10.4. For various non-zero real numbers $a, b$, and $c$, the following equalities are satisfied:

$$
a+\frac{2}{b}=b+\frac{2}{c}=c+\frac{2}{a}=d
$$

Find all possible values of the number $d$.
Answer: $\pm \sqrt{2}$.

Solution. Express $b$ and $c$ from the equations, we get $b=\frac{2}{d-a}, c=\frac{d a-2}{a}$. Hence, $b+\frac{2}{c}=\frac{2}{d-a}+\frac{2 a}{d a-2}=d$. Bring to a common denominator: $2(d a-2)+2 a(d-a)=d(d-a)(d a-2)$, or $2 d a-4-2 a^{2}-d^{3} a+a^{2} d^{2}+2 d^{2}=0$. Group the first term with the fourth, the second term with the sixth, and the third term with the fifth: $a d\left(2-d^{2}\right)+2\left(d^{2}-2\right)+a^{2}\left(d^{2}-2\right)=0$, we get the factorization $\left(d^{2}-2\right)\left(a^{2}-a d+2\right)=0$. Hence, $d^{2}=2$, or $a^{2}-a d+2=0$. But in the latter case, $a=\frac{d \pm \sqrt{d^{2}-8}}{2}$, and $c=d-\frac{2}{a}=d-\frac{4}{d \pm \sqrt{d^{2}-8}}=d-\frac{4\left(d \mp \sqrt{d^{2}-8}\right)}{8}=\frac{d \pm \sqrt{d^{2}-8}}{2}$, that is, $a=c$, but by the condition $a$ and $c$ are distinct.

Comment. A correct and justified solution - 7 points. The factorization is obtained, $d^{2}=2$ is found, but the case $a^{2}-a d+2=0$ is not excluded - 5 points. Some progress is made - 2 points. The solution is started, but the progress is insignificant - 1 point. Only the answer is given - 0 points. The problem is not solved or solved incorrectly - 0 points.

5) The angle bisectors of angles $A, B$, and $C$ of an acute triangle $ABC$ intersect the circumcircle of this triangle at points $A_{1}, B_{1}$, and $C_{1}$, respectively. Lines $AB$ and $B_{1}C_{1}$ intersect at point $M$, and lines $BC$ and $A_{1}B_{1}$ intersect at point $N$. Is it true that line $MN$ passes through the center of the inscribed circle of triangle $ABC$?

Answer: Yes, it is true.

Solution. Let $I$ be the center of the inscribed circle of triangle $ABC$ (point $I$ is the intersection of the angle bisectors $AA_{1}, BB_{1}$, and $CC_{1}$), and let line $B_{1}C_{1}$ intersect side $AC$ and angle bisector $AA_{1}$ at points $P$ and $Q$, respectively. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_caadc799d4ad0fdd38b8g-2.jpg?height=482&width=482&top_left_y=2329&top_left_x=1481)

![](https://cdn.mathpix.com/cropped/2024_05_06_caadc799d4ad0fdd38b8g-3.jpg?height=100&width=1093&top_left_y=84&top_left_x=487)

Since $BB_{1}$ is the angle bisector, from the equality of angles $ABB_{1}$ and $B_{1}BC$ follows the equality of arcs $CB_{1}$ and $B_{1}A$, and hence the equality of angles $AC_{1}B$ and $B_{1}C_{1}C$, which subtend these arcs. Therefore, $B_{1}C_{1}$ is the angle bisector of angle $AC_{1}I$. Moreover, from the perpendicularity of lines $AI$ and $B_{1}C_{1}$, it follows that triangle $AC_{1}I$ is isosceles, $AC_{1}=C_{1}I$, and thus $CQ$ is a median. Since $MQ$ also bisects $AI$, this also implies the isosceles property of triangle $AMI$, $AM=MI$. Similarly, it is proved that $B_{1}C_{1}$ is the angle bisector of angles $AB_{1}I$ and the isosceles property of triangle $API$. Since $AI$ is the angle bisector of angle $A$, triangles $AMI$ and $API$ have a common side $AI$ and two equal adjacent angles to this side. Therefore, these triangles are congruent, and quadrilateral $AMIP$ is a rhombus, so $MI \parallel AC$. Similarly, it is established that $NI \parallel AC$, that is, points $M, I$, and $N$ lie on the same line.

Comment. A complete and justified solution is provided - 7 points. A generally correct reasoning is provided, containing minor gaps or inaccuracies - up to 5 points. If the solution is not completed, points are awarded for proving useful auxiliary statements - 1-2 points. The problem is not solved or solved incorrectly - 0 points.