Offtopic: Math Puzzle

[Wacky]

Originally Posted by sdbardwick

hallucinations (like imagining 3 or 4 measurements)

Actually, that is the correct "off-the-cuff" first guess. Many recreational math problems have trivial answers such as the one that you imagined.

I'll give another hint. Kevin is certainly on the right track. The solution is a root of a quartic (at least similar in form to Kevin's -- I didn't bother to check his details -- The first three terms are correct.).

But you really should pay attention to Matthes' PS.

If you observe carefully, and use his suggested representation, similar triangles give a simple relationship.

You can easily reduce the quartic equation to a quadratic equation based on the roots of another quadratic equation.

[Jwb52z]

I don't remember enough math to do this anymore. LOL

[priwo]

using only a peace of paper and a pencil (and a ruler)
I get two solutions (4.85 and 1.25)

[Matthes]

And we have a winner. Nicely done andi314.

Matthes

[ewmayer]

I had my solution already typed up when I logged back in today and saw that andi314 had posted his. But damned if I'm going to waste all that work. :)

It's been too long since I fiddled with quartics, so I wound up using a numerical method to solve the quartic - kudos to andi for getting the solution(s) in closed form.

---------

We let x = length of the horizontal leg of the lower-right triangle (i.e. the distance along the x-axis between the right side of the box and the base of the ladder) and y = length of the vertical leg of the upper-right triangle (i.e. the distance along the y-axis between the top of the box and the tip of the ladder). Then, using Pythagoras, we have

(x+1)^2 + (y+1)^2 = x^2 + 2x + 1 + y^2 + 2y + 1 = 25. (1)

Let the hypotenuse of the lower-right triangle have length a. Then,

x^2 + 1 = a^2, (2)

y^2 + 1 = (5 - a)^2. (3)

Since the unknown a doesn't appear in (1) it might look like the system is underdetermined, but we have one more piece of information we can use, namely we can use that the three right triangles in our sketch are all similar to express the unknown a in terms of x and y. Using that the upper-left and lower-right triangles are similar, equality if the ration of their hypotenuses to their vertical sides gives

(5 - a)/y = a ==> 5 - a = a*y ==> 5 = a*(y+1) ==> a = 5/(y+1). (4)

Subbing this into (2) gives

(x^2 + 1)*(y+1)^2 = 25, or

(y+1)^2 = 25/(x^2 + 1). (2')

Subbing into (3) gives

y^2 + 1 = (5 - 5/(y+1))^2 = 5*(1 - 1/(y+1))^2 = 5*(y/(y+1))^2, or

(y^2 + 1)*(y+1)^2 = 25*y^2. We can manipulate this further, but it will not give us any additional information besides what we already have, which is sufficient to solve the problem.

Subbing (2') into (1) we get

(x+1)^2 + 25/(x^2 + 1) - 25 = 0.

Multiplying both sides by (x^2 + 1) we get a quartic in x:

(x^2 + 1)*(x+1)^2 - 25*(x^2 + 1) + 25 = x^4 + 2*x^3 - 23*x^2 + 2*x + 1= 0.

Not sure how to solve this exactly, but we see there is at least one (hence at least 2, since the polynomial has even degree) real solutions: set f(x) := (x^2 + 1)*(x+1)^2 - 25*(x^2 + 1) + 25, then observe that f(0) = 1, f(1) = -17, i.e. there is a zero in (0,1). Also, f(-1) = -25 so there is a second zero in (-1,0) - probably this is the symmetric counterpart of the foregoing solution, with the ladder leaning against a "wall" on the right. There is another zero in (3,4), which is the the x <==> y symmetry counterpart of the one in (0,1).

Use Newton-Raphson to find zero in (0,1):

Taylor series about x is f(x+dx) = f(x) + dx*f'(x) + ..., set f(x+dx) = 0, solve for dx = -f(x)/f'(x), set dx = x_(n+1) - x_n, get

x_(n+1) = x_n - f(x_n)/f'(x_n).

f'(x) = 4*x^3 + 6*x^2 - 46*x + 2, so

x_(n+1) = x_n - (x^4 + 2*x^3 - 23*x^2 + 2*x + 1)/(4*x^3 + 6*x^2 - 46*x + 2),

Iteration proceeds as follows, and converges quadratically on the solution in (0,1):

x_0 = 1,
x_1 = 0.319...
x_2 = 0.266...
x_3 = 0.26059...
x_4 = 0.26051836...
x_5 = 0.260518352903236...
x_6 = 0.2605183529032357542752121468785...

This corresponds to y = 5/sqrt(x^2 + 1) - 1 = 3.8385011606895490757530119621445..., which should also be a zero of f, and indeed it is. Thus the ladder touches the wall at height either = 1.2605183529032357542752121468785... or 4.8385011606895490757530119621445... .

[Axel Fox]

Here's a new math puzzle :

Take a regular clock, with an hour hand of 3 (cm or inches, doesn't really matter) and a minute hand of 4.

What is the distance between the tips of both hands at the point when the hands are moving the fastest towards each other.

I hope I translated this question correctly in English.

Axel Fox.

[eepiccolo]

Originally Posted by Axel Fox

Here's a new math puzzle :

Take a regular clock, with an hour hand of 3 (cm or inches, doesn't really matter) and a minute hand of 4.

What is the distance between the tips of both hands at the point when the hands are moving the fastest towards each other.

I hope I translated this question correctly in English.

Axel Fox.

When you refer to the time when the hands are moving fastest toward each other, do you mean when the line-of-sight velocity vector between the two points is negitive and at a maximum?

[Axel Fox]

Don't understand what you mean, but I think so, yes.

I mean it to be the point when the distance between the tips of the hands decreases the fastest.

[eepiccolo]

Yeah, that's what I mean.

I'll be back with a solution later if I get a chance to work on it.

[toferc]

sqrt(7)

[Axel Fox]

Correct, now show me how you got there and, if you have one, give us a new math puzzle.