A square piece of cardboard is divided into three equal…

A square piece of cardboard is divided into three equal strips and folded into a regular triangular prism. Find the angle between two adjacent segments into which the diagonal of the square is broken by the folding.

Explanation of Method

Angle between the segments of a square's diagonal after folding into a triangular prism.

Step 1: The unfolded square

Let the side be $a=3\ell$. The fold lines $x=\ell$ and $x=2\ell$ cut the diagonal from $(0,0)$ to $(a,a)$ at $Q_1=(\ell,\ell)$ and $Q_2=(2\ell,2\ell)$. On every strip the diagonal piece goes $\ell$ across and $\ell$ up.

$Q_0$ $Q_1$ $Q_2$ $Q_3$ $\ell$ $\vec s_1$ $\ell$ $\vec s_2$ $\ell$ $\vec s_3$ $a$

Unfolded square of side $a=3\ell$. The dashed fold lines become lateral edges; on every strip the diagonal runs $\ell$ across and $\ell$ up.

Step 2: The prism

Folding gives a regular triangular prism with base side $\ell$ and height $3\ell$. Take the base $A(0,0,0)$, $B(\ell,0,0)$, $C\!\left(\tfrac{\ell}{2},\tfrac{\sqrt3\,\ell}{2},0\right)$. The horizontal edge vectors traversed around the prism are

\[ \overrightarrow{AB}=\ell\,\vec e_1,\quad \overrightarrow{BC}=\ell\,\vec e_2,\quad \overrightarrow{CA}=\ell\,\vec e_3, \]

\[ \vec e_1=(1,0,0),\quad \vec e_2=\left(-\tfrac12,\tfrac{\sqrt3}2,0\right),\quad \vec e_3=\left(-\tfrac12,-\tfrac{\sqrt3}2,0\right). \]

Adjacent edge vectors make $120^\circ$: $\vec e_1\cdot\vec e_2=-\tfrac12$.

Step 3: The segment vectors

Each piece moves $\ell$ along one face horizontally and rises $\ell$: \[\vec s_k=\ell\,(\vec e_k+\vec k),\qquad \vec s_1=\ell(1,0,1),\quad \vec s_2=\ell\left(-\tfrac12,\tfrac{\sqrt3}{2},1\right).\]

Thus $Q_0=A$, $Q_1=(\ell,0,\ell)$ on edge $B$, $Q_2=\left(\tfrac{\ell}2,\tfrac{\sqrt3\ell}2,2\ell\right)$ on edge $C$, and $Q_3=(0,0,3\ell)$ back on edge $A$.

$Q_0$ $Q_1$ $Q_2$ $Q_3$ $\vec s_1$ $\vec s_2$ $\vec s_3$ $B$ $C$

Folded prism (side $\ell$, height $3\ell$). The diagonal becomes the broken line $Q_0Q_1Q_2Q_3$ winding once around the prism; $\vec s_3$ runs on the back face (dashed).

Step 4: The angle

\[ \vec s_1\cdot\vec s_2=\ell^2\left(-\tfrac12+0+1\right)=\frac{\ell^2}{2},\qquad |\vec s_1|=|\vec s_2|=\ell\sqrt2, \]

\[ \cos\varphi=\frac{\ell^2/2}{2\ell^2}=\frac14. \]

\[\varphi=\arccos\frac14\approx75.5^\circ\ (75^\circ31')\]

By symmetry the same angle appears between $\vec s_2$ and $\vec s_3$.

Result

\[\boxed{\,\varphi=\arccos\frac14\approx75.5^\circ\ (75^\circ31')\,}\]

Remark

Correction. The source places the base at $(\ell,0,0)$, $\left(-\tfrac{\ell}{2},\pm\tfrac{\sqrt3\ell}{2},0\right)$. That triangle has side $\ell\sqrt3$, not $\ell$. The edge directions $\vec e_k$ it uses are still correct, so the final answer is unaffected.

Which angle? $75.5^\circ$ is the angle between the two segments as lines, i.e. between the direction vectors $\vec s_1$ and $\vec s_2$. The angle at the vertex $Q_1$ of the broken line, $\angle Q_0Q_1Q_2$, is formed by $-\vec s_1$ and $\vec s_2$, so it equals $180^\circ-75.5^\circ\approx104.5^\circ$ ($\cos=-\tfrac14$).

Note on the source's comparison. Without the vertical rise the directions would be $\vec e_1,\vec e_2$, which make $120^\circ$; as lines that is $60^\circ$, the angle of the base triangle. The vertical component $\vec k$, common to both segments, changes the angle to $\arccos\tfrac14$.