Math Is Fun Forum

Hey!
I'm probably going to ignore the approximation that pi = 22/7. I've always just solved the equation as if pi was a variable and then given some significant figures. I like to be precise, and if I did it with pi = 22/7 I'd just be curious about how much it differed anyway

Oh, I know what happened with part (i). I took alpha to be the smaller root and beta to be the larger, whereas you took them to be the other way around. Since it wasn't specified, I suppose we are both right

Hello!

It will not make a difference, since the relevant date still does repeat yearly. If you wanted to work it out with the year included, you would have to know (or have some idea of) the variance in age of the 29 people, which would require estimates of the average age and the spread of ages.

I'd be more interested in including February 29th into this problem. (:

Hello

Hey!

Hi!

This one was a challenge

Hello;

I don't think there is enough information, or only enough to make guesses based on assumptions about x. But I'm looking forward to being wrong, haha

That's good! I think von Neumann is quite underappreciated.

I once played around with various lists of names for a bit of fun, in order to create an amateur scale of "revolutionary genius".
From memory, the top five names in order were:
Isaac Newton
Albert Einstein
Euclid
Aristotle
Galileo Galilei

I believe the problem is now about to be solved.
I will lay out the formulas.

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 = √(2x^2 - 2x(1-y) + (1-y)^2)

4. The distance from C to any point on PR: CX = √(CY^2 + XY^2) where x = 1 - y
CX = √(2x^2 - 2x(1-y) + y^2 + (1-y)^2 + 1/4)

For the fly: AC = √(1^2+1^2) = √2 ~ 1.414213562

For the termite: The minimum possible AX + CX where x=1-y, i.e.
Minimise: (√(8x^2 + 1) + √(8x^2 - 8x + 5)) / 2 (y=1-x)

The minimum is precisely 1/2 * √(3-√3) * (1+√3) at
x = (√3 - 1) / 4 and
y = (5 - √3) / 4

AX = 1/2 * √(3-√3)
CX = 1/2 * √(9-3√3)

In approximate terms:
The termite travels 1.538189001 to a point 18.30127019% along PR from P, and tunnels to C. He travels 0.5630162503 to get to that point (36.603% of the journey), and 0.9751727510 to complete the route (63.397%).

For the ant: The smallest possible AX + XY + CY, i.e.
Minimise: √(2x^2 + 1/4)+√(2x^2 - 2x(1-y) + (1-y)^2)+√(y^2 + 1/4) where 0<=x,y<=1

The minimum is precisely 1/2 * √(6+3√2) at
x = 3/(14√2) - 1/14 and
y = (3 - √2) / 2

AX = 1/7 * √(15 - 3/√2)
XY = 1/14 * √(102 - 69√2)
CY = √(3 - 3/√2)

In approximate terms:
The ant travels a total of 1.600206290 by walking to a point 8.009431025% along PR from P, then up to a point 20.71067812% along PQ from P, then to C. He travels 0.5126696764 to the line PQ (32.038% of the journey), 0.1501574717 up to PQ (9.3836%), and 0.9373791423 to C (58.579%)