6.90. What is the maximum number of acute angles a convex polygon can have

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6.90. Let a convex $n$-gon have $k$ acute angles. Then the sum of its angles is less than $k \cdot 90^{\circ} + (n-k) \cdot 180^{\circ}$. On the other hand, the sum of the angles of an $n$-gon is $(n-2) \cdot 180^{\circ}$. Therefore, $(n-2) \cdot 180^{\circ} < k \cdot 90^{\circ} + (n-k) \cdot 180^{\circ}$, i.e., $k < 4$. Since $k$ is an integer, $k \leqslant 3$.

For any $n \geqslant 3$, there exists a convex $n$-gon with three acute angles (Fig. 6.17).

![](https://cdn.mathpix.com/cropped/2024_05_21_22b0ba5e8c5a1f49bd81g-167.jpg?height=369&width=398&top_left_y=371&top_left_x=538)

Fig. 6.17

![](https://cdn.mathpix.com/cropped/2024_05_21_22b0ba5e8c5a1f49bd81g-167.jpg?height=331&width=393&top_left_y=412&top_left_x=991)

Fig. 6.18