<s>[INST] Divide an equilateral triangle into at least $100$ regions? A piece of paper has the shape of an equilateral triangle. What is the minimum number of straight lines parallel to its sides that divide the triangle into at least $100$ regions?
The answer said it is $16$. But. But. Don't know how to solve it. Thank you.
 [/INST] Let us start with one line, parallel to each side of the triangle. Consider the following diagram:
$$n=3$$
By drawing 3 lines, we split it up into 7 segments.
Now what we can do is split up the inner triangle with the same logic:
$$ n = 6  $$
Note that this splits up the inner triangle into another 7 parts. In addition to this, it also adds 6 other segments into the triangle, between the first and seconds sets of lines.
This means that with $6$ lines, we have $7 + 7 + 6  = 20 $ segments
It is obvious that if we add another 3 lines, one each parallel to the sides that it will split up the inner triangle into another 7 segments.
$$n=9$$

In addition to this 7, we see again, it adds another 6 segments, next to the 6 created after the previous step. And another 6 is also added, inside 1st set of lines but outside the second set.
Therefore, here we have $20 + 7 + 6 + 6 =39$ segments.
At this point we can put together a formula for this. Let $n$ be the number of lines divided by 3 and $f(n)$ be the number of segments. (We will work in sets of 3 as it is easier. Then we have $$f(1) = 7,\quad f(2)=20\quad f(3) = 39$$
For each addition of 1 in $n$, the inner triangle is split into 7. We also see an increase in multiples of 6.
We know that when $n=2$, we have $7 + 7 + 6$ segments.
When $n=3$, we have $7+7+7+6+6+6$ segments.
We can see a pattern here.
Thus, we can make the following equation:
$$ f(n) = 7n + 6n + 6(n-1) + 6(n-2) + ... 6,\quad n>0$$
$$\implies f(n) = 7n + 6n + 6n +... + 6n -6 - 12- 18 -...-6(n-1)$$
The $-6-12...$ bit can be written as $$-1(6 + 12 + 18 + ... + n-1) = -3(1+2+3+...+(n-1) = -3\left(\frac{(n-1)((n-1)+1)}{2}\right) = \frac{-2}{3}n(n-1)$$
So, now we have $$f(n) = 7n + 6n(n-1) -\frac{2}{3}n(n-1) = 7n + 6n^2 - 6n -\frac{2}{3}n + \frac{2}{3} = 6n^2 + \frac{1}{3}n + \frac{2}{3}$$
Now we want to investigate the case with 100 segments.
Let $f(n) = 100$
Then $$ 100 = 6n^2 + \frac{1}{3}n + \frac{2}{3} \implies n \approx 4.04$$
Therefore, we need at least $n=5$ for 100 segments (as $ n$ must be a whole number). 

Therefore, we need $ 3(5) = 15$ lines for at least 100 segments in the triangle.
</s>