Several figures can be made by attaching two equilateral triangles to the regular pentagon $ABCDE$ in two of the five positions shown. How many non-congruent figures can be constructed in this way?


To solve the problem of determining how many non-congruent figures can be made by attaching two equilateral triangles to a regular pentagon $ABCDE$, let's break down the problem step by step.

### Step 1: Understanding the problem
We are given a regular pentagon $ABCDE$ and five possible positions (one for each side of the pentagon) to attach an equilateral triangle. We need to find out how many non-congruent figures can be constructed by attaching two such equilateral triangles to any two of the five sides.

### Step 2: Counting the ways to select two sides
First, calculate the number of ways to choose two distinct sides from the five sides of the pentagon. The number of such pairs is given by the combination formula:
\[
\text{Number of pairs} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10
\]
So, there are 10 ways to select two sides to attach the equilateral triangles.

### Step 3: Identifying non-congruent figures
To determine the number of non-congruent figures, we need to consider the symmetry of the pentagon. A regular pentagon has rotational symmetry of order 5 (rotating by multiples of $72^\circ$). Two figures are congruent if one can be obtained from the other by rotation.

### Step 4: Analyzing different cases
1. **Case 1: Opposite sides (Sides separated by two other sides):**
   - There are two pairs of opposite sides in the pentagon: $ \{AB, CD\} $ and $ \{BC, DE\} $.
   - These two pairs are congruent under rotation. Therefore, they correspond to exactly 1 unique figure.
   
2. **Case 2: Adjacent sides:**
   - There are five possible pairs of adjacent sides: $ \{AB, BC\}, \{BC, CD\}, \{CD, DE\}, \{DE, EA\}, \{EA, AB\} $.
   - All of these pairs are congruent under the pentagon's rotational symmetry. Thus, they correspond to exactly 1 unique figure.
   
3. **Case 3: Sides with one side in between:**
   - There are three such pairs: $ \{AB, CD\}, \{BC, DE\}, \{CD, EA\} $.
   - Again, due to rotational symmetry, these pairs are congruent and thus correspond to exactly 1 unique figure.

### Step 5: Counting the non-congruent figures
Each of the cases above contributes 1 unique non-congruent figure. Hence, the total number of non-congruent figures is the sum of these unique cases:
\[
\text{Total non-congruent figures} = 1 + 1 + 1 = 3
\]

### Final Answer
Therefore, the total number of non-congruent figures that can be constructed by attaching two equilateral triangles to the pentagon $ABCDE$ is $ \boxed{3} $.

If you have any further questions or need additional details, feel free to ask. Here are some related problems to consider:

1. How many non-congruent figures can be formed by attaching one equilateral triangle to the pentagon?
2. What if the pentagon was regular but had different angles or side lengths?
3. How does the problem change if we are attaching squares instead of triangles?
4. How many distinct ways can you attach three triangles to the pentagon?
5. Explore a similar problem with a regular hexagon.

**Tip:** When dealing with problems involving symmetry, considering rotational and reflective symmetries can greatly simplify counting distinct cases.

Explore how to construct non-congruent figures by attaching two equilateral triangles to a regular pentagon. Understand the symmetrical properties of pentagons and the combinatorial methods used. Suitable for advanced high school students and beyond.

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