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TalentedTom    Canada. Jun 11 2014 16:49. Posts 20070
$20 to whoever can post a solution to these two or $10 per



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Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us and as we let our own lights shine we unconsciously give other people permision to do the same 

morph1   Sierra Leone. Jun 11 2014 17:18. Posts 2352

about tree fiddy

Always Look On The Bright Side of Life 

devon06atX   Canada. Jun 11 2014 17:20. Posts 5458

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


Gnarly   United States. Jun 11 2014 17:46. Posts 1723

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.

Diversify or fossilize! 

TalentedTom    Canada. Jun 11 2014 17:48. Posts 20070


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



dem fighting words

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us and as we let our own lights shine we unconsciously give other people permision to do the same 

TalentedTom    Canada. Jun 11 2014 17:54. Posts 20070


  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

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us and as we let our own lights shine we unconsciously give other people permision to do the same 

QuirkyEric   Slovakia. Jun 11 2014 17:57. Posts 308

6

Je ti 31 let a umíš akorát klikat myší, vzpamatuj se -Daniel Havlík 

napoleono   Romania. Jun 11 2014 17:59. Posts 771

93%


TalentedTom    Canada. Jun 11 2014 18:00. Posts 20070

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

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us and as we let our own lights shine we unconsciously give other people permision to do the same 

TalentedTom    Canada. Jun 11 2014 18:02. Posts 20070

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

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us and as we let our own lights shine we unconsciously give other people permision to do the same 

NewbSaibot   United States. Jun 11 2014 18:46. Posts 4943

Is this how sports betting is done?

bye now 

bigredhoss   Cook Islands. Jun 11 2014 19:28. Posts 8646

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.

Truck-Crash Life 

devon06atX   Canada. Jun 11 2014 19:33. Posts 5458

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.


gitpush   Australia. Jun 11 2014 20:22. Posts 62

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

 Last edit: 11/06/2014 20:34

LemOn[5thF]   Czech Republic. Jun 11 2014 20:22. Posts 15163

6ft? is that like 7kilometres or?

93% Sure!  

gitpush   Australia. Jun 11 2014 20:33. Posts 62

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

 Last edit: 11/06/2014 21:45

Into Infinity   United States. Jun 11 2014 20:37. Posts 1884

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

 Last edit: 11/06/2014 20:58

Daut    United States. Jun 11 2014 20:52. Posts 8955

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

NewbSaibot: 18 TIMES THE SPEED OF LIGHT. Because FUCK YOU, DautLast edit: 11/06/2014 20:53

GoTuNk   Chile. Jun 11 2014 21:26. Posts 2860


  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


ggplz   Sweden. Jun 11 2014 23:07. Posts 16784

really? daut is exactly who i expected to solve this

if poker is dangerous to them i would rank sports betting as a Kodiak grizzly bear who smells blood after you just threw a javelin into his cub - RaiNKhAN 

 
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