3. Miguel is making an obstacle course for field day. At the end of every sixth of the course, there is a tire. At the end of every third of the course, there is a cone. At the end of every half of the course, there is a hurdle. At which locations of the course will people need to go through more than one obstacle?

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To solve this problem, we need to determine the locations on the obstacle course where multiple obstacles coincide. The obstacles are placed at the following intervals:

1. Tires are placed at every sixth of the course: $$ \frac{1}{6} $$
2. Cones are placed at every third of the course: $$ \frac{1}{3} $$
3. Hurdles are placed at every half of the course: $$ \frac{1}{2} $$

### Step 1: Find the Least Common Multiple (LCM)

To find the locations where more than one obstacle is present, we need to find the least common multiple (LCM) of the denominators of the fractions representing the obstacles.

The denominators are:
- 6 (for tires)
- 3 (for cones)
- 2 (for hurdles)

### Step 2: Calculate the LCM

The LCM of 6, 3, and 2 can be calculated as follows:

- The prime factorization of each number:
  - $$ 6 = 2 \times 3 $$
  - $$ 3 = 3 $$
  - $$ 2 = 2 $$

The LCM is found by taking the highest power of each prime factor:
- The highest power of 2 is $$ 2^1 $$
- The highest power of 3 is $$ 3^1 $$

Thus, the LCM is:
$$
\text{LCM} = 2^1 \times 3^1 = 6
$$

### Step 3: Identify the Locations

Now that we have the LCM, we can find the locations where the obstacles coincide. The obstacles will be present at the following fractions of the course:

- Tires: $$ \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \frac{6}{6} $$
- Cones: $$ \frac{1}{3}, \frac{2}{3}, \frac{3}{3} $$ (which is the same as $$ \frac{6}{6} $$)
- Hurdles: $$ \frac{1}{2}, \frac{2}{2} $$ (which is the same as $$ \frac{6}{6} $$)

### Step 4: List the Coinciding Locations

Now we can list the locations where more than one obstacle is present:

1. At $$ \frac{1}{6} $$: Tire only
2. At $$ \frac{2}{6} $$ (or $$ \frac{1}{3} $$): Tire and Cone
3. At $$ \frac{3}{6} $$ (or $$ \frac{1}{2} $$): Tire and Hurdle
4. At $$ \frac{4}{6} $$ (or $$ \frac{2}{3} $$): Cone only
5. At $$ \frac{5}{6} $$: Tire only
6. At $$ \frac{6}{6} $$ (or the end of the course): Tire, Cone, and Hurdle

### Conclusion

The locations where people will need to go through more than one obstacle are:
- $$ \frac{1}{3} $$ (Tire and Cone)
- $$ \frac{1}{2} $$ (Tire and Hurdle)
- $$ \frac{6}{6} $$ (Tire, Cone, and Hurdle)

Thus, the answer is that participants will encounter multiple obstacles at $$ \frac{1}{3} $$, $$ \frac{1}{2} $$, and $$ 1 $$ (the end of the course).
People will need to go through more than one obstacle at $$ \frac{1}{3} $$, $$ \frac{1}{2} $$, and the end of the course.

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