1. If we increase both the numerator and the denominator of a certain fraction by 1, we get a fraction that is $1 / 20$ greater in value. If we perform the same operation on the larger fraction, we get a fraction that is $1 / 12$ greater than the original fraction's value. Determine all three fractions.
1. Let $a / b$ be the original fraction. According to the problem, the following equations hold:

$$
\frac{a+1}{b+1}-\frac{a}{b}=\frac{1}{20} \quad \text { and } \quad \frac{a+2}{b+2}-\frac{a}{b}=\frac{1}{12} \quad(a, b \in \mathbb{N})
$$

These are equivalent to the relations

$$
20 b(a+1)-20 a(b+1)=b(b+1) \quad \text { and } \quad 12 b(a+2)-12 a(b+2)=b(b+2)
$$

which we can rewrite as $19 b-20 a=b^{2}$ and $22 b-24 a=b^{2}$. By subtracting these two equations, we find that $4 a=3 b$, which, upon substitution into the second equation, gives $22 b-18 b=b^{2}$ or $b^{2}=4 b$. Given the condition $b \neq 0$, it follows that $b=4$ and $a=3$.

The sought fractions are therefore $3 / 4, 4 / 5$, and $5 / 6$.

Alternative solution. Let $a / b$ be the original fraction. From the relations

$$
\frac{1}{20}=\frac{1}{4 \cdot 5} \quad \text { and } \quad \frac{1}{12}=\frac{1}{4 \cdot 3}=\frac{2}{4 \cdot 6}
$$

we can estimate that the solution might be $b=4$. Then

$$
\frac{4(a+1)-5 a}{4 \cdot 5}=\frac{1}{20} \quad \text { and } \quad \frac{4(a+2)-6 a}{4 \cdot 6}=\frac{1}{12}
$$

which simplifies to $a=3$. However, we must still verify that there is no other solution. The conditions of the problem lead to the relations

$$
\frac{b-a}{b(b+1)}=\frac{1}{4 \cdot 5} \quad \text { and } \quad \frac{2(b-a)}{b(b+2)}=\frac{2}{4 \cdot 6}
$$

Taking the ratio of their left and right sides, we get

$$
\frac{b+2}{b+1}=\frac{6}{5}
$$

which is satisfied only by $b=4$.

Note. In a complete solution, the possibility $b \neq 4$ must be ruled out. For example, from similar equations $1 / 20=30 / 24 \cdot 25$ and $1 / 12=52 / 24 \cdot 26$, we might guess that $b=24$, which is not a solution.

For a complete and correctly justified solution, award 6 points, of which up to 3 points for setting up and appropriately simplifying the equations (typically to two equations with two unknowns with the fractions eliminated). If the solver identifies the fraction 3/4 as a solution but does not justify why no other solution exists, award only 1 point, or 2 points if the solver attempts some algebraic justification.