The pattern in the figure below continues inward infinitely. The base of the biggest triangle is 1. All triangles are equilateral. Find the shaded area.

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1. **Identify the area of the largest equilateral triangle:**
   The side length of the largest equilateral triangle is 1. The area \( A \) of an equilateral triangle with side length \( s \) is given by:
   \[
   A = \frac{\sqrt{3}}{4} s^2
   \]
   For \( s = 1 \):
   \[
   A = \frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}
   \]

2. **Determine the area of the first set of shaded triangles:**
   The first set of shaded triangles are three smaller equilateral triangles, each with a side length of \( \frac{1}{2} \). The area of each of these triangles is:
   \[
   A_1 = \frac{\sqrt{3}}{4} \left(\frac{1}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{1}{4} = \frac{\sqrt{3}}{16}
   \]
   Since there are three such triangles:
   \[
   \text{Total area of first set of shaded triangles} = 3 \cdot \frac{\sqrt{3}}{16} = \frac{3\sqrt{3}}{16}
   \]

3. **Identify the pattern and common ratio:**
   Each subsequent set of shaded triangles has a side length that is half of the previous set. Therefore, the area of each subsequent set of shaded triangles is:
   \[
   \text{Area of next set} = \left(\frac{1}{4}\right) \cdot \text{Area of previous set}
   \]
   This forms a geometric series with the first term \( a = \frac{3\sqrt{3}}{16} \) and common ratio \( r = \frac{1}{4} \).

4. **Sum the infinite geometric series:**
   The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
   \[
   S = \frac{a}{1 - r}
   \]
   Substituting the values:
   \[
   S = \frac{\frac{3\sqrt{3}}{16}}{1 - \frac{1}{4}} = \frac{\frac{3\sqrt{3}}{16}}{\frac{3}{4}} = \frac{\frac{3\sqrt{3}}{16}}{\frac{3}{4}} = \frac{3\sqrt{3}}{16} \cdot \frac{4}{3} = \frac{\sqrt{3}}{4}
   \]

5. **Calculate the total shaded area:**
   The total shaded area is the sum of the areas of all the shaded triangles:
   \[
   \text{Total shaded area} = \frac{\sqrt{3}}{4}
   \]

The final answer is \(\boxed{\frac{\sqrt{3}}{4}}\).