LP Geniuses

about tree fiddy

Doesn't matter, Crosby chokes in high-pressure games. The end.

What's even going on in the first one? I'm guessing the second one would be an equation. Measure volume of a sphere, minus that from the volume of a cone, and voila? I don't even know what inscibed means... I know what inscribed means, but I don't see how that would reduce any volume of the cone considering that the "sphere" would simply be written on the inside of the cone... which doesn't make much sense. So, there is only one possible answer, which you've already provided: the volume of a cone.

On June 11 2014 16:18 morph1 wrote: about tree fiddy

dem fighting words

On June 11 2014 16:46 Gnarly wrote: I don't even know what inscibed means... I know what inscribed means, .

spelling error ^^, expect many more

6

93%

im pretty sure the crosby one is 6 feet to the right since that will create an equilateral triangle and this give the highest possible angle, im just not sure how I can prove this statement mathematically

PO is the net, is it 6ft wide, he is 3 feeet outside the net but moving towards it (or away) horizontally

Is this how sports betting is done?

in the top half of your pic, i think the reason the various letters and numbers are gathered around the triangle but afraid to go inside is because there's a ninja turtle inside the triangle. hope this helps.

It would be 6ft if he was moving across the goal line. Since he's moving up towards the front of the crease, he'll have to travel a longer distance to get to the middle of the net mark.

I think the answer to the first problem: x = 3 * sqrt(3) = 5.196 ft

http://www.wolframalpha.com/input/?i=...t%28x%5E2+%2B+9%5E2%29%29%29%29+%3D+0

let SO = a, SP = b, SG = x, the shooting angle = A

a^2 = x^2 + 3^2
b^2 = x^2 + 9^2
cos A = (a^2 + b^2 - 6^2) / (2 * a * b) ----> source: http://www.mathsisfun.com/algebra/trig-solving-sss-triangles.html
cos A = (2*x^2 + 3^2 + 9^2 - 6^2) / (2 * sqrt(x^2 + 3^2) * sqrt(x^2 + 9^2))

so we can plot angle A like this:
http://www.wolframalpha.com/input/?i=...%29+*+sqrt%28x%5E2+%2B+9%5E2%29%29%29

seems like max angle is pi / 6 which is 30 degrees.

differentiate that and solve for 0, we get the distance you need to maximize angle is about 3 * sqrt(3) ft

6ft? is that like 7kilometres or?

2nd problem seems slightly easier (at the beginning) and more classic than the first problem so I am borrowing some pics off of the internet:



There is a similar triangle here, so we have a formula between r, h and R:



solve that for r^2 (we need to use the squared numbers first), we get r^2 = (h * R^2) / (h - 2 * R)
(in the wolfram alpha link, s = r^2)
http://www.wolframalpha.com/input/?i=...8s%29+%2F+sqrt%28s+%2B+h%5E2%29+for+s

substitute r^2 into our cone formula V = (1/3) * pi * r^2 * h, we get: V = pi * h^2 * R^2 / (3 * (h - 2R))
http://www.wolframalpha.com/input/?i=...+pi+*+h+*+R%5E2+%2F+%28h+-+2*R%29+*+h

from this point on we just need to substitute theta back into this equation and minimize it with some calculus..

lets go back to the similar triangle again, notice the smaller triangle with h-R and R? cos(theta) = R / (h - R). solve this equation for h we get h = R * (1 / cos(theta) + 1)
in this link sec(theta) = 1 / cos(theta)
wolfram alpha link here

substitute h back into V, we get V = (pi * R^4 * (sec(theta)+1)^2)/(3 * (R * (sec(theta)+1)-2 * R)) = (pi * (sec(theta)+1)^2)/(3 * ((sec(theta)+1)-2)) * R^3 ------> factoring out R so we can do some dirty stuff, since R is irrelevant to our calculations
wolfram alpha link for this substitution step

so if we can minimize (pi * (sec(theta)+1)^2)/(3 * ((sec(theta)+1)-2)), we can minimize V.

plotting it: http://www.wolframalpha.com/input/?i=...+%28sec%28theta%29%2B1%29-2+%29%29%29

differentiate that then solve for 0, we get some periodic answers, since this function is periodic.
http://www.wolframalpha.com/input/?i=...9+sec%5E2%28%CE%B8%29+%3D+0+for+theta

so we pick the smallest one (from the plot above, it seems like the smallest theta is around 0.5 to 1.5), so we solve that with the power of wolfram alpha:
http://www.wolframalpha.com/input/?i=...for+theta+where+0.5+%3C+theta+%3C+1.5

found that theta is about arccos(1/3) = 1.231 rad, which we can substitute cos(theta) = 1/3 (which means sec(theta) = 3) back into the V equation to solve for V, we finally get

V = (pi * R^4 * (sec(theta)+1)^2)/(3 * (R * (sec(theta)+1)-2 * R))
V = (pi * R^4 * (3+1)^2)/(3 * (R * (3+1)-2 * R)) ----> substituting sec(theta) = 3 here
V = (8/3) * pi * R^3


so at the end, when cos(theta) = 1/3 (theta = arccos(1/3)), V is smallest. V = (8/3) * pi * R^3

http://www.wolframalpha.com/input/?i=...%283+*+%28R+*+%283%2B1%29-2+*+R%29%29

i'm very confused by the wording of the first question, i take it the point S is moving to the right?

if so, the answer is 6 feet. you can prove this by mirroring the image, and making S a 0 on the X axis. as X -> 6 from the left hand side, theta approaches 60. as X -> 6 from the right hand side (starting at 12), theta approaches 60

edit - nm, looks like S is moving up the Y axis, you mentioned S was going to the right in one of your replies but the problem stated that that wasn't what it was looking for

i got 5.196 as well, slightly different method.

basically, try to maximize [ sin^-1 (9/(x^2+81)) - sin^-1 (3/(x^2+9)) ]. max angle at pi/6

https://www.dropbox.com/s/0u0j96436txi095/Screenshot%202014-06-11%2017.52.59.png

On June 11 2014 19:52 Daut wrote: i got 5.196 as well, slightly different method. basically, try to maximize [ sin^-1 (9/(x^2+81)) - sin^-1 (3/(x^2+9)) ]. max angle at pi/6 https://www.dropbox.com/s/0u0j96436txi095/Screenshot%202014-06-11%2017.52.59.png

I'm shocked you solved this

really? daut is exactly who i expected to solve this