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 5, to vertex $B$ is 7, and to vertex $C$ is 3. Find the area of triangle $A B C$.
Answer: $\frac{29}{2}+\frac{5}{2} \sqrt{17}$.

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, $BD=CO=3$. We obtain quadrilateral $OADB$, in which $OA=AD=5, BD=3, OB=7$, $\angle OAD=90^{\circ}$ (see drawing). We can reason in several ways from here.

![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-14.jpg?height=532&width=397&top_left_y=208&top_left_x=1572)

First method. Consider a coordinate system in which point $O$ has coordinates $(0,0)$, point $A$ has coordinates $(5,0)$, and point $D$ has coordinates $(5,-5)$. We will find the coordinates of point $B(x, y)$, taking into account that $OB=7$ and $DB=3$, i.e.,

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=49 \\
(x-5)^{2}+(y+5)^{2}=9
\end{array}\right.
$$

Subtracting the second equation from the first, we get $10x-10y-50=40$, hence $x-y=9$. Substituting $x=y+9$ into the first equation, we get $2y^{2}+18y+32=0, y^{2}+9y+16=0$, from which $y=\frac{-9 \pm \sqrt{17}}{2}$, $x=\frac{9 \pm \sqrt{17}}{2}$. Since point $B$ must lie on the same side of $AD$ as point $O$, then $y=\frac{-9-\sqrt{17}}{2}, x=\frac{9-\sqrt{17}}{2}$.

Finally, $S=\frac{AB^{2}}{2}=\frac{(x-5)^{2}+y^{2}}{2}=\frac{((9-\sqrt{17}) / 2-5)^{2}+((9+\sqrt{17}) / 2)^{2}}{2}=\frac{29}{2}+\frac{5}{2} \sqrt{17}$.

Second method. First, note that $OD=5 \sqrt{2}, \angle ODA=45^{\circ}$. Let $\angle ODB=\varphi$. Then, by the cosine theorem for triangle $ODB$, we have

$$
\cos \varphi=\frac{OD^{2}+BD^{2}-OB^{2}}{2 \cdot OD \cdot BD}=\frac{(5 \sqrt{2})^{2}+3^{2}-7^{2}}{2 \cdot 5 \sqrt{2} \cdot 3}=\frac{10}{2 \cdot 5 \sqrt{2} \cdot 3}=\frac{\sqrt{2}}{6}
$$

and then $\sin \varphi=\sqrt{1-\frac{2}{36}}=\frac{\sqrt{34}}{6}$. Now, by the cosine theorem for triangle $ADB$, we get

$$
S=\frac{AB^{2}}{2}=\frac{AD^{2}+BD^{2}-2 \cdot AD \cdot BD \cdot \cos \left(\varphi+45^{\circ}\right)}{2}=\frac{25}{2}+\frac{9}{2}-5 \cdot 3 \cdot \frac{\cos \varphi-\sin \varphi}{\sqrt{2}}=\frac{29}{2}+\frac{5}{2} \sqrt{17}
$$

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

$$
\begin{aligned}
& 7^{2}=x^{2}+5^{2}-10 x \cos \varphi \\
& 3^{2}=x^{2}+5^{2}-10 x \sin \varphi
\end{aligned}
$$

from which

$$
\begin{aligned}
10 x \cos \varphi & =x^{2}-24 \\
10 x \sin \varphi & =x^{2}+16
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-14.jpg?height=437&width=434&top_left_y=2012&top_left_x=1531)

Squaring these inequalities and adding them, we get a quadratic equation in $x^{2}: 100 x^{2}=2 x^{4}-16 x^{2}+832, x^{4}-58 x^{2}+416=0$. The roots of this equation are

$$
x_{1,2}^{2}=29 \pm 5 \sqrt{17}
$$

Note that $29-5 \sqrt{17}<25$, and in this case $x<AO$, meaning point $O$ will not lie inside the triangle, so $S=\frac{x^{2}}{2}=\frac{29}{2}+\frac{5}{2} \sqrt{17}$.

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 figure) 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 $XYCZB$ consists 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-15.jpg?height=739&width=719&top_left_y=190&top_left_x=1248)