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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    22 Feb '14 00:00
    A football field is 100 yards long, though up to 130 yards is acceptable. A goal is eight yards wide. So the angle at which goal is seen from the very center of the field is fairly small, atan(4/50), which is about 4.6 degrees.

    From what other points on the field can the goal be seen at the exact same angle of atan(4/50) ?
  2. Standard member talzamir
    Art, not a Toil
    23 Feb '14 10:37
    Making that a bit more challenging.. find all those points geometrically. That is.. using a compass and a straight line only.
  3. Subscriber joe shmo On Vacation
    Strange Egg
    25 Feb '14 01:01
    Wouldn't the angle from the center of the field be 2*atan(4/50) = 9.2 deg.?
  4. Standard member talzamir
    Art, not a Toil
    01 Mar '14 09:18
    Oops. Yes, it would.
  5. 01 Mar '14 15:37
    We want angles with base K, where the opposite angle is A

    Since all internal angles of a triangle add up to 180 degrees, the construction method is easy.

    Pick any angle B between 0 and 180
    Set angle C to be 180-B-A

    draw lines with internal angles B and C from each side of the goal, and they will meet at the required angle A
  6. Standard member talzamir
    Art, not a Toil
    01 Mar '14 20:52
    If I understand you correctly, that method gives valid points one at a time, or, two at a time, as the mirror image on the other side of the center line of the field would of course also be valid?
  7. 01 Mar '14 22:28 / 3 edits
    Yes, it is possible to give an equation for the x/y coordinate of each point using a single angle b:
    Imagine that the goal is at the top of the paper, and the "aperture angle" we want is a, the goal width is A

    Pick a point P, where we want to have the aperture angle A
    Draw a line C from the left hand side of the goal to P, this line meets the goal line at angle b
    So, the line from the other side of the goal to P meets the goal at (pi-b-a)

    The length of C, meets the equation, from the sine rule:

    C/sin(pi-b-a) = A/sin(a)

    C = A.sin(pi-b-a)/sin(a)

    So we have chosen the angle b, and we have the line length.

    If we want the answer in cartesian we can define:

    x (horizontal distance from left edge of the goal to P)
    y (vertical distance from goal to P)

    sin(b) = y/(A.sin(pi-b-a)/sin(a))
    cos(b) = x/(A.sin(pi-b-a)/sin(a))

    so
    y = sin(b)(A.sin(pi-b-a)/sin(a))
    x = cos(b)(A.sin(pi-b-a)/sin(a))

    where A = 8
    a = 2*atan(4/50)
    and b is any chosen angle between 0 and pi

    Hmm, something like that technique anyway, but the graph for that looks a bit weird, I'll have to check my maths carefully
  8. Standard member talzamir
    Art, not a Toil
    04 Mar '14 14:23
    Certainly a very different curve from the one I had in mind, which is
    a circle that passes through the goal posts and the center point of the field
    ?
  9. Subscriber Ponderable
    chemist
    04 Mar '14 16:56 / 1 edit
    Originally posted by joe shmo
    Wouldn't the angle from the center of the field be 2*atan(4/50) = 9.2 deg.?
    wouldn't it be rather atan (8/50) =9.09° ?

    In the sine correlation a/sin(alpha)=b/sin(beta) is then for beta= 9.09° and b= 8 b/sin(beta) a constant (50.637). We now need for any alpha an A and can calculate how the line develops a/50.637=sin(alpha). Alpha starts at 85.455° and an a of 50.478 (for the record: gamma and c have here the same starting valuesas alpaha and a since we look at an isoscle triangle with heigth 50 and base line 8).
  10. Subscriber Ponderable
    chemist
    04 Mar '14 18:21
    Aftert some though the 2*atan(4/50) is correct. so the angle is 9.2°. That makes alpha =45.4° and a still 50.478. b/sin(beta) is 50.039

    Now we get for every angle of alpha a value for a as alpha approaches 0 the length of a also approaches zero 0 (which is unattainable in a triangle) witha slight curvature, but certainly not a circle.
  11. Standard member talzamir
    Art, not a Toil
    06 Mar '14 10:01
    I wonder if we see the situation differently. I was thinking of this.. choose a circle with center point O and two points on the circumference, points A and B. Those are the goal posts. If you choose any point C on the circumference of the circle, except one that is in the field of the angle AOB, and then form the angle, ACB, the angle is the same regardless of where on the circle point C is, and angle ACB = angle AOB / 2. So if C is the middle point of the field, the challenging is simply to find point O, which is simple enough with a compass and a straight line, and then draw the circle with center point O and radius OC. It doesn't fall completely on the field so the 9.4 degree angles would be near the corners to either side of the goal, or then near the center line of the field.

    Is this reasoning flawed? Awesome. =) Tell me where I went wrong?
  12. Standard member smw6869
    Granny
    19 Mar '14 17:48
    Originally posted by Ponderable
    wouldn't it be rather atan (8/50) =9.09° ?

    In the sine correlation a/sin(alpha)=b/sin(beta) is then for beta= 9.09° and b= 8 b/sin(beta) a constant (50.637). We now need for any alpha an A and can calculate how the line develops a/50.637=sin(alpha). Alpha starts at 85.455° and an a of 50.478 (for the record: gamma and c have here the same starting valuesas alpaha and a since we look at an isoscle triangle with heigth 50 and base line 8).
    Please Stop it! And who the hell is ATAN ?


    GRANNY.