Subfield: Geometry


Question: Parallelogram $A B C D$ is rotated about $A$ in the plane, resulting in $A B^{\prime} C^{\prime} D^{\prime}$, with $D$ on $\overline{A B^{\prime}}$. Suppose that $\left[B^{\prime} C D\right]=\left[A B D^{\prime}\right]=\left[B C C^{\prime}\right]$. Compute $\tan \angle A B D$.

Solution:
Editor's Note: It was pointed out that the conditions of the problem determine two possible values of $\tan \angle A B D$ : one based on $\mathrm{m} \angle A<90^{\circ}$, and the other based on $\mathrm{m} \angle A>90^{\circ}$. A complete solution is provided below. We thank Matthew Babbitt and Silas Johnson for their contributions to this solution.



Let $A B=x, B C=y$, and $\mathrm{m} \angle A=\alpha$.

<img_3267>

It then follows that

<img_3933>

$$
\left[A B D^{\prime}\right]=\left\{\begin{array}{ll}
\frac{x y \sin 2 \alpha}{2} & \text { if } \alpha<90^{\circ} \\
\frac{-x y \sin 2 \alpha}{2} & \text { if } \alpha>90^{\circ}
\end{array} \quad \text { and } \quad\left[B^{\prime} C D\right]=\frac{x(x-y) \sin \alpha}{2}\right.
$$

Because $\overline{B C}, \overline{A B^{\prime}}$, and $\overline{D^{\prime} C^{\prime}}$ are all parallel, it follows that $\triangle B C C^{\prime}$ and $\triangle B C D^{\prime}$ have the same height with respect to base $\overline{B C}$, and thus $\left[B C C^{\prime}\right]=\left[B C D^{\prime}\right]$. Therefore $\left[B C D^{\prime}\right]=\left[A B D^{\prime}\right]$, and it follows that triangles $B C D^{\prime}$ and $A B D^{\prime}$ have the same height with respect to base $\overline{B D^{\prime}}$. Thus $A$ and $C$ are equidistant from $\overleftrightarrow{B D^{\prime}}$. Let $M$ be the midpoint of $\overline{A C}$. Consider the following two cases.

Case 1: Suppose that $\alpha<90^{\circ}$. Because $A$ and $C$ are equidistant from $\overleftrightarrow{B D^{\prime}}$, it follows that $M$ lies on $\overleftrightarrow{B D^{\prime}}$. But $\overleftrightarrow{B D}$ also passes through the midpoint of $\overline{A C}$ by parallelogram properties, so it follows that $D$ must lie on $\overline{B D^{\prime}}$. This implies that $\left[A B D^{\prime}\right]$ must also equal $\frac{y^{2} \sin \alpha}{2}+\frac{x y \sin \alpha}{2}=\frac{\left(x y+y^{2}\right) \sin \alpha}{2}$.

Thus $x(x-y) \sin \alpha=x y \sin 2 \alpha=\left(x y+y^{2}\right) \sin \alpha$, which implies $x: y=\sqrt{2}+1$ and $\sin \alpha=\cos \alpha=\frac{\sqrt{2}}{2}$. Finally, from right triangle $D^{\prime} A B$ with legs in the ratio $1: \sqrt{2}+1$, it follows that $\tan (\angle A B D)=\tan \left(\angle A B D^{\prime}\right)=$ $\sqrt{2}-1$.

Case 2: Suppose that $\alpha>90^{\circ}$. The points $D$ and $D^{\prime}$ lie on opposite sides of $\overleftrightarrow{A B}$. Because $B C=A D^{\prime}$ and points $A$ and $C$ are equidistant from $\overleftrightarrow{B D^{\prime}}$, it follows that $A C B D^{\prime}$ is either a parallelogram or an isosceles trapezoid. It cannot be the former because that would imply that $\overleftrightarrow{D^{\prime} A}\|\overleftrightarrow{B C}\| \overleftrightarrow{A D}$. Thus $A C B D^{\prime}$ is an isosceles trapezoid. Then $\left[B A D^{\prime}\right]=\left[B M D^{\prime}\right]$. Because $B, M$, and $D$ are collinear and $B D: B M=2$, it follows that $\left[B D D^{\prime}\right]=2 \cdot\left[B M D^{\prime}\right]$. Moreover, $\left[B D D^{\prime}\right]=\left[B A D^{\prime}\right]+[B A D]+\left[D A D^{\prime}\right]$, so $\left[B A D^{\prime}\right]=[B A D]+\left[D A D^{\prime}\right]$. Thus $\left[B A D^{\prime}\right]=\frac{x y \sin \alpha}{2}+\frac{y^{2} \sin \alpha}{2}=\frac{\left(x y+y^{2}\right) \sin \alpha}{2}$.

Thus $x(x-y) \sin \alpha=-x y \sin 2 \alpha=\left(x y+y^{2}\right) \sin \alpha$, which implies $x: y=\sqrt{2}+1, \sin \alpha=\frac{\sqrt{2}}{2}$, and $\cos \alpha=-\frac{\sqrt{2}}{2}$, so $\alpha=135^{\circ}$. Let $H$ be the foot of the perpendicular from $D$ to $\overleftrightarrow{A B}$. Then $A D H$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with $H A=H D=\frac{y}{\sqrt{2}}$. Thus

$$
\begin{aligned}
\tan \angle A B D & =\frac{D H}{B H}=\frac{D H}{B A+A H} \\
& =\frac{y / \sqrt{2}}{x+y / \sqrt{2}}=\frac{y}{x \sqrt{2}+y} \\
& =\frac{y}{y(\sqrt{2}+1)(\sqrt{2})+y} \\
& =\frac{1}{(\sqrt{2}+1)(\sqrt{2})+1} \\
& =\frac{\mathbf{3}-\sqrt{\mathbf{2}}}{\mathbf{7}}
\end{aligned}
$$
Let $x, y$, and $\alpha$ be as defined in the first solution. Then $C D=x$ because $A B C D$ is a parallelogram. Also note that $A B^{\prime}=x, B^{\prime} C^{\prime}=y$, and $A D^{\prime}=y$ because $A B C D$ and $A B^{\prime} C^{\prime} D^{\prime}$ are congruent. Thus $D B^{\prime}=A B^{\prime}-A D=x-y$. Let $E$ be the intersection of $\overleftrightarrow{A B}$ and $\overleftrightarrow{C^{\prime} D^{\prime}}$, as shown in both configurations below.
<img_3701>

Because $E$ lies on $\overleftrightarrow{A B}$, it follows that $\angle B^{\prime} A E=180^{\circ}-\angle B A D=180^{\circ}-\alpha$. Thus, in quadrilateral $A B^{\prime} C^{\prime} E$, $\overline{A B^{\prime}} \| \overline{C^{\prime} E}$ and $\angle A B^{\prime} C^{\prime}=\angle B^{\prime} A E=180^{\circ}-\alpha$. Therefore quadrilateral $A B^{\prime} C^{\prime} E$ is an isosceles trapezoid. Hence $A E=B^{\prime} C^{\prime}=y$. It follows that $B E=B A+A E=x+y$. Therefore, from the sine area formula with respect to $\angle C B E$,

$$
[B C E]=\frac{1}{2} x(x+y) \sin \left(180^{\circ}-\alpha\right)=\frac{1}{2} x(x+y) \sin \alpha .
$$

Because $\overline{E C^{\prime}} \| \overline{B C}$, it follows that $\left[B C C^{\prime}\right]=[B C E]=\frac{1}{2} x(x+y) \sin \alpha$. From the sine area formula with respect to $\angle B A D^{\prime}$ and $\angle B^{\prime} D C$, respectively,

$$
\left[B A D^{\prime}\right]=\frac{1}{2} x y|\sin (2 \alpha)|, \quad\left[B^{\prime} D C\right]=\frac{1}{2} x(x-y) \sin \alpha
$$

Thus

$$
\frac{1}{2} x(x+y) \sin \alpha=\frac{1}{2} x y|\sin (2 \alpha)|=\frac{1}{2} x(x-y) \sin \alpha .
$$


Because $\overline{B C}, \overline{A B^{\prime}}$, and $\overline{D^{\prime} C^{\prime}}$ are all parallel, it follows that $\triangle B C C^{\prime}$ and $\triangle B C D^{\prime}$ have the same height with respect to base $\overline{B C}$, and thus $\left[B C C^{\prime}\right]=\left[B C D^{\prime}\right]$. Therefore $\left[B C D^{\prime}\right]=\left[A B D^{\prime}\right]$, and it follows that triangles $B C D^{\prime}$ and $A B D^{\prime}$ have the same height with respect to base $\overline{B D^{\prime}}$. Thus $A$ and $C$ are equidistant from $\overleftrightarrow{B D^{\prime}}$. Let $M$ be the midpoint of $\overline{A C}$. Consider the following two cases.

Case 1: Suppose that $\alpha<90^{\circ}$. Because $A$ and $C$ are equidistant from $\overleftrightarrow{B D^{\prime}}$, it follows that $M$ lies on $\overleftrightarrow{B D^{\prime}}$. But $\overleftrightarrow{B D}$ also passes through the midpoint of $\overline{A C}$ by parallelogram properties, so it follows that $D$ must lie on $\overline{B D^{\prime}}$. This implies that $\left[A B D^{\prime}\right]$ must also equal $\frac{y^{2} \sin \alpha}{2}+\frac{x y \sin \alpha}{2}=\frac{\left(x y+y^{2}\right) \sin \alpha}{2}$.

Thus $x(x-y) \sin \alpha=x y \sin 2 \alpha=\left(x y+y^{2}\right) \sin \alpha$, which implies $x: y=\sqrt{2}+1$ and $\sin \alpha=\cos \alpha=\frac{\sqrt{2}}{2}$. Finally, from right triangle $D^{\prime} A B$ with legs in the ratio $1: \sqrt{2}+1$, it follows that $\tan (\angle A B D)=\tan \left(\angle A B D^{\prime}\right)=$ $\sqrt{2}-1$.

Case 2: Suppose that $\alpha>90^{\circ}$. The points $D$ and $D^{\prime}$ lie on opposite sides of $\overleftrightarrow{A B}$. Because $B C=A D^{\prime}$ and points $A$ and $C$ are equidistant from $\overleftrightarrow{B D^{\prime}}$, it follows that $A C B D^{\prime}$ is either a parallelogram or an isosceles trapezoid. It cannot be the former because that would imply that $\overleftrightarrow{D^{\prime} A}\|\overleftrightarrow{B C}\| \overleftrightarrow{A D}$. Thus $A C B D^{\prime}$ is an isosceles trapezoid. Then $\left[B A D^{\prime}\right]=\left[B M D^{\prime}\right]$. Because $B, M$, and $D$ are collinear and $B D: B M=2$, it follows that $\left[B D D^{\prime}\right]=2 \cdot\left[B M D^{\prime}\right]$. Moreover, $\left[B D D^{\prime}\right]=\left[B A D^{\prime}\right]+[B A D]+\left[D A D^{\prime}\right]$, so $\left[B A D^{\prime}\right]=[B A D]+\left[D A D^{\prime}\right]$. Thus $\left[B A D^{\prime}\right]=\frac{x y \sin \alpha}{2}+\frac{y^{2} \sin \alpha}{2}=\frac{\left(x y+y^{2}\right) \sin \alpha}{2}$.

Thus $x(x-y) \sin \alpha=-x y \sin 2 \alpha=\left(x y+y^{2}\right) \sin \alpha$, which implies $x: y=\sqrt{2}+1, \sin \alpha=\frac{\sqrt{2}}{2}$, and $\cos \alpha=-\frac{\sqrt{2}}{2}$, so $\alpha=135^{\circ}$. Let $H$ be the foot of the perpendicular from $D$ to $\overleftrightarrow{A B}$. Then $A D H$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with $H A=H D=\frac{y}{\sqrt{2}}$. Thus

$$
\begin{aligned}
\tan \angle A B D & =\frac{D H}{B H}=\frac{D H}{B A+A H} \\
& =\frac{y / \sqrt{2}}{x+y / \sqrt{2}}=\frac{y}{x \sqrt{2}+y} \\
& =\frac{y}{y(\sqrt{2}+1)(\sqrt{2})+y} \\
& =\frac{1}{(\sqrt{2}+1)(\sqrt{2})+1} \\
& =\frac{\mathbf{3}-\sqrt{\mathbf{2}}}{\mathbf{7}}
\end{aligned}
$$

Final answer: $\sqrt{2}-1$,$\frac{3-\sqrt{2}}{7}$