Pangolin14

$$12/3 = 4, 12/9 ≈ 1.$$

There are 5 powers of 3 in $12!$

$$15/3=5, 15/9 ≈1.$$

There are 6 powers of 3 in $15!$

$$18/3 = 6, 18/9 =2.$$

There are 8 powers of 3 in $18!$

You can likewise figure out that there are 9 in $21!$ and 10 in $24!.$ Thus, as there are 5 numbers, there will be no multiplying of 3 within non-3-divisible numbers, so the answer will relate to the sum $3^5 + 3^6 + 3^8 + 3^9 + 3^{10}.$ This is just $3^5(1 + 3 + 27 + 81 + 243,)$ or $3^5(355),$ so the answer is $\boxed{5}.$

We might also take the discriminant, b^2 - 4ac.

We have that (-a)^2 - 4 * 2016 * 1 > 1, so a^2 - 8064 > 0. Thus, a^2 > 8064, and a > ±√8064, so a> about 89.8 and a< about -89.8. However, we would like to find the integer solutions, which means a must be an integer, which means that a ≤ -90.

(By the way CPhill, you have a typo on the first line)

This problem is unanswerable, because you have not included a diagram.

That's an elegant solution, Melody. Definitely better than bashing with incircle coordinates 

Or, would it just be 24 * 25/300?

I didn't use trig- I'll explain.

The incenter coordinates are $\left(\frac{1}{2} (-a + b + c), \frac{\sqrt{-(a - b - c) (a + b - c) (a - b + c) (a + b + c)}}{{2 (a + b + c)}} \right)$ where a,b,and c are the side lengths of a triangle. Then we have that the incenter's coordinates, if we let a=8 and b = c, are $\left(\frac{1}{2} (-8 + b + b), \frac{\sqrt{-(8 - b - b) (8 + b - b) (8 - b + b) (8 + b + b)}}{2 (8 + b + b)}\right)$ which simplifies to $\left(\frac{1}{2} (2 b - 8), 2 \frac{\sqrt{(2 b - 8) (2 b + 8)}}{ b + 4}\right).$ If we set the right side of the equation equal to 3, we can easily cross multiply and get $b = \frac{100}{7}.$ Thus, AB = $\frac{100}{7},$ and we may want to use heron's formula, but for a more convenient solution we can say that A = (0,a), and using the distance formula with B = (-4,0), we have $\sqrt{a^2-16} = 100/7,$ so $a = \frac{96}{7},$ and the height is $\frac{96}{7}$

I meant 2√7

sqrt(8^2 - 6^2) = 4 √ 3.

We might also say that AB = 8 sqrt (13), and as OD=8, BD = sqrt(OB^2 - OD^2) = sqrt(208 - 64) = 12.

Extend BD to meet the circle at C. Then, BC = 2 BD = 24. Then, 832 -576 = 256, so AC = 16. Thus, AD = sqrt(AC^2 + DC^2) = sqrt(256+144) = sqrt(400) = 20.

I can tell you that $cos(x) + sin(x) = \frac{5}{\sqrt{13}},$ so $2 \cdot cos(x) + 2\cdot sin(x) = \frac{10}{\sqrt{13}}.$ Subtracting, $sin(x) = \frac{3\sqrt{13}}{13}.$ Can you find $cos(x)$ now?