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

why does daut have a purple pigeon by his name anyway

I believe they call it "origami"

i just googled "origami" seems pretty gay i guess that's why it's purple

(no offense daut!)

origami (because he's so good at folding)

thanks a lot everyone, managed to solve the cone one eventually but the Crosby one i could not (had a doh moment when I read the beginning of the solution)
gitpush your answers were very insightful thanks a lot, shoot me a PM ill send you the prize


thanks everyone who participated, including bigbob

On June 11 2014 23:32 TalentedTom wrote: thanks a lot everyone, managed to solve the cone one eventually but the Crosby one i could not (had a doh moment when I read the beginning of the solution) gitpush your answers were very insightful thanks a lot, shoot me a PM ill send you the prize thanks everyone who participated, including bigbob

just PM'ed

not too sure how everyone got the first question but i got the same answer as everyone else

theta = atan(9/x) - atan(3/x)
dtheta/dx = 3/(x^2+9) - 9/(x^2+81)
find maxima by setting dtheta/dx to 0
1/(x^2+9) = 3/(x^2+81)
x^2+81 = 3x^2+27
2x^2 = 54
x^2 = 27
x = 5.196

That blows my mind. I mean, the best angle to get a shot from is at the middle of the net, right? Net is 6ft wide, so 72 inches. So the ideal angle regarding the net is 36 inches.

He's 36 inches *away* from the net, so he has to travel that just to get to the net.

36+36 = 72 inches (6 feet)

So, travelling along the goalline, he would have to travel 6 feet to reach the best angle to shoot from (the middle). But* he's going at an angle from the goal line... therefore he would have to travel further, no?

Does this reasoning make sense to anyone?

I mean, the optimal point to shoot from is the very middle of the net. Even if you travelled along the goalline, you would need to travel exactly 6ft to get to the optimal spot to shoot.

However, since you're travelling up and over, the distance would increase significantly.

That's why I say 5.196 feet has got to be wrong. By quite a bit.

Someone please explain to me why I'm wrong. Cuz I just don't see it.

You can't ever get to the middle of the net given the rules you are just sliding north or south in route always perpendicular to the goalposts.

To get to your desired position would require parallel or diagonal movement. I mean if you could go anywhere the optimal point would be between the posts as close to the goal line as possible making the shooting angle approach 180 degrees. That is against the directions tho

Did you read my posts? I don't get what you're trying to accomplish by what you wrote. Don't mean to sound like a dick, but I read your post and was basically *facepalm. That's why I threw in the 'Even' part in the "Even if you travelled along the goalline".

Yes, I'm aware that moving a puck through a metal post isn't possible. I'm also aware that it's against the rules.

All I stated was the shortest possible distance to get to the desired angle (middle of the net) would be 6 feet. I understand that isn't the answer, as per the fact I can't move through a post, and that it also breaks the rules.

What I'm saying is, if the literal shortest distance to the required angle is 6 feet, how is this 'correct' answer, where you have to travel significantly further, a shorter distance than the literal shortest possible distance?

Oh, I think I understand what you're trying to say. And honestly man? You realize you can be shooting from the middle of the net... 17000 km away, right? wtf trax...

When someone shoots from the centre ice dot, that's the middle of the net.

I would like someone to answer my question though, because as is, I'm pretty damn positive that answer is wrong.

On June 12 2014 12:56 devon06atX wrote: I mean, the optimal point to shoot from is the very middle of the net. Even if you travelled along the goalline, you would need to travel exactly 6ft to get to the optimal spot to shoot. However, since you're travelling up and over, the distance would increase significantly. That's why I say 5.196 feet has got to be wrong. By quite a bit. Someone please explain to me why I'm wrong. Cuz I just don't see it.

this was almost exactly my original thought process, it makes a lot of logical sence to shoot from exactly the middle, but apparently math says otherwise, kinda akward
optimized cosine law dont lie,

On June 12 2014 10:18 Rhaegar99 wrote: not too sure how everyone got the first question but i got the same answer as everyone else theta = atan(9/x) - atan(3/x) dtheta/dx = 3/(x^2+9) - 9/(x^2+81) find maxima by setting dtheta/dx to 0 1/(x^2+9) = 3/(x^2+81) x^2+81 = 3x^2+27 2x^2 = 54 x^2 = 27 x = 5.196

i am looking for a more simplified compressed solution, something that would not require technological (other then a calculator) intervention, will check this out later when I get home.
Gonna go vote on provincial ellections

ITT people try to solve a cut and dried math problem with no math and wonder why they are wrong

On June 12 2014 12:56 devon06atX wrote: I mean, the optimal point to shoot from is the very middle of the net. Even if you travelled along the goalline, you would need to travel exactly 6ft to get to the optimal spot to shoot. However, since you're travelling up and over, the distance would increase significantly. That's why I say 5.196 feet has got to be wrong. By quite a bit. Someone please explain to me why I'm wrong. Cuz I just don't see it.

as the height increases, the angles all change. not going to draw a picture, but basically the bottom left angle and the top angles decrease and the angle on the right side increases as the height increases. so moving further away just makes the angle smaller.

essentially this is a quadratic, something is increasing, other things are decreasing, you have to find the maximum point along which 2 conflicting things are changing

i didnt read it as a math problem, the numbers and stuff are just added details to enrich the story the picture is telling. the real problem lies in the ninja turtle that's inside of the triangle, and his psychological intimidation of the outside letters and numbers.

On June 12 2014 15:49 devon06atX wrote: Oh, I think I understand what you're trying to say. And honestly man? You realize you can be shooting from the middle of the net... 17000 km away, right? wtf trax...

Right. But it's obviously not the same angle at any distance from the goal. The further away from the goal you go the more acute the angle becomes.

You are WTFing me but it seems as though you don't understand the correct answer people arrived at or what I am saying.

They are talking about the angle created from the shooting point to each goal post. If you are right in front of the net that is a very wide angle; nearly 180 degrees. If you are 1700km from the net that angle will be less than 1 degree given the posts are only 6 feet wide.

Looking at the original picture all the question asks is what optimal distance S-G (perpendicular to the goaline meaning the line segment created between the points S & G will always form a right angle with the goaline) will create the largest angle theta (no symbol for that in iPhone but talented tom draws the symbol on the angle in question in the problem.

Don't know how to state it any more clearly than that. The fact that the truly optimal place to shoot is the center of the goalposts has nothing to do with the problem presented.

On June 12 2014 17:15 TalentedTom wrote: Show nested quote + On June 12 2014 12:56 devon06atX wrote: I mean, the optimal point to shoot from is the very middle of the net. Even if you travelled along the goalline, you would need to travel exactly 6ft to get to the optimal spot to shoot. However, since you're travelling up and over, the distance would increase significantly. That's why I say 5.196 feet has got to be wrong. By quite a bit. Someone please explain to me why I'm wrong. Cuz I just don't see it. this was almost exactly my original thought process, it makes a lot of logical sence to shoot from exactly the middle, but apparently math says otherwise, kinda akward optimized cosine law dont lie,

This is incorrect. If you were exactly in the middle of the two posts, the optimal distance will be when you and the two posts form an equilateral triangle, not when your 6ft away. Surprisingly, the height of a 6ft equilateral triangle is also 5.196ft.

On June 12 2014 17:19 TalentedTom wrote: Show nested quote + On June 12 2014 10:18 Rhaegar99 wrote: not too sure how everyone got the first question but i got the same answer as everyone else theta = atan(9/x) - atan(3/x) dtheta/dx = 3/(x^2+9) - 9/(x^2+81) find maxima by setting dtheta/dx to 0 1/(x^2+9) = 3/(x^2+81) x^2+81 = 3x^2+27 2x^2 = 54 x^2 = 27 x = 5.196 i am looking for a more simplified compressed solution, something that would not require technological (other then a calculator) intervention, will check this out later when I get home. Gonna go vote on provincial ellections

I believe this is as simple of a solution that you will get. This is a basic optimization question which you learn in early calculus.

Firstly you want to look for the function of theta (let us use t) with respect to x (our distance); that is, the angle S of triangle SGP minus the angle S of triangle SGO.
Both of these are right angle triangles so we can use the tan trigonometry function tan(t) = opposite/adjacent for each triangle
To get t by itself we inverse tan (aka atan), the equation to get t = atan(opposite/adjacent)
Therefore our t function is: t = atan(9/x) - atan(3/x)

On our next step, we need to derive the function.
The derivative of our function is: dt/dx = 3/(x^2+9) - 9/(x^2+81)

We calculate the derivative of the function (dt/dx) so that we get the rate at which t changes at any point with respect to x.
A positive dt/dx indicates that t is increasing when x increases, and a negative dt/dx indicates that t is decreasing when x increases.
Therefore to get the the maximum value of t, it is at the point where dt/dx is neither increase nor decreasing, which is when dt/dx = 0
This is also called the maxima of the function.
Therefore to get the maximum distance, we solve for d when dt/dx = 0
The rest is just basic algebra

0 = 3/(x^2+9) - 9/(x^2+81)
1/(x^2+9) = 3/(x^2+81)
x^2+81 = 3x^2+27
2x^2 = 54
x^2 = 27
x = 5.196

nah, trax, you're right. I (stupidly) thought crosby was skating at an angle towards the net. aka - the right angle would be at the middle of the net, not at the goal-line 3 ft outside the post.

Even thought it clearly states in the question that he's skating perpendicular, I stupidly thought he was skating at 45 degrees... and the diagram just fucked me all up haha.

My idiotic mistake.

I hope that clears up why I had an impossible time accepting the correct answer. I was sorta embarassed once I figured out my own error heh.



edit - sorry for being '*facepalmy' yesterday heh

man everyone got super hung up on these two classic calculus problems :O

On June 12 2014 17:19 TalentedTom wrote: Show nested quote + On June 12 2014 10:18 Rhaegar99 wrote: not too sure how everyone got the first question but i got the same answer as everyone else theta = atan(9/x) - atan(3/x) dtheta/dx = 3/(x^2+9) - 9/(x^2+81) find maxima by setting dtheta/dx to 0 1/(x^2+9) = 3/(x^2+81) x^2+81 = 3x^2+27 2x^2 = 54 x^2 = 27 x = 5.196 i am looking for a more simplified compressed solution, something that would not require technological (other then a calculator) intervention, will check this out later when I get home. Gonna go vote on provincial ellections

I think this solution is slightly simpler than the one i had originally (i am gitpush lol), since if you remember what the derivative of arctan(x) is you can do this by hand. (my solution based on the cosine law requires computer to solve the derivative and optimizations)

i remember i had to memorize derivative of arctan(x) when i learned calculus.. but most of the time we use Maple / Mathematica (Wolfram Alpha) for most of our school work.

On June 11 2014 22:07 ggplz wrote: really? daut is exactly who i expected to solve this

Np all good