784. Find such a point in an isosceles right triangle for which the sum of the squares of the distances to its vertices is the smallest.
Solution. Let's choose a rectangular coordinate system $x O y$ as shown in Fig. 144, then the coordinates of the vertices of the triangle will be $A(0,0), B(a, 0), C(0, a)$. We take an arbitrary point $M(x, y)$ inside the triangle

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-258.jpg?height=332&width=323&top_left_y=550&top_left_x=862)

Fig. 144 and determine the sum of the squares of the distances from it to the vertices of the triangle $u=M A^{2}+M B^{2}+M C^{2}=3 x^{2}+$ $+3 y^{2}-2 a y-2 a x+2 a^{2}$. It depends on two variables $x$ and $y$, which, according to the condition, can take any values from the closed region of the triangle $A B C$.

Next, according to the rule indicated at the beginning of this paragraph, we find the minimum value of the function $u(x, y)$ in the triangle $A B C$:
A. $u_{x}^{\prime}=6 x-2 a ; u_{y}^{\prime}=6 y-2 a$.

From the system of equations $u_{x}^{\prime}=0, u_{y}^{\prime}=0$ we find the only critical point $K\left(\frac{a}{3}, \frac{a}{3}\right)$, lying inside the triangle $A B C$. The value of $u$ at this point $u(K)=\frac{4}{3} a^{2}$.

B. On the side $A B$ we have: $y=0, u(x, 0)=u_{1}=3 x^{2}-2 a x+$ $+2 a^{2}$, where $0 \leqslant x \leqslant a$.
I. $u_{1}^{\prime}=6 x-2 a ; u_{1}^{\prime}=0$ at $x=\frac{a}{3}$ (at point $M_{1}$); $u_{1}\left(\frac{a}{3}\right)=\frac{5}{3} a^{2}$. II. $u_{1}(0)=2 a^{2} ; u_{1}(a)=3 a^{2}$.