88 of 100: The Second Rule of Octagon Club Is Don't Talk About Octagon Club

Let's find an upper bound by going to the extreme. Assume that $n$ internal angles are $90^\circ$ and $8-n$ angles are $360^\circ$ . So,

$n \times 90^\circ + (8-n) \times 360^\circ = (8-2) \times 180^\circ \\ \implies n \times 90^\circ + (32-4n) \times 90^\circ = 12 \times 90^\circ \\ \implies 32-3n=12 \\ \implies n=\frac{20}{3}$

What we really need is $n$ angles each less than $90^\circ$ . So, there's no way of attaining the upper bound we just found. Hence, we'll be taking the floor function of $n$ . Or in other words, we'll take the largest integer less than $n$ .

$\lfloor{n}\rfloor=\lfloor{\frac{20}{3}}\rfloor=6$

Now, it's only left to show that $\boxed{6}$ acute internal angles are indeed attainable!

We know, for a 'n' sided polygon the sum of all internal angles=2×(n–2)π/2.
So for an octagon this.
sum=2×(8–2)π/2.
=6π.
Now if, 8 angles are acute, then the sum will be<{8×π/2}=4π.
So, it can not be true.
If 7 angles are acute then the sum of those 7 angles is<{7×π/2}=3.5π.
Then, the rest angle must be greater than (6π–3.5π)=2.5π which can't exist.
If 6 angles are acute, then the sum of those 6 angles is<{6×π/2}=3π and the sum of other 2 angles>(6π--3π)=3π.
So it can exist.
Hence the answer is 6

Let the internal angles be $\alpha_i$ . Then $\sum_{i=1}^{8} 180-\alpha_i = 360$ . Hence, $\sum \alpha_i = 8\times180 - 360 = 6\times 180 = 12\times 90$ . We also know $\alpha_i < 360$ .

If $k$ angles are less than $90$ , then this must be 'compensated' by the other $8-k$ angles. Thus, we get $12\times 90 = \sum \alpha_i < k 90 + (8-k) 360 = (32 - 3k) 90$ . Hence, $12 < 32 - 3k$ , or $k < 20 / 3 < 7$ .

Maximum $k = \fbox{\$6\$}$ . Six angles just under 90, can be compensated by two angles just over 270.

For this maximum number, the acute angles can even be just over 60 (see picture): $6\times 60 + 2\times 360 = 6\times 180$ .

Total angles = 4(180) + 360 = $1080^{o}$

The figure has at least 4 acute angles, so

Sum of remaining angles > 1080 - 4(90) = $720^{o}$

If it has another acute angle, then

Sum of remains angles > 720 - 90 = $630^{o}$

And if it has another acute angle, then

Sum of remaining angles > 630 - 90 = $540^{o}$

And if it has another acute angle, then,

Sum of remaining angles > 540 - 90 = (450^{o}_

So this shows that it can not have another acute angle because the last angle,

will be greater than 360 degrees. So if it has 6 acute angles, Here is a picture what it looks like:

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