Subfield: Geometry


Question: Points $P$ and $Q$ are located inside the square $A B C D$ such that $D P$ is parallel to $Q B$ and $D P=Q B=P Q$. Determine the minimum possible value of $\angle A D P$.

<img_3459>

Solution:
Placing the information on the coordinate axes, the diagram is indicated to the right.

We note that $P$ has coordinates $(a, b)$.

By symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.

Since $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$

$$
\begin{aligned}
& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\
& \left(a-\frac{4}{3}\right)^{2}+\left(b-\frac{4}{3}\right)^{2}=\frac{8}{9}
\end{aligned}
$$

<img_3483>

$P$ is on a circle with centre $O\left(\frac{4}{3}, \frac{4}{3}\right)$ with $r=\frac{2}{3} \sqrt{2}$.

The minimum angle for $\theta$ occurs when $D P$ is tangent to the circle.

So we have the diagram noted to the right.

Since $O D$ makes an angle of $45^{\circ}$ with the $x$-axis then $\angle P D O=45-\theta$ and $O D=\frac{4}{3} \sqrt{2}$.

Therefore $\sin (45-\theta)=\frac{\frac{2}{3} \sqrt{2}}{\frac{4}{3} \sqrt{2}}=\frac{1}{2}$ which means $45^{\circ}-\theta=30^{\circ}$ or $\theta=15^{\circ}$.

Thus the minimum value for $\theta$ is $15^{\circ}$.

<img_3977>
Let $A B=B C=C D=D A=1$.

Join $D$ to $B$. Let $\angle A D P=\theta$. Therefore, $\angle P D B=45-\theta$.

Let $P D=a$ and $P B=b$ and $P Q=\frac{a}{2}$.



We now establish a relationship between $a$ and $b$.

In $\triangle P D B, b^{2}=a^{2}+2-2(a)(\sqrt{2}) \cos (45-\theta)$

$$
\text { or, } \quad \cos (45-\theta)=\frac{a^{2}-b^{2}+2}{2 \sqrt{2} a}
$$

<img_3307>

In $\triangle P D R,\left(\frac{a}{2}\right)^{2}=a^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}-2 a \frac{\sqrt{2}}{2} \cos (45-\theta)$

or, $\cos (45-\theta)=\frac{\frac{3}{4} a^{2}+\frac{1}{2}}{a \sqrt{2}}$

Comparing (1) and (2) gives, $\frac{a^{2}-b^{2}+2}{2 \sqrt{2} a}=\frac{\frac{3}{4} a^{2}+\frac{1}{2}}{a \sqrt{2}}$.

Simplifying this, $a^{2}+2 b^{2}=2$

$$
\text { or, } \quad b^{2}=\frac{2-a^{2}}{2}
$$

Now $\cos (45-\theta)=\frac{a^{2}+2-\left(\frac{2-a^{2}}{2}\right)}{2 a \sqrt{2}}=\frac{1}{4 \sqrt{2}}\left(3 a+\frac{2}{a}\right)$.

Now considering $3 a+\frac{2}{a}$, we know $\left(\sqrt{3 a}-\sqrt{\frac{2}{a}}\right)^{2} \geq 0$

$$
\text { or, } \quad 3 a+\frac{2}{a} \geq 2 \sqrt{6}
$$

Thus, $\cos (45-\theta) \geq \frac{1}{4 \sqrt{2}}(2 \sqrt{6})=\frac{\sqrt{3}}{2}$

$$
\cos (45-\theta) \geq \frac{\sqrt{3}}{2}
$$

$\cos (45-\theta)$ has a minimum value for $45^{\circ}-\theta=30^{\circ}$ or $\theta=15^{\circ}$.
Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\angle A D P=\theta$.

By interior alternate angles, $\angle P=\angle Q$ and $\angle P D M=\angle Q B M$.

Thus $\triangle P D M \cong \triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.

So $M$ is the midpoint of $B D$ and the centre of the square.



Without loss of generality, let $P M=1$. Then $P D=2$.

Since $\theta+\alpha=45^{\circ}$ (see diagram), $\theta$ will be minimized when $\alpha$ is maximized.

<img_3749>

Consider $\triangle P M D$.

Using the sine law, $\frac{\sin \alpha}{1}=\frac{\sin (\angle P M D)}{2}$.

To maximize $\alpha$, we maximize $\sin \alpha$.

But $\sin \alpha=\frac{\sin (\angle P M D)}{2}$, so it is maximized when $\sin (\angle P M D)=1$.

In this case, $\sin \alpha=\frac{1}{2}$, so $\alpha=30^{\circ}$.

Therefore, $\theta=45^{\circ}-30^{\circ}=15^{\circ}$, and so the minimum value of $\theta$ is $15^{\circ}$.
We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.

Let $P D=P Q=Q B=a$, and $\angle A D P=\theta$.

Drop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.

Then $P X=a \sin \theta, D X=a \cos \theta$.

Therefore the coordinates of $P$ are $(a \sin \theta, a \cos \theta)$.

Since $P D \| B Q$, then $\angle Q B C=\theta$.

So by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \sin \theta, 1+a \cos \theta)$.

<img_3706>

Now $(P Q)^{2}=a^{2}$, so $(1-2 a \sin \theta)^{2}+(1-2 a \cos \theta)^{2}=a^{2}$

$$
2+4 a^{2} \sin ^{2} \theta+4 a^{2} \cos ^{2} \theta-4 a(\sin \theta+\cos \theta)=a^{2}
$$



$$
\begin{aligned}
2+4 a^{2}-a^{2} & =4 a(\sin \theta+\cos \theta) \\
\frac{2+3 a^{2}}{4 a} & =\sin \theta+\cos \theta \\
\frac{2+3 a^{2}}{4 \sqrt{2} a} & =\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\cos \left(45^{\circ}\right) \sin \theta+\sin \left(45^{\circ}\right) \cos \theta \\
\frac{2+3 a^{2}}{4 \sqrt{2} a} & =\sin \left(\theta+45^{\circ}\right)
\end{aligned}
$$

$$
\text { Now } \begin{aligned}
\left(a-\sqrt{\frac{2}{3}}\right)^{2} & \geq 0 \\
a^{2}-2 a \sqrt{\frac{2}{3}}+\frac{2}{3} & \geq 0 \\
3 a^{2}-2 a \sqrt{6}+2 & \geq 0 \\
3 a^{2}+2 & \geq 2 a \sqrt{6} \\
\frac{3 a^{2}+2}{4 \sqrt{2} a} & \geq \frac{\sqrt{3}}{2}
\end{aligned}
$$

and equality occurs when $a=\sqrt{\frac{2}{3}}$.

So $\sin \left(\theta+45^{\circ}\right) \geq \frac{\sqrt{3}}{2}$ and thus since $0^{\circ} \leq \theta \leq 90^{\circ}$, then $\theta+45^{\circ} \geq 60^{\circ}$ or $\theta \geq 15^{\circ}$.

Therefore the minimum possible value of $\angle A D P$ is $15^{\circ}$.

Final answer: $15^{\circ}$