5. The border of a square with a side of 9, cut out of white cardboard, is painted red. It is necessary to cut the square into 6 equal-area parts, the boundaries of which contain segments painted red with the same total length.

## Solutions

Option 1

Problem 1
Answer: 2 km.

Solution

$S-$ the length of the path, $S_{1}$ - the length of the path on descents, $S_{2}$ - the length of the path on ascents, $S / 2=S_{1}+S_{2}$

$t=15-8-6=1-$ the time of the journey there and back. Then

$1=\frac{S_{1}}{6}+\frac{S_{2}}{3}+\frac{S}{2 \cdot 4}+\frac{S_{1}}{3}+\frac{S_{2}}{6}+\frac{S}{2 \cdot 4}=\left(\frac{1}{6}+\frac{1}{3}\right)\left(S_{1}+S_{2}\right)+\frac{S}{4}=\left(\frac{1}{6}+\frac{1}{3}\right) \frac{S}{2}+\frac{S}{4}=\frac{S}{2} \rightarrow S=2$

Problem 2 Answer: $\frac{1}{6}$

Solution.

By the condition $\frac{p}{q}5 p$. Considering the naturality of $p$ and $q$, and the condition $p+q<8$ we have

$6 p<p+q<8 \rightarrow p=1 \rightarrow q=6$

Problem 3 Answer: $a_{\min }=15876$

Solution

Among the divisors of an even number divisible by 21 and being a square of an integer, there are $7^{2}, 3^{2}, 2^{2}$, i.e.

$a=7^{2} \cdot 3^{2} \cdot 4 \cdot k^{2}$. Considering the five-digit nature, the inequality holds:

$10000 \leq 7^{2} \cdot 3^{2} \cdot 4 \cdot k^{2} \leq 99998 \rightarrow 6 \leq k^{2} \leq 57 \rightarrow k_{\min }^{2}=9 \rightarrow a_{\min }=7^{2} \cdot 3^{2} \cdot 4 \cdot 9=15876$

Problem 4 Answer: $x=\frac{3}{2}$

Solution

$x=\frac{4-[x]}{2} \rightarrow\{x\}=x-[x]=\frac{4-[x]}{2}-[x]=\frac{4-3[x]}{2}$

Since $0 \leq\{x\}<1 \rightarrow 0 \leq \frac{4-3[x]}{2}<1 \rightarrow \frac{2}{3}<[x] \leq \frac{4}{3}$ and the number $[x]-$ is an integer, the only valid value is $[x]=1$. Then $x=\frac{3}{2}$

Problem 5

Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_9f2b6f89403dbc5f8f95g-2.jpg?height=400&width=434&top_left_y=1882&top_left_x=662)

O - the center of the square, the polygon $O B C D$ - one of the six desired pieces, $B C=x, C D=y, a=9$.

$S_{\text {OBCD }}=\frac{a^{2}}{6}=\frac{a}{4} \cdot(x+y) \rightarrow x+y=\frac{2 a}{3}=6$ - the length of the red part of the boundary of the polygon.

A way of cutting (one of ...)

![](https://cdn.mathpix.com/cropped/2024_05_06_9f2b6f89403dbc5f8f95g-3.jpg?height=322&width=308&top_left_y=136&top_left_x=680)

Variant 2

Problem 1 Answer: 3 km.

Problem 2 Answer: $\frac{1}{4}$

Problem 3 Answer: $a_{\max }=5625$

Problem 4 Answer: $x=\frac{5}{2}$

Problem 5 Answer:

![](https://cdn.mathpix.com/cropped/2024_05_06_9f2b6f89403dbc5f8f95g-3.jpg?height=355&width=349&top_left_y=1116&top_left_x=685)

Variant 3

Problem 1 Answer: 1.5 km.

Problem 2 Answer: $\frac{1}{4}, \frac{1}{5}$

Problem 3 Answer: $a_{\text {min }}=327184$

Problem 4 Answer: $x=-\frac{7}{2}$

Problem 5 Answer:

![](https://cdn.mathpix.com/cropped/2024_05_06_9f2b6f89403dbc5f8f95g-3.jpg?height=460&width=439&top_left_y=2157&top_left_x=657)

Variant 4

Problem 1 Answer: 1.4 km.

Problem 2 Answer: $\frac{1}{6}, \frac{1}{7}$

Problem 3 Answer: $a_{\max }=74529$

Problem 4 Answer: $x=-\frac{3}{2}$

Problem 5 Answer:

![](https://cdn.mathpix.com/cropped/2024_05_06_9f2b6f89403dbc5f8f95g-4.jpg?height=348&width=340&top_left_y=597&top_left_x=661)