## 1st grade - high school - $B$ category,  May 3, 2007.

## Solutions

Problem 1B-1. What relation connects the numbers $a, b$, and $c$ if for some $x$ and $y$ the equalities

$$
a=x-y, \quad b=x^{2}-y^{2}, \quad c=x^{3}-y^{3} ?
$$
Solution. Considering the first equality, the second and third equalities can be simplified as follows:

$$
\begin{aligned}
& b=x^{2}-y^{2}=(x-y)(x+y)=a(x+y) \\
& c=x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=a\left(x^{2}+x y+y^{2}\right)
\end{aligned}
$$

Now from the system

$$
\begin{aligned}
x-y & =a \\
x+y & =\frac{b}{a} \\
x^{2}+x y+y^{2} & =\frac{c}{a}
\end{aligned}
$$

we want to eliminate \(x\) and \(y\).

From the first two equations, we find \(x=\frac{b+a^{2}}{2 a}\) and \(y=\frac{b-a^{2}}{2 a}\). Substituting these into the third equation gives

$$
\left(\frac{b+a^{2}}{2 a}\right)^{2}+\frac{b+a^{2}}{2 a} \cdot \frac{b-a^{2}}{2 a}+\left(\frac{b-a^{2}}{2 a}\right)^{2}=\frac{c}{a}
$$

After simplification, this equation takes the final form

$$
a^{4}+3 b^{2}-4 a c=0
$$

Problem 1B-2. Given a right triangle \(\triangle ABC\) with a right angle at vertex \(C\). On the leg \(\overline{BC}\), choose point \(A_1\), and on the leg \(\overline{AC}\), choose point \(B_1\). Prove that

$$
\left|A A_{1}\right|^{2}+\left|B B_{1}\right|^{2}=|A B|^{2}+\left|A_{1} B_{1}\right|^{2}
$$

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_b8d514dad6d399995164g-05.jpg?height=288&width=439&top_left_y=2283&top_left_x=817)

From the right triangle \(A A_{1} C\), we have

$$
\left|A_{1} A\right|^{2}=|A C|^{2}+\left|C A_{1}\right|^{2}
$$

and from the right triangle \(B B_{1} C\), we have

$$
\left|B B_{1}\right|^{2}=\left|B_{1} C\right|^{2}+|C B|^{2}
$$

Adding these equations, we get

$$
\begin{aligned}
\left|A A_{1}\right|^{2}+\left|B B_{1}\right|^{2} & =|A C|^{2}+|C B|^{2}+\left|C A_{1}\right|^{2}+\left|B_{1} C\right|^{2} \\
& =|A B|^{2}+\left|A_{1} B_{1}\right|^{2}
\end{aligned}
$$

since triangles \(A B C\) and \(A_{1} B_{1} C\) are right triangles.

Problem 1B-3. Prove that the number

$$
\underbrace{111 \ldots 1}_{2 n}-\underbrace{222 \ldots 2}_{n \text { digits }}
$$

is the square of a natural number.

Solution.

$$
\begin{aligned}
\underbrace{111 \ldots 1}_{2 n}-\underbrace{222 \ldots 2}_{n} & =\underbrace{111 \ldots 1}_{2 n}-2 \cdot \underbrace{111 \ldots 1}_{n} \\
& =(\underbrace{111 \ldots 1}_{n} \underbrace{000 \ldots 0}_{n}+\underbrace{111 \ldots 1}_{n})-2 \cdot \underbrace{111 \ldots 1}_{n} \\
& =\underbrace{111 \ldots 1}_{n} \underbrace{000 \ldots 0}_{n}-\underbrace{111 \ldots 1}_{n} \\
& =\underbrace{111 \ldots 1}_{n} \cdot 10^{n}-\underbrace{111 \ldots 1}_{n} \\
& =\underbrace{111 \ldots 1}_{n} \cdot\left(10^{n}-1\right) \\
& =\underbrace{111 \ldots 1}_{n} \cdot \underbrace{999 \ldots 9}_{n} \\
& =\underbrace{111 \ldots 1}_{n} \cdot(9 \cdot \underbrace{111 \ldots 1}_{n}) \\
& =9 \cdot(\underbrace{111 \ldots 1}_{n})^{2}=(3 \cdot 111 \ldots 1)^{2} \\
& =(\underbrace{333 \ldots 3}_{n})^{2} .
\end{aligned}
$$

Problem 1B-4. A circle is inscribed in a square with side length 1, and four circles are constructed with the sides of the square as diameters. Calculate the perimeter of the "propeller" in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_b8d514dad6d399995164g-07.jpg?height=531&width=548&top_left_y=334&top_left_x=754)

## Solution.

Draw a quarter of the square and the corresponding circular arcs. Point \(A\) is the center of the square, and \(B\) and \(E\) are the midpoints of its sides.

![](https://cdn.mathpix.com/cropped/2024_05_30_b8d514dad6d399995164g-07.jpg?height=374&width=400&top_left_y=1201&top_left_x=834)

Triangles \(A B D\) and \(A C E\) are equilateral with side length 1, so triangle \(A C D\) is isosceles with legs \(\overline{A C}\) and \(\overline{A D}\) of length 1 and angle \(\angle C A D = 30^\circ\). Therefore, the perimeter of a quarter of the "propeller," which is the sum of the lengths of arcs \(\overparen{A C}\), \(\overparen{A D}\), and \(C D\), is

$$
2 \cdot \frac{\pi}{3} + \frac{\pi}{6} = \frac{5 \pi}{6}
$$

The total perimeter of the propeller is

$$
4 \cdot \frac{5 \pi}{6} = \frac{10 \pi}{3}
$$

Problem 1B-5. Find all natural numbers \(m, n \in \mathbf{N}\) such that

$$
\frac{1}{m} + \frac{1}{n} - \frac{1}{m n} = \frac{2}{5}
$$

Solution. First solution. Without loss of generality, assume \(m \geq n\). Then

$$
\frac{2}{5} = \frac{1}{m} + \frac{1}{n} - \frac{1}{m n} \leq \frac{2}{n} - \frac{1}{m n} < \frac{2}{n}
$$

Thus,

$$
\frac{2}{5} < \frac{2}{n} \Longleftrightarrow n < 5
$$

Rearrange the equation for \(m\):

$$
\begin{aligned}
& n + m - 1 = \frac{2}{5} m n \\
& 5 n - 5 = 2 m n - 5 m \\
& m = \frac{5 n - 5}{2 n - 5}
\end{aligned}
$$

For \(n = 1\) and \(n = 2\), we get \(m \leq 0\), so there is no solution.

For \(n = 3\), we get \(m = \frac{5 \cdot 3 - 5}{2 \cdot 3 - 5} = 10\).

For \(n = 4\), we get \(m = \frac{5 \cdot 4 - 5}{2 \cdot 4 - 5} = \frac{15}{3} = 5\).

The solutions are \(m = 10, n = 3\); \(m = 5, n = 4\); \(m = 4, n = 5\); \(m = 3, n = 10\).

Second solution. The equation can be written as

$$
\begin{gathered}
\frac{m + n - 1}{m n} = \frac{2}{5} \\
\frac{m n - m - n + 1}{m n} = 1 - \frac{2}{5} \\
\frac{(m - 1)(n - 1)}{m n} = \frac{3}{5} \\
5(m - 1)(n - 1) = 3 m n
\end{gathered}
$$

Without loss of generality, assume \(m \geq n\). The number \(m - 1\) is relatively prime to \(m\), and the number \(n - 1\) is relatively prime to \(n\). Therefore, only two cases are possible:

(a) \(m - 1 = n\) and \(5(n - 1) = 3 m\), or (b) \(m - 1 = 3 n\) and \(5(n - 1) = m\).

In case (a), we get \(5(n - 1) = 3(n + 1)\), so \(n = 4\) and then \(m = 5\).

In case (b), we get \(5(n - 1) = 3 n + 1\), so \(n = 3\) and then \(m = 10\).

## NATIONAL MATHEMATICS COMPETITION