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Do you take requests? P:
I would add John von Neumann, and John Nash, even if bobbym thinks game theory is boring xD
Using my equations I obtained a minimum length for the termite, but because Bob disagrees with my result of 1.60759 for the ant I am not entirely sure of them.
I obtained a length of about 1.58451 for the termite. He walks to a point about 12.3929% along PR from P.
To sum up, these are the results I got.
Fly: √2 ~ 1.4142
Termite: ~1.58451
Ant: ~1.60759
If these results are not correct it is because I made a mistake on one of the following equations:
1. The distance from C to any point on PQ: CY = √(y^2 + 1/4) where y = YQ/PQ
2. The distance from A to any point on PR: AX = √(2x^2 + 1/4) where x = PX/PR
3. The distance from any point on PR to any point on PQ (using the Law of Cosines): XY = √(x^2 - √2*x(1-y) + (1-y)^2)
And finally,
4. The distance from C to any point on PR: CX = √(CY^2 + XY^2) where x = 1 - y
Or the long version: CX = √(√(x^2 - √2*x(1-y) + (1-y)^2) + y^2 + 1/4)
So my results are based on the minimisation of these systems of equations.
For the ant: AX + XY + CY
For the termite: AX + CX
If the equations are correct, I consider this minimisation to be the final proof.
They were √5 / 2 and √2.
1.118033988749894848204586834365638117720309179805762862135448...
and
1.414213562373095048801688724209698078569671875376948073176679...
I tried to solve for the straight line along the net while looking at the net, and came up with 1.5584 :S I need another break or a good explanation lol
What distances do you need? We already know the diagonals are √2
This was a fun problem (constructing and solving simultaneous equations for a and r)
Hello,
Hello!
The equations are also true for 2, 0, -2.
No worries
I am back to it.
Notice that bob's net is properly to scale, magnified to 4x. It makes the mathematics easy to visualise
The equation for the distance CY where Y is a point on PQ is:
CY = sqrt(y^2 + 1/4) where y = YQ/PQ
You can use the cosine rule to create a formula for any XY!
So XY = √(x^2 - √2*x(1-y) + (1-y)^2) where, again, x = PX/PR and X is any point on PR.
To prove the ant's path, once and for all, we just have to minimise AX + XY + CY. Which is this formula: √(2x^2 + 1/4) + √(x^2 - √2*x(1-y) + (1-y)^2) + √(y^2 + 1/4)
It looks like I was wrong about the ant. There is a solution to that equation which gives a marginally smaller distance than AP + PC ~ 1.6180.
I don't have the exact answers, but the distance is approximately 1.60759 if the ant walks from A to a point roughly 0.0939912 from P along PR, and then from there to a point roughly 0.128874 from P along PQ, and from there to C. That is, x is about 0.0664618 and y is about 0.871126 (where these are fractions along the lines PR from P and PQ from Q respectively).
This doesn't look to be quite the line drawn in bob's net to me (it looks like the climb up the wall is a little bit steeper), but all the equations are given to analyse it and why it is the way it is.
The termite proof will use a right triangle with small sides CY and XY and hypotenuse CX, where x = 1 - y (that equation ensures that the points referred to are in the same vertical line). I will continue this at some point, but that is simple enough now; it is not even a three-dimensional problem anymore
Just passing through,
For 5., bob has it, and the range is
Agreed with bobbym (:
Well that's not a very exciting profit! (:
24 - x = x/100
Oh, I should have paid more attention to the wording (: five MORE books, not five books. I agree with your solution, although the negative one seems counterintuitive (how is it that if they pay you $15 for every book, you could buy 5 more for $300 if they gave you $20 instead, or something? haha)
Hi;
1/4 * (-√(p^2 + 120) - p) = -5
p = 7
k = 7/4
First equation has roots of -5 and 3/2, second has root of -1/2
Every movie lover knows Casablanca!
Hello;
I only just wrote the sums and stared at them, but I think:
I agree with bobbym
-2n + 11 = -3n + 18
I will get back to this problem within another day, and play around with general rules for the distance from C to any point on PQ, C to any point on PR, and between points on PQ and PR. Notice also that I didn't yet irrefutably prove that the ant will prefer AP + PC to moving to PR and then climbing to PQ; I strongly suspect it is true though.
It looks like the termite will obey the path of bob's line, but determining that distance will take more work. But time for sleep first! (:
That's a pretty good reason (: Much worse ways to spend free time, and it is free for a reason.
But just to adjust for waking hours, it's really once every 23 minutes 38 seconds if you average 8 hours sleep. Just in case you were wondering P:
I just brushed up on these now (:
#5587
25 - 4k^2 = 0
k = +/-5/2
Gives root of +/-1
I don't know if it makes sense to solve for different combinations of solutions of k substituted in, but doing so gives further x-values of +/-1 +/- √2
bobbym, how do you have so many posts? You've averaged 40.6 posts per day for the last six and a half years as of today. One every 35 and a half minutes. For six and a half years. It's obscene. lol