Square $A C D E$ is inscribed in a circle centred at $O$ with radius $\sqrt{2}$. Point $B$ is on the circumference of the circle so that $B E$ and $B D$ have the same length. Line segment $A C$ intersects $B E$ and $B D$ at $F$ and $G$, respectively. Determine the ordered pair $(a, b)$ of integers so that the area of $\triangle B F G$ equals $a+b \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d49dd011068fc0fd28d1g-2.jpg?height=433&width=415&top_left_y=415&top_left_x=1408)
Connect $O$ to each of $A, C, D$, and $E$ as shown.

Each of $O A, O C, O D$, and $O E$ are radii of the circle and so are equal.

As well, $A C=C D=D E=E A$ since they are the sidelengths of a square, which means $\triangle A O C, \triangle C O D, \triangle D O E$, and $\triangle E O A$ are all congruent.

This means

$$
\angle A O C=\angle C O D=\angle D O E=\angle E O A
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-05.jpg?height=455&width=436&top_left_y=168&top_left_x=1495)

and since the sum of these angles must be $360^{\circ}$,

$$
\angle A O C=\angle C O D=\angle D O E=\angle E O A=\frac{360^{\circ}}{4}=90^{\circ} .
$$

By the Pythagorean theorem, $O E^{2}+O D^{2}=E D^{2}$.

Since the radius of the circle is $\sqrt{2}$, we have $(\sqrt{2})^{2}+(\sqrt{2})^{2}=E D^{2}$ or $2+2=4=E D^{2}$, so $E D=2$ since $E D>0$.

Now draw a line from $B$ through $O$ and label by $H$ and $J$ its intersections with $E D$ and $A C$, respectively.

We know that $\angle E O D=90^{\circ}$, which means

$$
\angle E O B+\angle D O B=360^{\circ}-90^{\circ}=270^{\circ} \text {. }
$$

Furthermore, $O E, O D$, and $O B$ are all radii of the circle and hence are equal, and $B E=B D$ by construction, so $\triangle E O B$ and $\triangle D O B$ are congruent.

This means $\angle E O B=\angle D O B$ and since their sum is $270^{\circ}$,

we must have $\angle E O B=\frac{270^{\circ}}{2}=135^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-05.jpg?height=536&width=491&top_left_y=1025&top_left_x=1465)

Since $B O H$ is a straight line, this means $\angle E O H=180^{\circ}-135^{\circ}=45^{\circ}$.

We previously established that $\triangle E O D$ is right and isosceles, so $\angle O E D=45^{\circ}$ as well.

This means $\angle O H E=180^{\circ}-45^{\circ}-45^{\circ}=90^{\circ}$.

We now have that $\triangle E B H$ and $\triangle D B H$ are right triangles having a common hypotenuse length and a shared side, which means they are congruent.

Therefore, $E H=H D$ and it follows that $E H=1$ because $E D=2$.

Line segments $A C$ and $D E$ are parallel because they are opposite sides of a square.

Therefore, $\angle B F J=\angle B E H$ and $\angle B J F=\angle B H E$. It follows that $\triangle B F J$ and $\triangle B E H$ are similar.

Since $\angle B H E=90^{\circ}$, we also have that $A E$ and $B H$ are parallel which means $\angle A E F=\angle J B F$ by alternating angles.

Angles $\angle A F E$ and $\angle J F B$ are opposite and hence equal, and both $\angle E A F$ and $\angle B J F$ are right, which means $\triangle E F A$ and $\triangle B F J$ are similar.

By similar reasoning, $\triangle D G C$ is also similar to $\triangle B F J, \triangle B E H$, and $\triangle E F A$.

Since $\triangle D G C$ and $\triangle E F A$ are similar and $A E=C D$, we have that $\triangle D G C$ and $\triangle E F A$ are congruent, which means $A F=G C$.

Since $\triangle E F A$ is similar to $\triangle B E H$, we get that

$$
\frac{A F}{A E}=\frac{H E}{H B}=\frac{H E}{H O+O B}=\frac{1}{1+\sqrt{2}} .
$$

Since $A E=2$, this means

$$
A F=\frac{2}{1+\sqrt{2}}=\frac{2}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}=2 \sqrt{2}-2 .
$$

Since $A F=G C$, we have that $G C=2 \sqrt{2}-2$ as well, and using that $A C=2$, we have

$$
F G=A C-A F-G C=2-(2 \sqrt{2}-2)-(2 \sqrt{2}-2)=6-4 \sqrt{2} .
$$

Since $A C$ and $E D$ are parallel, we have that $\triangle B F G$ is similar to $\triangle B E D$.

This means their heights have the same ratio as their sides, so

$$
\frac{B J}{B H}=\frac{F G}{E D}
$$

Since $B H=1+\sqrt{2}, F G=6-4 \sqrt{2}$, and $E D=2$, this means

$$
B J=B H \frac{F G}{E D}=(1+\sqrt{2}) \frac{6-4 \sqrt{2}}{2}=(1+\sqrt{2})(3-2 \sqrt{2})=-1+\sqrt{2}
$$

Therefore, the area of $\triangle B F G$ is

$$
\frac{1}{2} \times F G \times B J=\frac{1}{2}(6-4 \sqrt{2})(-1+\sqrt{2})=(3-2 \sqrt{2})(-1+\sqrt{2})=-7+5 \sqrt{2}
$$

The answer is $(-7,5)$.

ANSWER: $(-7,5)$