Subfield: Geometry


Question: In the diagram, $P Q R S$ is a square with sides of length 4. Points $T$ and $U$ are on sides $Q R$ and $R S$ respectively such that $\angle U P T=45^{\circ}$. Determine the maximum possible perimeter of $\triangle R U T$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b301ea0eb47ddd66a1f3g-1.jpg?height=398&width=414&top_left_y=1926&top_left_x=1300)

Solution:
Rotate a copy of $\triangle P S U$ by $90^{\circ}$ counterclockwise around $P$, forming a new triangle $P Q V$. Note that $V$ lies on the extension of $R Q$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4d95b82665f44d964a68g-1.jpg?height=585&width=420&top_left_y=1285&top_left_x=950)

Then $P V=P U$ by rotation.

Also, $\angle V P T=\angle V P Q+\angle Q P T=\angle U P S+\angle Q P T=90^{\circ}-\angle U P T=90^{\circ}-45^{\circ}$.

This tells us that $\triangle P T U$ is congruent to $\triangle P T V$, by "side-angle-side".

Thus, the perimeter of $\triangle R U T$ equals

$$
\begin{aligned}
U R+R T+U T & =U R+R T+T V \\
& =U R+R T+T Q+Q V \\
& =U R+R Q+S U \\
& =S U+U R+R Q \\
& =S R+R Q \\
& =8
\end{aligned}
$$

That is, the perimeter of $\triangle R U T$ always equals 8 , so the maximum possible perimeter is 8 .
Let $\angle S P U=\theta$. Note that $0^{\circ} \leq \theta \leq 45^{\circ}$.

Then $\tan \theta=\frac{S U}{P S}$, so $S U=4 \tan \theta$.

Since $S R=4$, then $U R=S R-S U=4-4 \tan \theta$.

Since $\angle U P T=45^{\circ}$, then $\angle Q P T=90^{\circ}-45^{\circ}-\theta=45^{\circ}-\theta$.

Thus, $\tan \left(45^{\circ}-\theta\right)=\frac{Q T}{P Q}$ and so $Q T=4 \tan \left(45^{\circ}-\theta\right)$.

Since $Q R=4$, then $R T=4-4 \tan \left(45^{\circ}-\theta\right)$.

But $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, so $\tan \left(45^{\circ}-\theta\right)=\frac{\tan \left(45^{\circ}\right)-\tan \theta}{1+\tan \left(45^{\circ}\right) \tan \theta}=\frac{1-\tan \theta}{1+\tan \theta}$, since $\tan \left(45^{\circ}\right)=1$.

This gives $R T=4-4\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\frac{4+4 \tan \theta}{1+\tan \theta}-\frac{4-4 \tan \theta}{1+\tan \theta}=\frac{8 \tan \theta}{1+\tan \theta}$.

By the Pythagorean Theorem in $\triangle U R T$, we obtain

$$
\begin{aligned}
U T & =\sqrt{U R^{2}+R T^{2}} \\
& =\sqrt{(4-4 \tan \theta)^{2}+\left(\frac{8 \tan \theta}{1+\tan \theta}\right)^{2}} \\
& =4 \sqrt{(1-\tan \theta)^{2}+\left(\frac{2 \tan \theta}{1+\tan \theta}\right)^{2}} \\
& =4 \sqrt{\left(\frac{1-\tan ^{2} \theta}{1+\tan \theta}\right)^{2}+\left(\frac{2 \tan \theta}{1+\tan \theta}\right)^{2}} \\
& =\sqrt[4]{\frac{1-2 \tan ^{2} \theta+\tan ^{4} \theta+4 \tan ^{2} \theta}{\left(1+\tan ^{2}\right)^{2}}} \\
& =\sqrt[4]{\frac{1+2 \tan ^{2} \theta+\tan ^{4} \theta}{\left(1+\tan ^{2}\right.}} \\
& =4 \sqrt{\frac{\left(1+\tan ^{2} \theta\right)^{2}}{\left(1+\tan ^{2}\right.}} \\
& =4\left(\frac{1+\tan ^{2} \theta}{1+\tan ^{2}}\right)
\end{aligned}
$$

Therefore, the perimeter of $\triangle U R T$ is

$$
\begin{aligned}
U R+R T+U T & =4-4 \tan \theta+\frac{8 \tan \theta}{1+\tan \theta}+4\left(\frac{1+\tan ^{2} \theta}{1+\tan \theta}\right) \\
& =4\left(\frac{1-\tan ^{2} \theta}{1+\tan \theta}+\frac{2 \tan \theta}{1+\tan \theta}+\frac{1+\tan ^{2} \theta}{1+\tan \theta}\right) \\
& =4\left(\frac{2+2 \tan \theta}{1+\tan \theta}\right) \\
& =8
\end{aligned}
$$

Thus, the perimeter is always 8 , regardless of the value of $\theta$, so the maximum possible perimeter is 8 .

Final answer: 8