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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    25 Apr '13 09:22
    Triangle first..

    * I have a stick one meter long, reaching from point A to D. I break it at two mutually independent spots and call those B and C, with B closer to A than to D, and then try to form the triangle with side length AB, BC, and CD. Find the likelihoods that
    * I can't form a triangle at all (e.g. side length 1 cm, 1 cm, 98 cm)
    * I get an acute angle triangle,
    * I get an obtuse angle triangle.

    * instead of a stick I use a rope, one meter long, tied as a closed loop. I mark two points along it, pull the rope taut between those points, and then try to pull the rest of the rope also taut so I'd get a triangle with a perimeter length equal to that of the rope. What are the likelihoods now for no triangle, acute angle triangle, and obtuse angle triangle?

    * I have a trapezoid ABCD with a perimeter length of one meter. One of the sides is 35 cm long and another 25 cm long. How long can the other two sides be?
  2. Standard member talzamir
    Art, not a Toil
    17 May '13 12:03 / 1 edit
    Marking the first point of breaking the stick with x, and the second with y, there are two cases, x > y and y > x. If 1 > x > y > 0, the stick lengths are y, x-y, and 1-x. "No triangle" happens if one of the sticks is longer than the other two put together, so

    1 > x > y > 0 and
    y > x-y+1-x
    x-y > 1-x+y or
    1-x > y+x-y

    or

    1 > y > x > 0 and
    x > y-x+1-y
    y-x > 1-y+x or
    1-y > x+y-x

    The easiest way I can think of to solve the likelihood is by geometric probability, by finding the matching areas in a 1x1 rectangle where point (x,y) either gives or doesn't give a triangle. It makes a rather pretty picture too.

    As for acute or not, it requires the Pythagoras, finding whether c^2 is less than or more than a^2 + b^2.

    The rope always forms a triangle, unless two points at exactly one half length of the rope are chosen. Picking the third point suitably, the triangle that is formed can always be made obtuse.

    As for the trapezoid, I think it is best tackled as three separate cases - whether the given sides are parallel and opposite, not opposite, and not parallel so the remaining two sides are the parallel ones.
  3. 02 Jun '13 15:20 / 2 edits
    Triangle:
    Pick a random point A along the stick this will be on one side or the other of the middle, and on average 0.25m from the end it is closest to.
    Label the end of the stick A is furthest from X.

    now pick point B at some random point on the stick.
    if B is on the same side of the middle of A then XA will be bigger than 0.5 and a triangle cannot be formed. The chance of this is 1/2.

    If B is on the other side of the middle from A, but BA is bigger than 0.5, again we cannot form a triangle. Since the average position of A is 0.25 from the edge, this danger zone for B is then on average from X to X+0.25, the chance of B being in it is 1/4.

    So the total chance of failing to form a triangle is 1/2+1/4 = 3/4, and the chance of being able to form some triangle is therefore 1/4.
  4. Standard member talzamir
    Art, not a Toil
    03 Jun '13 07:58 / 2 edits
    Aye. Or, as equation,

    A and B are points along the stick. there is a 50% chance for each to be on the first half of the stick, and 50% on the latter half. if both are on the former, or both on the latter.. probability of .5 x .5 = .25 each, for a total of .50 , no triangle can be formed.

    If one is on the first half and other on the latter half, the distance from the start of the stick is .25 + a for the first, where a is between -.25 and .25, and 0.75 + b for the other, where b is between -.25 and 0.25. the distance between the two points is

    .75 + b - (.25 + a) = .5 + (a - b)

    which by symmetry has equal chance of being less than .5 (triangle is possible) or more than .5 (no triangle), giving the .75 chance that a triangle is impossible to form.

    Whether the triangle is acute or not, though, takes a bit more math and is probably not very pretty to solve. A numerical solution for it would of course be easy enough to find, but exact a bit harder.