Home › Off Topic › Forum Games

Please keep our forum rules in mind: we aspire to provide a safe environment for our users, so will not tolerate discriminatory, hateful, inflammatory or threatening posts. Thank you. http://mcgam.es/forum-rules

Please note that registration for the forum is age-restricted.

Home › Off Topic › Forum Games

## Comments

280Pool ProA point inside an equilateral triangle has distances 3,4, and 5 from the three vertices of the triangle. What is the area of the triangle?

4,114modAh! Could have been more clear with the equation next time!

I know the answer and solution for this problem, but I am going to leave this for other users to work out.

Sets of Possible Methods:4,114modSince the problems aren't solved, I am going to remove them.I have moved this thread to

Forum Gamescategory...280Pool ProAh! Could have been more clear with the equation next time!

Here's how us guys who are better at programming than geometry solve this kind of thing.

If a side of the equilateral triangle is x, you can use Heron's formula four times so triangle (3,4,x)+(3,5,x) + (4,5,x) = (x,x,x), which gives a horrible unsolvable equation. Solve it numerically and plug into Wolfram to give the closed-form solution x=sqrt(25+12*sqrt(3)).

2,239Pool Championheron's formula is the answer to life, no questions asked

1. for math

2. most people don't know it idk why

96Pool Player9x-7i>3(3x-7u).

280Pool Pro280Pool Pro280Pool Pro280Pool Pro280Pool Pro280Pool Pro280Pool ProI double-checked the math on my solution and it looks okay, so to verify I solved it numerically for the principal branch of n=0, giving

x= 0.000619235135215395 - i * 9.64910373153867e-007

Plugging it back into the left and right sides of your equation, they are both equal.

left = pow(1621.0*x, 2016.0) = -2015.9999999954 - i * 0.412858624584096

right = pow(2016.0, left+2017.0) = -2016.00002831113 - i * 0.412858634433167

280Pool ProI double-checked the math on my solution and it looks okay, so to verify I solved it numerically for the principal branch of n=0, giving

x= 0.000619235135215395 - i * 9.64910373153867e-007

Plugging it back into the left and right sides of your equation, they are both equal.

left = pow(1621.0*x, 2016.0) = -2015.9999999954 - i * 0.412858624584096

right = pow(2016.0, left+2017.0) = -2016.00002831113 - i * 0.412858634433167

280Pool ProOh, I see. Then it is just the same thing except a=2016^2017/1621^2015 and z=1621*x^2016, so

x= [(Wn(-2016^2017/1621^2015*log(2016)))/(-1621*log(2016))]^(1/2016)

280Pool ProOh, I see. Then it is just the same thing except a=2016^2017/1621^2015 and z=1621*x^2016, so

x= [(Wn(-2016^2017/1621^2015*log(2016)))/(-1621*log(2016))]^(1/2016)

280Pool Pro280Pool Pro280Pool Pro280Pool Pro