Square $P Q R S$ has side length $4 \mathrm{~m}$. Point $U$ is on $P R$ with $P R=4 U R$. A circle centered at $U$ touches two sides of the square. $P W$ is a tangent to the circle, with $W$ on $Q R$. The length of $P W$, to the nearest thousandth of a metre, is
(A) $4.123 \mathrm{~m}$
(B) $4.472 \mathrm{~m}$
(C) $4.685 \mathrm{~m}$
(D) $4.726 \mathrm{~m}$
(E) $4.767 \mathrm{~m}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-095.jpg?height=455&width=469&top_left_y=786&top_left_x=1294)
Throughout this solution, we suppress the units (metres) until the very end. All lengths until then are given in metres.

Since square $P Q R S$ has side length 4 , then its diagonal $P R$ has length $4 \sqrt{2}$.

Since $P R=4 U R$, then $P U=\frac{3}{4} P R=\frac{3}{4}(4 \sqrt{2})=3 \sqrt{2}$ and $U R=\frac{1}{4} P R=\sqrt{2}$.

Suppose that the circle touches $W R$ at $Y, R S$ at $Z$, and $P W$ at $X$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-188.jpg?height=455&width=469&top_left_y=865&top_left_x=882)

Since $R S$ is tangent to the circle at $Z$, then $\angle U Z R=90^{\circ}$.

Since $\angle P R S=45^{\circ}$ (because $P R$ is the diagonal of a square), then $\triangle U Z R$ is isosceles and right-angled.

Thus, $U Z=\frac{1}{\sqrt{2}} U R=\frac{1}{\sqrt{2}}(\sqrt{2})=1$. That is, the radius of the circle is 1 .

Therefore, $U Y=U X=U Z=1$.

Now since $P W$ is tangent to the circle at $X$, then $\angle P X U=90^{\circ}$.

By the Pythagorean Theorem, $P X=\sqrt{P U^{2}-U X^{2}}=\sqrt{(3 \sqrt{2})^{2}-1^{2}}=\sqrt{18-1}=\sqrt{17}$.

Also, $\sin (\angle U P X)=\frac{U X}{U P}=\frac{1}{3 \sqrt{2}}$ so $\angle U P X \approx 13.63^{\circ}$.

Since we know the length of $P X$, then to determine the length of $P W$, we must determine the length of $X W$.

Since $W X$ and $W Y$ are tangents to the circle from the same point $W$, then $W X=W Y$, which tells us that $\triangle U W X$ and $\triangle U W Y$ are congruent, so $\angle U W X=\angle U W Y$.

Looking at the angles in $\triangle P W R$, we have

$$
\begin{aligned}
\angle W P R+\angle P W R+\angle W R P & =180^{\circ} \\
2(\angle U W X) & \approx 180^{\circ}-45^{\circ}-13.63^{\circ} \\
2(\angle U W X) & \approx 121.37^{\circ} \\
\angle U W X & \approx 60.68^{\circ}
\end{aligned}
$$

In $\triangle U W X$, we have $\tan (\angle U W X)=\frac{U X}{X W}$ so $X W \approx \frac{1}{\tan \left(60.68^{\circ}\right)} \approx 0.5616$.

Therefore, $P W=P X+X W \approx \sqrt{17}+0.562 \approx 4.6847 \mathrm{~m}$.

To the thousandth of a metre, this equals $4.685 \mathrm{~m}$.