Solid Geometry – Problem 2: Prove that

In the triangular frustum $ABC-A_1B_1C_1$, plane $ABB_1A_1\perp$ plane $ABC$, $E$ is the midpoint of $CC_1$, $CA=CB$, and $$AB=2,\quad 2AA_1=2BB_1=2,\quad A_1B_1=4.$$

(1) Prove that $AB_1\perp A_1C$.

(2) When the volume of $ABC-A_1B_1C_1$ is $\dfrac{7\sqrt3}{3}$, find $\sin\theta$, where $\theta$ is the angle between line $CC_1$ and plane $A_1B_1E$.

(1) Let $O$ be the midpoint of $AB$, and connect $OC,OA_1,AB_1$. From $CA=CB$, we get $OC\perp AB$. Since plane $ABB_1A_1\perp$ plane $ABC$ and their intersection is $AB$, it follows that $OC\perp$ plane $ABB_1A_1$, hence $OC\perp AB_1$.

Also, $A_1B_1\parallel AO$ and $A_1B_1=A_1A=AO$, so quadrilateral $A_1AOB_1$ is a rhombus, hence $AB_1\perp A_1O$. Since $OC\cap A_1O=O$ and both $OC,A_1O\subset$ plane $A_1OC$, we have $AB_1\perp$ plane $A_1OC$, therefore $$AB_1\perp A_1C.$$

(2) Set up a 3D rectangular coordinate system with the midpoint $O$ of $AB$ as origin. From the volume condition $$V=\frac{h}{3}(S_1+S_2+\sqrt{S_1S_2})=\frac{7\sqrt3}{3}$$ one obtains $h=\sqrt3$. Take $$A(-1,0,0),\ B(1,0,0),\ C(0,2,0),\ A_1(-1,0,\sqrt3),\ B_1(1,0,\sqrt3),\ C_1(0,1,\sqrt3),$$ so $E=\dfrac{C+C_1}{2}=\left(0,\dfrac32,\dfrac{\sqrt3}{2}\right)$.

Then $$\overrightarrow{CC_1}=(0,-1,\sqrt3).$$ Compute a normal vector of plane $A_1B_1E$: $$\overrightarrow{A_1B_1}=(2,0,0),\quad \overrightarrow{A_1E}=\left(1,\frac32,-\frac{\sqrt3}{2}\right),$$ $$\vec n=\overrightarrow{A_1B_1}\times\overrightarrow{A_1E}=(0,\sqrt3,3).$$ Hence $$\sin\theta=\frac{|\overrightarrow{CC_1}\cdot\vec n|}{|\overrightarrow{CC_1}|\,|\vec n|}=\frac{|-\sqrt3+3\sqrt3|}{\sqrt{1+3}\cdot\sqrt{3+9}}=\frac{2\sqrt3}{2\cdot2\sqrt3}=\frac12.$$

(1) By proving $AB_1$ is perpendicular to both $OC$ and $A_1O$, we get $$AB_1\perp A_1C.$$

(2) After determining the missing metric from the volume and using a vector normal to plane $A_1B_1E$, we find $$\sin\theta=\frac12.$$ So the angle between $CC_1$ and plane $A_1B_1E$ has sine $\frac12$.