by Gemini

To prove that only a triangle ($n=3$) can have all acute interior angles at a university level, we utilize the Exterior Angle Sum Theorem for convex polygons or the Interior Angle Sum Formula.

1. The Direct Proof (Using Interior Angles)

Let $P$ be a convex polygon with $n$ vertices. Let the interior angles be denoted by $\alpha_1, \alpha_2, \dots, \alpha_n$. The sum of the interior angles is given by: $$\sum_{i=1}^{n} \alpha_i = (n-2)\pi$$

If we assume the polygon is strictly acute, then every angle $\alpha_i$ must satisfy $\alpha_i \lt \frac{\pi}{2}$. Therefore, the sum of all angles must satisfy:
$$\sum_{i=1}^{n} \alpha_i \lt n\left(\frac{\pi}{2}\right)$$

For the polygon to exist, the actual sum must be less than the theoretical maximum for acute angles: $$(n-2)\pi \lt \frac{n\pi}{2}$$ Dividing both sides by $\pi$ (since $\pi > 0$): $$n - 2 \lt \frac{n}{2}$$ $$2n - 4 \lt n$$ $$n \lt 4$$

Since a polygon must have $n \ge 3$ sides, the only integer solution for $n$ is $3$.

2. The Exterior Angle Proof (Topological Perspective)

A more elegant proof involves exterior angles. For any convex polygon, the sum of the exterior angles $\theta_i$ is always $2\pi$: $$\sum_{i=1}^{n} \theta_i = 2\pi$$

An interior angle $\alpha_i$ and its corresponding exterior angle $\theta_i$ are supplementary ($\alpha_i + \theta_i = \pi$).
• If an interior angle is acute ($\alpha_i \lt \frac{\pi}{2}$), then the exterior angle must be obtuse ($\theta_i \gt \frac{\pi}{2}$).

If all $n$ interior angles are acute, then all $n$ exterior angles must be greater than $\frac{\pi}{2}$. This leads to the inequality: $$\sum_{i=1}^{n} \theta_i \gt n\left(\frac{\pi}{2}\right)$$

Substituting the known sum of exterior angles: $$2\pi > \frac{n\pi}{2}$$ $$4 > n$$

Again, we find that $n$ must be less than 4.

3. Conclusion and Geometric Interpretation

For $n=3$ (a triangle), the average interior angle is $\frac{(3-2)180^\circ}{3} = 60^\circ$. Since $60^\circ \lt 90^\circ$, acute triangles (like equilateral triangles) are possible.

For $n=4$ (a quadrilateral), the average interior angle is $\frac{(4-2)180^\circ}{4} = 90^\circ$. For all angles to be acute, their sum would be $\lt 360^\circ$, which violates the Euclidean plane geometry requirement that the sum must be exactly $360^\circ$. For any $n \ge 4$, the "angular deficit" required to keep the angles acute prevents the polygon from closing.

QED!!!