Tuesday, July 19, 2011

Fox 9

This was an old one. I think a solution was found but then lost. Yeah, it sounds stupid :)
Here, we received a claim that a solution was found again. It should not be terribly hard. Enjoy...




http://www.8foxes.com/

7 comments:

  1. Hey, thanks for reposting!

    I bet the solution is lame now compared to some of the other problems posted, but here is what I found:

    Let's call the tip of the angle O. The point where the upper segment meets the upper horizontal line is A and the one on the bottom is B.

    Also, let's introduce the parameter Alpha as the angle between the center horizontal line and [OA]. This is the parameter that varies, and there is a specific value of Alpha that gives us the minimum value of x+y.

    Now let's trace a vertical line passing by O.

    If we take the symmetrical line of [OB] with respect to the vertical line, we get a segment [OB']. We can state that OB = OB', meaning y=OB'.

    So, x+y = x+OB'.

    when Alpha changes, the angle between [OA] and [OB'] changes. When that angle is 180 degrees - A, O, B' aligned - the distance x+OB' will be minimal (Straight path distance).

    So the problem consists of finding the value of Alpha for which A, O and B' are aligned.

    Under these conditions, by studying the angles, we get that Alpha = Theta/2.

    We can also easily show that x=a/sin(Alpha) and y=b/sin(Alpha).

    We get: min(x+y)=(a+b)/sin(Theta/2)
    ---------------------------

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  2. Using cosinus theorem:

    sqrt(x^2+y^2-2xycos(theta)) >= a+b
    (x+y) >= sqrt((a+b)^2+2xy(cos(theta)+1))

    If theta goes to 180, cos(theta)=-1, x+y get the minimum value which is a+b.

    It is not the answer, but if we know while theta=180, x+y is equal to a+b, do we need write a function with parameter theta?

    ReplyDelete
  3. The solution of 'Anonimous' can't be right, because Fox 7 is a special case of this problem with a = 1, b = 2 and Theta = 90 degrees.
    The correct answer to Fox 7 is (1+2^(2/3))^(3/2).
    And that's NOT the same result as I get with his formula...
    :-(

    ReplyDelete
  4. Hey guys,

    Henkie is right. I realize now where my (not-so-smart) mistake was: when Alpha changes, O is fixed but both A and B' change, so the straight line argument does not hold since the ends of the segments are moving.

    I spent some time trying to transform the problem to a more approachable one geometrically, but so far, the best I got is prolonging the angle's legs to intercept the opposite horizonal lines. x+y will be proportional to the sum of the two full lines, which are now the diagonals of a trapeze.

    After that, all of a sudden, I decided to switch for a frontal assault by simply writing the trigonometric equality then differentiating in order to find the value of Alpha. Except that the equation is too difficult to be solved for a general Theta...

    Any ideas? Have I completely lost it?

    Thanks!

    PS. I did not quite understand what Contact said...

    ReplyDelete
  5. If the angle between x and the middle parallel line is A:
    x = a/sin(theta) and y = b/sin(theta-A)
    x+y = asin(theta)cos(A) - asin(A)cos(theta) + bsin(A)
    The minimum is a stationary point where d(x+y)/dA = 0.
    d(x+y)/dA = -asin(theta)sin(A) - acos(A)cos(theta) + bcos(A) = 0
    b - a(cos(theta)cos(A) + sin(theta)sin(A)) = 0
    cos(A-theta ) = b/a
    A = cos^-1(b/a) - theta

    x + y = asin(theta)cos(cos^-1(b/a) - theta) - acos(theta)sin(cos^-1(b/a) - theta) + bsin(cos^-1(b/a) - theta)

    ReplyDelete
  6. 3rd line is wrong.
    (6th line was also wrong by the way)

    another anon

    ReplyDelete
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