Problem 313

Proof by contradiction involves assuming the negation of the statement we want to prove. The direct negation of the statement "In a triangle, at least one of the angles is not greater than 60 degrees" is:

"All angles of the triangle are greater than 60 degrees."

Thus, for the proof by contradiction, we should assume that "All three angles of the triangle are greater than 60 degrees."

Starting with this assumption, we can proceed with the enhanced proof:
1. Assume, for the sake of contradiction, that all three angles in the triangle are greater than 60 degrees.
2. Let the three angles be $A$, $B$, and $C$, then according to the assumption, we have $A > 60^\circ$, $B > 60^\circ$, and $C > 60^\circ$.

3. The sum of the angles in any triangle is $180^\circ$. So, we can write an equation for the sum of angles:

$$A + B + C = 180^\circ$$

4. Since each angle is greater than 60 degrees, adding them together would give a sum greater than $180^\circ$, which can be expressed as:

$$A + B + C > 60^\circ + 60^\circ + 60^\circ$$

5. Simplifying the right side, we get:

$$A + B + C > 180^\circ$$

6. This is a contradiction because we previously stated that the sum of angles in a triangle must equal $180^\circ$. Therefore, our initial assumption must be false.

7. This leads to the conclusion that not all angles in a triangle can be greater than 60 degrees, which means that at least one angle must be $60^\circ$ or less.

8. Hence, the original proposition is proven to be true: "In a triangle, at least one of the angles is not greater than 60 degrees."

$$\boxed{\text{At least one angle in a triangle is} \leq 60^\circ}$$