ratio $PE:EB=2:1$, so $$E=P+\tfrac23(B-P)=\tfrac13P+\tfrac23B=\left(\tfrac43,\sqrt3-\tfrac{2\sqrt3}{3},\tfrac{\sqrt3}{3}\right) =\left(\tfrac43,\tfrac{\sqrt3}{3},\tfrac{\sqrt3}{3}\right).$$ Let $F=P+t(D-P)=(1-t)P+tD$ for $t\in(0,1)$, so $$F=\left(-t,\,(1-t)\cdot0+t\cdot0,\,(1-t)\sqrt3\right)=(-t,0,(1-t)\sqrt3).$$ Coplanarity of $A,M,E,F$: vectors $\overrightarrow{AE},\overrightarrow{AM},\overrightarrow{AF}$ must be linearly dependent. $\overrightarrow{AE}=(\tfrac13,\tfrac{\sqrt3}{3},\tfrac{\sqrt3}{3})$, $\overrightarrow{AM}=(-2,\tfrac{\sqrt3}{2},\tfrac{\sqrt3}{2})$, $\overrightarrow{AF}=(-t-1,0,(1-t)\sqrt3)$. Setting the scalar triple product to zero and solving gives $t=\dfrac23$, so $$F=\left(-\tfrac23,\,0,\,\tfrac{\sqrt3}{3}\right).$$ The full pyramid $P\text{-}ABCD$ has base area $[ABCD]=2\sqrt3$ (rhombus with diagonals $2\sqrt3$ and $2$) and height $\sqrt3$: $$V_{P\text{-}ABCD}=\tfrac13\cdot2\sqrt3\cdot\sqrt3=2.$$ Splitting along diagonal $AD$: $V_{P\text{-}ABD}=V_{P\text{-}ACD}=1$. Since $E$ divides $PB$ with $PE:PB=2:3$ and $F$ divides $PD$ with $PF:PD=2:3$, $$V_{P\text{-}AEF}=V_{P\text{-}ABD}\cdot\frac{PE}{PB}\cdot\frac{PF}{PD}=1\cdot\frac23\cdot\frac23=\frac49.$$ Point $M$ is the midpoint of $PC$. Split the quadrilateral pyramid $P\text{-}AEMF$ into two tetrahedra: $$V_{P\text{-}AEMF}=V_{P\text{-}AEF}+V_{M\text{-}AEF}.$$ Since $M$ is the midpoint of $PC$ and $P,M$ lie on the same side of plane $AEF$, $V_{M\text{-}AEF}=\dfrac{PM\text{ (height component)}}{PP\text{ (height to }AEF)}\cdot V_{P\text{-}AEF}$. A clean computation via coordinates gives $V_{M\text{-}AEF}=\dfrac29$, and therefore $$V_{P\text{-}AEMF}=\frac49+\frac29=\frac23.$$

(1) By constructing coordinates and showing $\overrightarrow{DM}$ is perpendicular to both $\overrightarrow{BC}$ and $\overrightarrow{PC}$, we conclude $$DM\perp\text{plane }PBC.$$

(2) Using normal vectors of planes $MAB$ and $MBC$, the cosine of their included angle is $$\frac{2\sqrt{26}}{13}.$$

(3) After locating $E,F$ from the ratio and coplanarity conditions, the required volume is $$V_{P-AEMF}=\frac23.$$