Problem 4. Point $O$ lies inside an isosceles right triangle $A B C$. The distance from it to vertex $A$ of the right angle is 6, to vertex $B$ is 9, and to vertex $C$ is 3. Find the area of triangle $A B C$.
Answer: $\frac{45}{2}+9 \sqrt{2}$.

First solution. Consider a rotation around point $A$ by an angle of $90^{\circ}$, which maps point $C$ to point $B$. Let point $O$ be mapped to point $D$ during this rotation; then segment $BD$ is the image of segment $CO$; since the length of segments does not change during rotation, $BD=CO=8$. We obtain quadrilateral $OADB$, in which $OA=AD=6, BD=3, OB=9$, $\angle OAD=90^{\circ}$ (see diagram). We can reason in several ways from here.

First method. Consider a coordinate system where point $O$ has coordinates $(0,0)$, point $A$ has coordinates $(6,0)$, and point $D$ has coordinates $(6,-6)$. We will find the coordinates of point $B(x, y)$, taking into account that $OB=9$ and $DB=3$, i.e.,

![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-20.jpg?height=648&width=471&top_left_y=207&top_left_x=1569)

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=81 \\
(x-6)^{2}+(y+6)^{2}=9
\end{array}\right.
$$

Subtracting the second equation from the first, we get $12x-12y-72=72$, hence $x-y=12$. Substituting $x=y+12$ into the first equation, we get $2y^{2}+24y+63=0$, from which $y=\frac{-12 \pm 3 \sqrt{2}}{2}, x=\frac{12 \pm 3 \sqrt{2}}{2}$. Since point $B$ must lie on the same side of $AD$ as point $O$, then $y=\frac{-12-3 \sqrt{2}}{2}, x=\frac{12-3 \sqrt{2}}{2}$. Finally, $S=\frac{AB^{2}}{2}=\frac{(x-6)^{2}+y^{2}}{2}=\frac{((12-3 \sqrt{2}) / 2-6)^{2}+((12+3 \sqrt{2}) / 2)^{2}}{2}=\frac{45}{2}+9 \sqrt{2}$. Second method. First, note that $OD=6 \sqrt{2}, \angle ODA=45^{\circ}$. Notice that $OD^{2}+BD^{2}=$ $72+9=81=OB^{2}$. Then, by the converse of the Pythagorean theorem, $\angle ODB=90^{\circ}$, and $\angle ADB=135^{\circ}$.

Then, by the cosine rule for triangle $ADB$:

$$
S=\frac{AB^{2}}{2}=\frac{AD^{2}+BD^{2}-2 \cdot AD \cdot BD \cdot \cos 135^{\circ}}{2}=\frac{36+9-36 \sqrt{2}}{2}=\frac{45}{2}+9 \sqrt{2}
$$

Second solution. Let $AB=AC=x, \angle OAB=\varphi$. By the cosine rule for triangles $OAB$ and $OAC$ we have:

$$
\begin{aligned}
& 9^{2}=x^{2}+6^{2}-12 x \cos \varphi \\
& 3^{2}=x^{2}+6^{2}-12 x \sin \varphi
\end{aligned}
$$

from which

$$
\begin{aligned}
& 12 x \cos \varphi=x^{2}-45 \\
& 12 x \sin \varphi=x^{2}+27
\end{aligned}
$$

Squaring these inequalities and adding them, we get a quadratic equation in $x^{2}: 144 x^{2}=2 x^{4}-36 x^{2}+2754, x^{4}-90 x^{2}+1377=0$. The roots of this equation are

$$
x_{1,2}^{2}=45 \pm 18 \sqrt{2}
$$

Note that $45-18 \sqrt{2}<36$, and in this case $x<AO$, meaning point $O$ will not lie inside the triangle, so $S=\frac{x^{2}}{2}=\frac{45}{2}+9 \sqrt{2}$.

Plan for the third solution. Reflect point $O$ symmetrically with respect to the sides $AB, AC$, and $BC$ of triangle $ABC$; denote the images by $X, Y$, and $Z$ respectively. (see diagram) Then $AY=AX=AO, CY=CZ=CO, BX=BZ=BO$. A simple calculation of angles shows that $\angle YCZ=2 \angle ACB=90^{\circ}, \angle XBZ=2 \angle ABC=90^{\circ}, \angle XAY=2 \angle BAC=180^{\circ}$.

The area of pentagon $XYCZB$ is twice the area of triangle $ABC$. On the other hand, the area of $XYCZB$ is composed of the areas of two right triangles $YCZ$ and $XBZ$, in which we know one leg, as well as triangle $XYZ$, for which we know all sides, so we can find its area, for example, using Heron's formula.

![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-21.jpg?height=737&width=719&top_left_y=188&top_left_x=1248)