Problem 1365

1. Understanding the Problem:
We need to prove that if one of the midlines of a triangle is longer than one of its medians, then the triangle must have an obtuse angle.

2. Definitions and Notations:
- A midline (or midsegment) of a triangle is a line segment connecting the midpoints of two sides of the triangle.
- A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
- Let $ABC$ be a triangle with $D$, $E$, and $F$ as the midpoints of sides $BC$, $CA$, and $AB$ respectively.
- Let $AD$, $BE$, and $CF$ be the medians of the triangle.

3. Key Property of Midlines:
- The midline connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. For example, $DE \parallel AB$ and $DE = \frac{1}{2} AB$.

4. Key Property of Medians:
- The length of a median can be found using Apollonius's theorem, which states that for a triangle $ABC$ with median $AD$:
$$AD^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4}$$

5. Proof by Contradiction:
- Assume the triangle $ABC$ is acute. This means all its angles are less than $90^\circ$.
- For an acute triangle, it is known that any side is less than twice the length of any median. This can be shown using the properties of medians and the triangle inequality.

6. Using the Given Condition:
- Suppose one of the midlines, say $DE$, is longer than one of the medians, say $AD$.
- Since $DE = \frac{1}{2} AB$, the condition $DE > AD$ implies:
$$\frac{1}{2} AB > AD$$
- Rearranging, we get:
$$AB > 2AD$$

7. Contradiction with Acute Triangle Property:
- For an acute triangle, we have:
$$AB < 2AD$$
- This contradicts our assumption that $AB > 2AD$.

8. Conclusion:
- Since assuming the triangle is acute leads to a contradiction, the triangle must have an obtuse angle.

$$\blacksquare$$