Solid Geometry – Problem 3: Prove that points are coplanar

In the right prism $ABCD-A_1B_1C_1D_1$, the base is a rhombus with $AA_1=AC=2$, $BD=4$. Points $E,F,P,Q$ move on edges $AA_1,BB_1,CC_1,DD_1$, respectively, and satisfy $$BF=DQ,\quad CP-BF=DQ-AE=1.$$

(1) Prove that points $E,F,P,Q$ are coplanar.

(2) If the cosine of the dihedral angle $B-PQ-E$ is $\dfrac{2\sqrt5}{5}$, find $|BF|$.

(3) If $F$ is the midpoint of $BB_1$, and $M$ moves on segment $FB_1$ (excluding endpoints), find the range of the surface area of the circumscribed sphere of tetrahedron $M-PQF$.

(1) Take points $M,N$ on $CP,DQ$, respectively, such that $PM=QN=1$. Then quadrilateral $MNQP$ is a parallelogram, so $MN\parallel PQ$. Also, from $AE=ND$ and $AE\parallel ND$, quadrilateral $ADNE$ is a rectangle, hence $AD\parallel NE$. Similarly, $FM\parallel BC\parallel AD$, and $NE=MF=AD$, so quadrilateral $FMNE$ is a parallelogram, thus $EF\parallel MN$. Therefore $EF\parallel PQ$, so points $E,F,P,Q$ are coplanar.

(2) Build a 3D rectangular coordinate system with origin at the intersection $O$ of the base diagonals. Take $$A(2,0,0),\ B(0,1,0),\ D(0,-1,0),\ C(-2,0,0),\ E(2,0,b).$$ From $BF=DQ$, $CP-BF=DQ-AE=1$, we get $$Q(0,-1,1+b),\ F(0,1,1+b),\ P(-2,0,2+b).$$ A normal vector of plane $EFPQ$ can be $\vec n_1=(1,0,2)$, and a normal vector of plane $BPQ$ can be $\vec n_2=(b+3,2+2b,4)$. Using the dihedral-angle condition $$\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=\frac{2\sqrt5}{5}$$ we solve $b=\dfrac{8\sqrt6-17}{19}$, hence $$|BF|=b+1=\frac{8\sqrt6+2}{19}.$$

(3) When $F$ is the midpoint of $BB_1$, we have $|BF|=1$, so from the constraint equations $AE=0$, $BF=1$, $CP=2$, $DQ=1$. With $b=AE=0$: $$F(0,1,1),\quad P(-2,0,2),\quad Q(0,-1,1).$$ Let $M=(0,1,1+2m)$ for $m\in(0,1)$ (moving from $F$ toward $B_1=(0,1,2)$). The tetrahedron $M\text{-}PQF$ has vertices $P(-2,0,2)$, $Q(0,-1,1)$, $F(0,1,1)$, $M(0,1,1+2m)$. By symmetry, the circumsphere center is $S=(x,0,z)$. From $SM^2=SF^2$: $$x^2+1+(z-1-2m)^2=x^2+1+(z-1)^2\Rightarrow z=m+1.$$ From $SP^2=SF^2$: $$(x+2)^2+(z-2)^2=x^2+1+(z-1)^2 \Rightarrow 4x+6-2z=0 \Rightarrow x=\frac{z-3}{2}=\frac m2-1.$$ Hence $$R^2=SF^2=x^2+1+(z-1)^2=\left(\frac m2-1\right)^2+1+m^2 =\frac54\left(m-\frac25\right)^2+\frac95.$$ For $m\in(0,1)$, the minimum is attained at $m=\frac25$, so $$R^2\in\left[\frac95,\frac94\right).$$ Therefore the circumsphere surface area satisfies $$S_{\text{sphere}}=4\pi R^2\in\left[\frac{36\pi}{5},9\pi\right).$$

(1) Through auxiliary points and parallel-line chains, we derive $EF\parallel PQ$, so points $E,F,P,Q$ are coplanar.

(2) Setting coordinates and applying the dihedral-cosine equation gives $$|BF|=\dfrac{8\sqrt6+2}{19}.$$ This is the unique value consistent with the given angle condition.

(3) Parameterizing $M$ and expressing the circumsphere radius as a quadratic in the parameter yield $$R^2\in\left[\frac95,\frac94\right).$$ Hence the surface-area range is $$\left[\dfrac{36\pi}{5},9\pi\right).$$