Maximally Acute Decagon

Let the number of acute angles be $A$ . Then, the number of non-actue angles is $10 - A$ .

Consider the sum of these angles. We have:

$8 \times 180 ^ \circ = \sum \text{internal angles } < A \times 90 ^ \circ + (10-A) \times 360^ \circ$

This in turn gives us $A < 8$ . Since $A$ is an integer, thus 7 is an upper bound.

It remains to show that 7 can indeed be achieved. I got this by starting out with making the acute angles close to (but just below) 90, and making the non-actue angles close to 360.

There is quite a lot of leeway in the inequality, since $7 \times 90^\circ + 3 \times 360^\circ = 1710^\circ$ vs $8 \times 180^\circ = 1440^\circ$ . So, there should be a lot of possible (and nicer) diagrams.

In addition to Calvin's solution, Such a shape will only have a maximum of 1 line of reflective symmetry.

The sum of the angles in the decagon is 1440, and there are 10 angles. The biggest possible angle inside is nearly to 360.

$\frac{1440}{10}$ = 144

$\frac{1080}{9}$ = 120

$\frac{720}{8}$ = 90

So it is clear that there just can be maximum 7 acute angles.