3d geometry problem with a trick, pyramid

Question

English is not my first language, and math terms are completely unkown to me, sorry in advance

In a pyramid SABCD in which the base ABCD has 4 equal sides (not necessarily a square) the length of each side is a, and the dihedral angle between the base and each face is $2\alpha$.

Through AB we create a plane, such that its angle with the base is $\alpha$ ; the plane intersects SC and SD resp. in C' and D'.

Prove that $$C'D' = a \sin(\alpha)/\sin(3 \alpha).$$

Calculate ABC'D'

if someone can provide drawings as well it will be great.

Thanks

Answer 1

Let $H$ be the foot of the altitude issued from $O$ (rhombus' center) onto side $AB$ (meaning that $OH \perp AB$). Let $H'$ be symmetrical of $H$ with respect to $O$.

SO, being assumed orthogonal to base plane, triangle $HSH'$ is isosceles with altitude SO and belongs to a vertical plane. Therefore, in this triangle, $\hat{H}=\hat{H'}=2 \alpha$ by definition of a dihedral angle ; consequently $\hat{S}=\pi-4 \alpha$.

Let $E$ be the intersection point of the oblique plane passing through $AB$ with $SH'$ (in particular, $E \in $ line segment $C'D'$).

Let us consider now triangle $HSE$. Its angles are :

$$\hat{H}=\alpha, \hat{S}=\pi - 4\alpha \ \text{implying} \ \hat{E}=3\alpha.$$

Let us apply the sine law to this triangle :

$$\dfrac{SH}{\sin 3 \alpha}=\dfrac{SE}{\sin \alpha}.$$

Otherwise said (because $SH=SH'$) :

$$\dfrac{\sin \alpha}{\sin 3 \alpha}=\dfrac{SE}{SH'}\tag{1}$$

Besides, line segment $C'D'$ is orthogonaly projected on the base plane onto line segment $C_1D_1$ with $C_1 \in OC$ and $D_1 \in OD$, with preservation of length because line segment $C'D'$ is parallel to the base plane.

Now, by considering triangles $SCD$ and $SC'D'$ one one hand and $OCD$ and $OC_1D_1$ (which are their projections onto the base plane), on the other, and using intercept theorem we can conclude that :

$$\dfrac{C'D'}{CD}=\dfrac{C_1D_1}{CD}=\dfrac{SE}{SH'}\tag{2}$$

Combining (1) and (2), using the fact that $CD=a$, we get the result.

Answer 2

With four congruent sides, $\square ABCD$ is a rhombus; let $O$ be its center. For dihedral angles along $\overline{AB}$ and $\overline{CD}$ to be congruent, the altitude from $S$ must meet the base halfway between the lines of those edges; likewise, the altitude must meet the base halfway between the lines of $\overline{BC}$ and $\overline{AD}$. Consequently, $\overline{SO}$ is itself the altitude.

Since $\overline{AB}\parallel\overline{CD}$, it follows that planes through $\overline{AB}$ cut $\triangle SCD$ in parallel lines. In particular, $\overline{C'D'}\parallel\overline{CD}$. By the Intercept Theorem (or simple similar triangles), we have $$\frac{|C'D'|}{|CD|}=\frac{|SC'|}{|SC|} \tag{1}$$

Now, let the perpendicular from $O$ meet $\overline{AB}$ at $R$; define $r := |OR|$. Let the cutting plane through $\overline{AB}$ meets the pyramid's altitude at $S'$. Then $\angle SRO$ and $\angle S'RO$ are the dihedral angles $2\alpha$ and $\alpha$, respectively; moreover, we have $$|OS|=r\tan 2\alpha \qquad |OS'|=r\tan\alpha \tag{2}$$

In the plane of (isosceles) $\triangle SAC$, define $$p:=|OA|=|OC|, \quad \theta:=\angle SCO=\angle SAO, \quad \phi:=\angle S'AO$$

so that $$|OS|=p\tan\theta \qquad |OS'|=p\tan\phi \tag{3}$$ Note that $A$, $S'$, $C'$ are collinear, so that we can evaluate the ratio in $(1)$ as follows: $$\begin{align} \frac{|SC'|}{|SC|}=\frac{|SC'|}{|SA|} &=\frac{\sin(\theta-\phi)}{\sin(\theta+\phi)}=\frac{\sin\theta\cos\phi-\cos\theta\sin\phi}{\sin\theta\cos\phi+\cos\theta\sin\phi} \tag{4}\\[4pt] &=\frac{\tan\theta-\tan\phi}{\tan\theta+\tan\phi} \tag{5}\\[4pt] &=\frac{|OS|-|OS'|}{|OS|+|OS'|} \tag{6}\\[4pt] &=\frac{\tan2\alpha-\tan\alpha}{\tan2\alpha+\tan\alpha} \tag{7}\\[4pt] &=\frac{\sin2\alpha\cos\alpha-\cos2\alpha\sin\alpha}{\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha} \tag{8}\\[4pt] &=\frac{\sin\alpha}{\sin 3\alpha} \tag{9} \end{align}$$ This gives the first result. $\square$

Finding the area of $\square ABC'D'$ is left as an exercise to the reader.