A three-movement piece of music takes 60 minutes to play. No single movement is longer than the combined duration of the other two movements. There is at least a 3-minute difference in length between any two movements. Within what limits can the duration of the shortest movement vary?
I. solution. Let the lengths of the 3 movements of the musical piece be $a, b, c$ such that $a < b < c$; according to $a + b + c = 60$, we have $c = 60 - a - b$. No movement is longer than the sum of the other two; in our case, this means that $60 - a - b = c \leq a + b$, i.e., $a + b \geq 30$. The differences are all at least 3, so $b \geq a + 3$ and $60 - a - b = c \geq b + 3$. Thus, the conditions of the problem can be written with the following inequalities:

$$
b \geq 30 - a \quad (1) \quad b \geq 3 + a \quad (2) \quad b \leq 28.5 - \frac{1}{2} a \quad (3)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_db6e2fefe46a426ebf0bg-1.jpg?height=458&width=529&top_left_y=454&top_left_x=771)

Let's represent these relationships in the $a ; b$ coordinate system (with $a$ on the horizontal axis). The possible pairs of $a, b$ values are the coordinates of the points falling into the intersection of the relevant regions (the marked triangle), so by calculating the coordinates of the $X, Y, Z$ points, we can obtain the limits of the shortest movement's length. It can be seen that the first coordinate of the $Z$ point provides the smallest possible value, while the first coordinate of the $Y$ point provides the largest possible value for $a$.

The coordinates of the $Y$ point are the solution to the system of equations:

$$
b = 3 + a, \quad b = 28.5 - 0.5 a
$$

which gives $Y(17 ; 20)$.

The coordinates of the $Z$ point are obtained from the solution to the system of equations:

$$
b = 30 - a \quad 2 b = 57 - a
$$

which gives $Z(3 ; 27)$.

Based on these, the length of the shortest movement can be at least 3 and at most 17 minutes, i.e., $3 \leq a \leq 17$.

II. solution. If the lengths of the three movements are $a < b < c$, and the differences between them are at least 3 for each pair, then $a + 3 \leq b \leq c - 3$ and from $a + 3 + b + c - 3 = 60$ it follows that $a + 3$ is at most 20, so $a \leq 17$.

We know that $b \leq c - 3$, i.e., $3 \leq c - b$. From the condition $c \leq a + b$ (no movement is longer than the sum of the other two), it follows that $c - b \leq a$, so $3 \leq c - b \leq a$, hence $3 \leq a$. Together with the previous result, we have $3 \leq a \leq 17$.

We still need to show that these extreme values are indeed bounds, not just limits, by finding an example for each: $a = 17, b = 20, c = 23$, and $a = 3, b = 27, c = 30$. (The examples are immediately given by the construction.)

Note. We did not use the trivial conditions $a \leq b + c$ and $b \leq c + a$ that follow from $(0 <) a < b < c$. For any $3 \leq a \leq 17$, possible values for $b$ and $c$ are: $b = \frac{57 - a}{2}, c = \frac{63 - a}{2}$.