In a special triangle formed by connecting points A, B, and C on a circle, each side is an integer length. The triangle has a perimeter of 25 units. If a red bead is placed at each vertex and a blue bead is placed at the midpoint of each side, find the probability that a randomly chosen set of three beads forms a triangle, assuming the triangle formed by the beads must have positive area.

We are given a special triangle formed by connecting points \( A \), \( B \), and \( C \) on a circle, where each side of the triangle has an integer length, and the perimeter of the triangle is 25 units. Additionally, a red bead is placed at each vertex, and a blue bead is placed at the midpoint of each side. We are tasked with finding the probability that a randomly chosen set of three beads forms a triangle, assuming the triangle formed by the beads must have positive area.

### Step 1: Understand the bead arrangement
There are three red beads at the vertices \( A \), \( B \), and \( C \), and three blue beads placed at the midpoints of the sides \( AB \), \( BC \), and \( CA \). In total, there are 6 beads (3 red and 3 blue).

### Step 2: Total number of ways to choose 3 beads
Since we need to choose 3 beads out of the 6, the total number of ways to select 3 beads is calculated using combinations:

\[
\text{Total number of ways} = \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]

### Step 3: Conditions for forming a triangle with positive area
To form a triangle, the three chosen points must not be collinear. The sets of three beads that are collinear will not form a valid triangle. Let's analyze the cases where three chosen beads are collinear.

#### Case 1: All three beads are collinear
- The three blue beads (at the midpoints of the sides) are collinear because they lie on the midlines of the triangle. If all three blue beads are chosen, they do not form a triangle, so this is one degenerate case.
- Any two blue beads and one red bead will also be collinear, as the blue beads lie on the same line as the red bead. For example, the blue beads on sides \( AB \) and \( BC \), along with the red bead at \( B \), are collinear. There are 3 ways to choose two blue beads and the corresponding collinear red bead, one for each vertex. So, there are 3 additional degenerate cases here.

Thus, there are \( 1 + 3 = 4 \) degenerate cases where the chosen points are collinear.

### Step 4: Number of valid triangles
The total number of ways to select 3 beads is 20, and 4 of these cases are degenerate. Therefore, the number of valid triangles is:

\[
\text{Valid triangles} = 20 - 4 = 16
\]

### Step 5: Probability
The probability that a randomly chosen set of three beads forms a triangle is the ratio of the number of valid triangles to the total number of ways to choose 3 beads:

\[
\text{Probability} = \frac{16}{20} = \frac{4}{5}
\]

Thus, the probability that a randomly chosen set of three beads forms a triangle is \( \frac{4}{5} \).

### Conclusion:
The probability that a randomly chosen set of three beads forms a triangle with positive area is \( \frac{4}{5} \).

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Do you want more details or have any questions? Here are 5 related questions to expand your understanding:

1. What are the properties of a triangle inscribed in a circle?
2. How do midpoints of sides in a triangle relate to the medians and centroid?
3. Can you prove why three collinear points cannot form a triangle with positive area?
4. How does the triangle inequality apply when the sides of a triangle are integers?
5. How can you generalize this problem to triangles with different perimeters?

**Tip**: Always verify if points are collinear before assuming they form a triangle in geometric probability problems.

Find the probability of forming a triangle by selecting three beads from six (three red at the vertices and three blue at the midpoints of the sides) on a triangle inscribed in a circle with integer side lengths and a perimeter of 25 units. This solution uses combinatorics and probability principles to arrive at the result.

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