China Mathematical Competition

Let the line through segment $AB$ be $x$-axis with the origin $O$ being the midpoint of $AB$ and let the line through segment $OC$ be $y$-axis to establish a space rectangular coordinate system shown in Fig. 7.1. Assuming the length of an edge is 2, we have $B(1,0,0)$, $B_1(1,0,2)$, $A_1(-1,0,2)$, $P(0,\sqrt{3},1)$. Then Fig. 7.1 $$\begin{array}{rl}\overrightarrow{B{A}_1}& =(-2,0,2),\overrightarrow{BP}=(-1,\sqrt{3},1),\\ \overrightarrow{B_1{A}_1}& =(-2,0,0),\overrightarrow{B_{1}P}=(-1,\sqrt{3},-1).\end{array}$$ Let vectors $\vec{m}=(x_1,y_1,z_1)$ and $\vec{n}=(x_2,y_2,z_2)$ be perpendicular to $B{A}_{1}P$ and $B_1{A}_{1}P$, respectively. We have $$\{\begin{array}{l}\vec{m}\cdot \overrightarrow{B{A}_1}=-2x_1+2z_1=0,\\ \vec{m}\cdot \overrightarrow{BP}=-x_1+\sqrt{3}{y}_1+z_1=0,\\ \vec{n}\cdot \overrightarrow{B_1{A}_1}=-2x_2=0,\\ \vec{n}\cdot \overrightarrow{B_{1}P}=-x_2+\sqrt{3}{y}_2-z_2=0.\end{array}$$ We can then assume that $\vec{m}=(1,0,1)$, $\vec{n}=(0,1,\sqrt{3})$. From $|\vec{m}\cdot \vec{n}|=|\vec{m}|\cdot |\vec{n}|\cos \alpha$, we have $$\sqrt{3}=\sqrt{2}\cdot 2\cos \alpha \Rightarrow \cos \alpha =\frac{\sqrt{6}}{4}.$$

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Alternative solution.

As seen in Fig. 7.2, we have $PC=P{C}_1$, $P{A}_1=PB$. Suppose $A_{1}B$ and $A{B}_1$ intersect at $O$. Then we get $O{A}_1=OB$, $OA=O{B}_1$, $A_{1}B\perp A{B}_1$. Since $PA=P{B}_1$, then $PO\perp A{B}_1$. Therefore $A{B}_1\perp$ plane $P{A}_{1}B$. On plane $P{A}_{1}B$ through $O$ draw line $OE\perp A_{1}P$ with foot point $E$. Fig. 7.2 Connecting $B_{1}E$, $\angle B_{1}EO$ is then the plane angle of the dihedral angle $B-A_{1}P-B_1$. Assuming $A{A}_1=2$, it is easy to find that $PB=P{A}_1=\sqrt{5}$, $A_{1}O=B_{1}O=\sqrt{2}$, $PO=\sqrt{3}$. In right triangle $△P{A}_{1}O$, we have $A_{1}O\cdot PO=A_{1}P\cdot OE$, or $\sqrt{2}\cdot \sqrt{3}=\sqrt{5}\cdot OE$. So $OE=\frac{\sqrt{6}}{\sqrt{5}}$. As $B_{1}O=\sqrt{2}$, we have $$B_{1}E=\sqrt{B_1{O}^2+O{E}^2}=\sqrt{2+\frac{6}{5}}=\frac{4\sqrt{5}}{5}.$$ Finally, $$\sin \alpha =\sin \angle B_{1}EO=\frac{B_{1}O}{B_{1}E}=\frac{\sqrt{2}}{\frac{4\sqrt{5}}{5}}=\frac{\sqrt{10}}{4}.\square$$