Let the point P lies in the interior of an equilateral triangle with side lengths 2 units, each. Find the sum of the shortest distances from P to the sides of the triangle.

Answer

To solve the problem, we need to find the sum of the shortest distances from point P to the sides of an equilateral triangle with side lengths of 2 units.

### Step-by-Step Solution:

1. **Identify the Triangle and Point P**:
- Let triangle ABC be an equilateral triangle with vertices A, B, and C, each having a side length of 2 units.
- Place point P inside triangle ABC.

2. **Draw Perpendiculars**:
- From point P, draw perpendiculars to each of the sides of the triangle. Let these perpendicular distances be denoted as P1 (from P to side AB), P2 (from P to side BC), and P3 (from P to side CA).

3. **Calculate the Area of Triangle ABC**:
- The area of an equilateral triangle can be calculated using the formula:
\[ \text{Area} = \frac{\sqrt{3}}{4} \times a^2 \]
where \( a \) is the length of a side. For our triangle, \( a = 2 \):
\[ \text{Area} = \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3} \]

4. **Express the Area in Terms of P1, P2, and P3**:
- The area of triangle ABC can also be expressed as the sum of the areas of the three smaller triangles formed by point P:
\[ \text{Area} = \text{Area of } \triangle APB + \text{Area of } \triangle BPC + \text{Area of } \triangle CPA \]
- The area of each smaller triangle can be calculated as:
\[ \text{Area of } \triangle APB = \frac{1}{2} \times AB \times P1 = \frac{1}{2} \times 2 \times P1 = P1 \]
\[ \text{Area of } \triangle BPC = \frac{1}{2} \times BC \times P2 = \frac{1}{2} \times 2 \times P2 = P2 \]
\[ \text{Area of } \triangle CPA = \frac{1}{2} \times CA \times P3 = \frac{1}{2} \times 2 \times P3 = P3 \]

5. **Combine the Areas**:
- Therefore, we can write:
\[ \text{Area of triangle ABC} = P1 + P2 + P3 \]

6. **Set Up the Equation**:
- We have established that:
\[ \sqrt{3} = P1 + P2 + P3 \]

7. **Conclusion**:
- The sum of the shortest distances from point P to the sides of the triangle is:
\[ P1 + P2 + P3 = \sqrt{3} \]

### Final Answer:
The sum of the shortest distances from point P to the sides of the triangle is \( \sqrt{3} \) units.