Problem 4. In triangle $X Y Z$, the lengths of the sides are 2, 7, and $5 \sqrt{3}$. Find the area of the figure consisting of those and only those points $A$ inside triangle $X Y Z$ for which the condition $A X^{2}+A Y^{2}+A Z^{2} \leqslant 43$ is satisfied.
Answer: $\pi / 9$.

Solution. Let $Y Z=a, X Z=b, X Y=c, \rho^{2}=43$.

Let $G$ be the intersection point of the medians of triangle $X Y Z$. Represent

$$
\overrightarrow{A X}=\overrightarrow{G X}-\overrightarrow{G A}, \overrightarrow{A Y}=\overrightarrow{G Y}-\overrightarrow{G A}, \overrightarrow{A Z}=\overrightarrow{G Z}-\overrightarrow{G A}
$$

then

$$
A X^{2}+A Y^{2}+A Z^{2}=G X^{2}+G Y^{2}+G Z^{2}-2 \cdot \overrightarrow{G A} \cdot(\overrightarrow{G X}+\overrightarrow{G Y}+\overrightarrow{G Z})+3 \cdot G A^{2}
$$

Since $G$ is the centroid of triangle $X Y Z$, then

$$
\overrightarrow{G X}+\overrightarrow{G Y}+\overrightarrow{G Z}=0
$$

and $G X^{2}+G Y^{2}+G Z^{2}=\frac{4}{9}\left(m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\right)=\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)$.

Thus,

$$
\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)+3 \cdot G A^{2} \leqslant \rho^{2}
$$

or

$$
G A^{2} \leqslant \frac{1}{9} \cdot\left(3 \rho^{2}-a^{2}-b^{2}-c^{2}\right)
$$

Therefore, the geometric locus of points $A$ that satisfy the given condition is a circle with radius $\frac{1}{3} \sqrt{3 \rho^{2}-a^{2}-b^{2}-c^{2}}$ centered at the intersection point of the medians of triangle $X Y Z$.

This circle belongs to the triangle if its radius is not greater than one third of the smallest height of $\triangle X Y Z$:

$$
\frac{1}{3} \sqrt{3 \rho^{2}-a^{2}-b^{2}-c^{2}} \leqslant \frac{2 S_{\triangle X Y Z}}{3 \max \{a, b, c\}}
$$

Thus, when the condition

$$
\frac{a^{2}+b^{2}+c^{2}}{3}<\rho^{2} \leqslant \frac{a^{2}+b^{2}+c^{2}}{3}+\frac{4}{3} \cdot\left(\frac{S_{\triangle X Y Z}}{\max \{a, b, c\}}\right)^{2}
$$

is satisfied, the desired area is $S=\frac{\pi}{9} \cdot\left(3 \rho^{2}-a^{2}-b^{2}-c^{2}\right)$.

Using Heron's formula, we find the area of the triangle:

$$
S_{\triangle X Y Z}=\frac{1}{4} \sqrt{(9+5 \sqrt{3})(9-5 \sqrt{3})(5 \sqrt{3}+5)(5 \sqrt{3}-5)}=\frac{5 \sqrt{3}}{2} .
$$

Calculate

$$
\frac{a^{2}+b^{2}+c^{2}}{3}=\frac{128}{3}, \frac{S_{\triangle X Y Z}}{\max \{a, b, c\}}=\frac{1}{2}
$$

Since $\rho^{2}=43$, the condition $(*)$ is satisfied:

$$
\frac{128}{3}<43 \leqslant \frac{128}{3}+\frac{1}{3}=43
$$

Thus, the answer is: $S=\frac{\pi}{9} \cdot(129-128)=\frac{\pi}{9}$.

Another solution. The height of the triangle, drawn to the side of length $5 \sqrt{3}$, is 1. The base of the height divides this side into segments of lengths $\sqrt{3}$ and $4 \sqrt{3}$. Introduce a coordinate system as shown in the figure. Then $X(-\sqrt{3} ; 0), Y(0 ; 1), Z(4 \sqrt{3} ; 0)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5f8906e33c6403142659g-09.jpg?height=329&width=1062&top_left_y=795&top_left_x=497)

$A X^{2}+A Y^{2}+A Z^{2}=(x+\sqrt{3})^{2}+y^{2}+x^{2}+(y-1)^{2}+(x-4 \sqrt{3})^{2}+y^{2}=3 x^{2}+3 y^{2}-6 \sqrt{3} x-2 y+52 \leqslant 43$.

Rewrite the inequality as:

$$
(x-\sqrt{3})^{2}+\left(y-\frac{1}{3}\right)^{2} \leqslant \frac{1}{9}
$$

This inequality defines a circle with radius $R=\frac{1}{3}$ centered at point $K(\sqrt{3} ; 1 / 3)$. We will show that all points of this circle belong to triangle $X Y Z$. For this, we find the distances from point $K$ to the sides of the triangle. The equation of side $X Y: x-y \sqrt{3}+\sqrt{3}=0$, the distance to it is $d_{1}=\frac{|\sqrt{3}-\sqrt{3} / 3+\sqrt{3}|}{\sqrt{1^{2}+(-\sqrt{3})^{2}}}=\frac{5 \sqrt{3}}{6}$. The equation of side $Y Z: x+4 \sqrt{3} y-4 \sqrt{3}=0$, the distance is $d_{2}=\frac{|\sqrt{3}+4 \cdot(\sqrt{3} / 3)-4 \sqrt{3}|}{\sqrt{1^{2}+(4 \sqrt{3})^{2}}}=\frac{5 \sqrt{3}}{21}$. The distance from point $K$ to side $X Z$ is obviously the smallest of the three and is $d_{3}=\frac{1}{3}=R$. Therefore, the circle belongs to the triangle, hence the desired area is $S=\pi R^{2}=\frac{\pi}{9}$.