\section*{Task 2 - 310932}

In a collection of curiosities, there was supposed to be a vessel with the following inscription:

Five Stroke Two Zero fits as a measure in me, after the first digit read "through" for the stroke! Or turn around, Zero Two Five you will find, after the first digit add a decimal point!

Indeed, \(\frac{5}{20}=0.25\).

Are there any other combinations of three digits where the rule, in the same or reversed order, of writing the division bar or the decimal point after the first digit, results in two identical numerical values?
}

Let \(a, b\) and \(c\) be the three digits. Thus, the equation \(\frac{a}{10 b+c}=c+\frac{10 b+a}{100}\) needs to be solved.

Case 1: \(b=0\).

Then the equation reduces to \(\frac{a}{c}=c+\frac{a}{100}\) or, after multiplying by \(c \neq 0\), to \(a=c^{2}+\frac{a c}{100}\). Since the left side is an integer, the right side must also be an integer, so \(a c\) must be divisible by 100. Given \(0 \leq a, c < 10\).

Then \(10 b+c \geq 10 > a\), so \(\frac{a}{10 b+c} < 1\). If \(c > 0\), then \(\frac{a}{10 b+c} < c\), which would imply \(\frac{a}{10 b+c} < c + \frac{10 b + a}{100}\). Therefore, both \(a\) and \(b\) must be positive, leading to \(a b > 0\) and \(10 a > 10 b^{2}\) or \(a > b^{2}\). This is not possible for \(b \geq 3\) because \(a \leq 9\), so \(b \in \{1,2\}\).

Case 2.1: \(b=1\). Then \(10 a = a + 10\) or \(9 a = 10\), which has no solution in integers.

Case 2.2: \(b=2\). Then \(10 a = 2 a + 40\) or \(8 a = 40\), so \(a = 5\). Indeed, the triplet \((a, b, c) = (5, 2, 0)\) mentioned in the problem statement is the only possible solution.