Square $A B C D$ has centre $O$. Points $P$ and $Q$ are on $A B, R$ and $S$ are on $B C, T$ and $U$ are on $C D$, and $V$ and $W$ are on $A D$, as shown, so that $\triangle A P W, \triangle B R Q, \triangle C T S$, and $\triangle D V U$ are isosceles and $\triangle P O W, \triangle R O Q, \triangle T O S$, and $\triangle V O U$ are equilateral. What is the ratio of the area of $\triangle P Q O$ to that of $\triangle B R Q$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_d13f9eaf1d614231018ag-4.jpg?height=393&width=401&top_left_y=389&top_left_x=924)
Solution 1

Let $X$ be the foot of the altitude of $\triangle P Q O$ from $O$ to $P Q$, and let $r=O Q$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c3c8a275c8ba62c19ff3g-14.jpg?height=597&width=609&top_left_y=1767&top_left_x=823)

It is given that $\triangle R B Q$ is isosceles and since $\angle Q B R=90^{\circ}$, we must have that $\angle B Q R=45^{\circ}$. As well, $\triangle R O Q$ is equilateral so $\angle R Q O=60^{\circ}$.

Points $X, Q$, and $B$ are on a line, so $\angle X Q O=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$.

By very similar reasoning, $\angle X P O=75^{\circ}$, which means $\triangle O P Q$ is isosceles with $O P=O Q$.

Since $\angle O Q X=75^{\circ}$, we have $\sin 75^{\circ}=\frac{O X}{O Q}=\frac{O X}{r}$ and so $O X=r \sin 75^{\circ}$.

Similarly, $Q X=r \cos 75^{\circ}$, and since $\triangle O P Q$ is isosceles, $O X$ bisects $P Q$, which means $P Q=2 r \cos 75^{\circ}$.

From the double-angle formula, $\sin 150^{\circ}=2 \sin 75^{\circ} \cos 75^{\circ}$. Using this fact and the fact that $\sin 150^{\circ}=\frac{1}{2}$, we can compute the area of $\triangle O P Q$ as

$$
\begin{aligned}
\frac{1}{2} \times P Q \times O X & =\frac{1}{2}\left(2 r \cos 75^{\circ}\right)\left(r \sin 75^{\circ}\right) \\
& =\frac{r^{2}}{2}\left(2 \sin 75^{\circ} \cos 75^{\circ}\right) \\
& =\frac{r^{2}}{2} \sin 150^{\circ} \\
& =\frac{r^{2}}{2} \times \frac{1}{2} \\
& =\frac{r^{2}}{4}
\end{aligned}
$$

Since $\triangle R O Q$ is equilateral, $Q R=O Q=r$, and since $\angle R Q B=45^{\circ}, \cos 45^{\circ}=\frac{B Q}{Q R}=\frac{B Q}{r}$.

Rearranging and using that $\cos 45^{\circ}=\frac{1}{\sqrt{2}}$, we get $B Q=\frac{r}{\sqrt{2}}$.

Since $B Q=B R$, we get $B R=\frac{r}{\sqrt{2}}$ as well.

Therefore, the area of $\triangle B R Q$ is

$$
\frac{1}{2} \times B Q \times B R=\frac{1}{2} \times \frac{r}{\sqrt{2}} \times \frac{r}{\sqrt{2}}=\frac{r^{2}}{4}
$$

The areas of $\triangle P Q O$ and $\triangle B R Q$ are equal, so the ratio is $1: 1$.

Solution 2

Let $O Q=r$, which implies $Q R=r$ as well since $\triangle R O Q$ is equilateral.

Since $\triangle B R Q$ has a right angle at $B$ and is isosceles, the Pythagorean theorem implies that $B Q^{2}+B R^{2}=r^{2}$ or $2 B Q^{2}=r^{2}$, so $B Q=\frac{r}{\sqrt{2}}$.

Therefore, the area of $\triangle B R Q$ is $\frac{1}{2} \times \frac{r}{\sqrt{2}} \times \frac{r}{\sqrt{2}}=\frac{r^{2}}{4}$.

Since $\triangle B R Q$ is isosceles and right angled at $B, \angle R Q B=45^{\circ}$. As well, $\triangle R O Q$ is equilateral, which means $\angle R O Q=60^{\circ}$. Using that $P, Q$, and $B$ lie on the same line, we can compute $\angle P Q O=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$.

By similar reasoning, $\angle Q P O=75^{\circ}$, which implies that $\triangle P Q O$ is isosceles with $O P=O Q=r$. Using the two known angles in $\triangle P Q O$, we get $\angle P O Q$ as $\angle P O Q=180^{\circ}-75^{\circ}-75^{\circ}=30^{\circ}$.

This means the area of $\triangle O P Q$ is $\frac{1}{2} \times O P \times O Q \times \sin 30^{\circ}=\frac{r^{2}}{4}$.

The areas of $\triangle P Q O$ and $\triangle B R Q$ are equal, so the ratio is $1: 1$.