5. On the sides of triangle $A B C$ with side lengths 2, 3, and 4, external squares $A B B_{1} A_{1}, B C C_{2} B_{2}, C A A_{3} C_{3}$ are constructed. Find the sum of the squares of the side lengths of the hexagon $A_{1} B_{1} B_{2} C_{2} C_{3} A_{3}$.

## Answers and Solutions

Problem 1 Answer: 165 m.
Solution

$m$ - the number of steps Petya takes, $n$ - the number of steps Vova takes

$0.75 m=0.55 n=L-$ path length

$k$ - the step number of Petya leading to the coincidence of the footprint,

$i$ - the step number of Vova leading to the coincidence of the footprint,

$0.75 k=0.55 i \rightarrow 15 k=11 i \rightarrow\left\{\begin{array}{l}k=11 t \\ i=15 t\end{array}, t=0,1, \ldots T\right.$

Total number of footprints: $(m+1)+(n+1)-(T+1)=15 T+11 T-T+1=25 t+1=501 \rightarrow T=20$ $m=11 \cdot T=220 \rightarrow L=0.75 \cdot 220=165$

Problem 2 Answer: $x=6, y=55$ or $x=55, y=6$

Solution

$y(x-5)=5(x+5) \rightarrow y=5+\frac{50}{x-5} \rightarrow x-5= \pm 1 ; \pm 2 ; \pm 5 ; \pm 10 ; \pm 25 ; \pm 50$

$y-5= \pm 50 ; \pm 25 ; \pm 10 ; \pm 5 ; \pm 2 ; \pm 1 \rightarrow x+y-10= \pm 51 ; \pm 27 ; \pm 15 ;$

The maximum value of $x+y$ is 61 and is achieved at $x=6, y=55$ or $x=55, y=6$

Problem 3 Answer: $a=-\frac{9}{8}, b=-6$

## Solution

When dividing the polynomial $P(x)=a x^{4}+3 x^{3}+b$ by $x-2$, we have the remainder $16 a+b+24=0$. Then

$P(x)=(x-2)\left(a x^{3}+(2 a+3) x^{2}+(4 a+6) x+(8 a+12)\right)=(x-2) P_{1}(x)$

When dividing the polynomial $P_{1}(x)$ by $x-2$, we have the remainder $32 a+36=0 \rightarrow a=-\frac{9}{8} \rightarrow b=-6$.

Problem 4 Answer: $a=2$

Solution

Condition for the existence of roots:

$\frac{D}{4}=16(a-1)^{2}+8(a-4) \geq 0 \rightarrow 2 a^{2}-3 a-2 \geq 0 \rightarrow a \in(-\infty ;-0.5] \cup[2 ;+\infty)$.

$x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(a-1)^{2}-\frac{4-a}{4}=a^{2}-\frac{7 a}{4}$

We need to find the minimum of this quadratic trinomial on the set $(-\infty ;-0.5] \cup[2 ;+\infty)$.

The abscissa of its vertex $a=\frac{7}{8}$ is closer to 2 than to $-0.5$, since $2-\frac{7}{8}=\frac{9}{8}<\frac{7}{8}+\frac{1}{2}=\frac{11}{8}$.

Therefore, the minimum possible value of the sum of the squares of the roots of the equation is achieved at $a=2$.

Problem 5 Answer: 116

Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_f48757316b8b60dacdebg-2.jpg?height=897&width=968&top_left_y=1276&top_left_x=178)

Additional construction: $B C=C K=a, M$ - midpoint of side $A B$.

$\square A C K=\square C_{2} C B_{3}=\square K C B_{3}=180^{\circ}-\square B C A$

$\square A C K={ }_{B_{3} C C_{2} \rightarrow C_{2} B_{3}=A K=2 C M}$

$C M^{2}=\frac{2 a^{2}+2 b^{2}-c^{2}}{4} \rightarrow C_{2} B_{3}{ }^{2}=2 a^{2}+2 b^{2}-c^{2}$

Similarly,

$$
B_{1} B_{2}{ }^{2}=2 a^{2}+2 b^{2}-c^{2}, \quad A_{1} A_{3}{ }^{2}=2 b^{2}+2 c^{2}-a^{2}
$$

$A_{1} A_{2}{ }^{2}=c^{2}, B_{2} C_{2}{ }^{2}=a^{2}, A_{3} B_{3}{ }^{2}=b^{2}$

Combining, we arrive at the sum $\sigma^{2}$ of the squares of the sides of the hexagon being $4\left(a^{2}+b^{2}+c^{2}\right) \cdot$ In the conditions of the variant $1 a=2, b=3, c=4$, therefore $\sigma^{2}=4(4+9+16)=116$