Pangolin14

You answered your own question 

Be careful when adding times, guest!

Let 1/T be the part of the whole work that each sibling does. We also have that the total time will be 1/T, so We have that 3/T = 1/3, so T = 9 hrs.

As they only worked for 115 minutes, and they were supposed to work for 3 hrs, they only got 115/180 = 23/36 of the job done.

Two people working together will get the job done in 4.5 hrs. So, as 2 people working at the same speed worked for 45 minutes, we have that they got 45/270 = 1/6 more, or 6/36, which adds to 29/36.

Next, we have to find how many hours it takes one person to finish 7/36 of a field. The answer is 9 * 7/36, or 105 minutes, or 1 hr 45 minutes, so 3:40 + 1:45  = 5:25.

Correct me if I made any mistakes.

By the way, that is not the correct answer nor the same question.

My solution: Draw the altitude BP. Then PA = PC = 10. So, we have that 20*BP/2 = 280, so BP = 28. As we have that 10^2+28^2 = 884, the perimeter of ABC is 4√221 + 20.

$$z^3 = 8i$$

$$ \sqrt[3]{z^3} = \sqrt[3]{8i} \Rightarrow \sqrt[3]{8i} = \sqrt[6]{-64}$$

$$= -2i, i - \sqrt{3}, \sqrt{3} + i$$

$\square$

$\frac{a}{1-r} = 80$

$\frac{a}{7/8} = 80$

$a = 70$

$$12/3 = 4, 12/9 ≈ 1.$$

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?