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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    24 Dec '07 03:58 / 1 edit
    I have been told by several people that the following problem "NEEDS" to be solved using Trigonometry. I feel I am close to a solution in some other way, well here it is....

    Slap a circle on a graph with a radius (A) centered on the origin

    eqtn: x^2 + y^2 = A^2

    Given that there is a specified number of degrees in a section of the circle, I want to find the length of the line underneath that ark.....

    I start by passing a line with an arbitrary slope through the origin

    then i pass a line perpendicular to the previous line

    this gives a 90 degree angle

    right there I could solve the system and find two intersections with the circle and they would be 90 degrees apart.

    then use the distance formula to determine the length of the line under that particular ark...

    My question: How could i find a solution given a number of degrees that isnt so easily determined

    is there a way to find, using the slopes of the lines the number of degrees in that angle.

    sorry for any confusion. I know it must be tough to decipher what im attempting to say since i don't know any terminology to simplify this problem
  2. Subscriber joe shmo On Vacation
    Strange Egg
    24 Dec '07 05:04
    Originally posted by joe shmo
    I have been told by several people that the following problem "NEEDS" to be solved using Trigonometry. I feel I am close to a solution in some other way, well here it is....

    Slap a circle on a graph with a radius (A) centered on the origin

    eqtn: x^2 + y^2 = A^2

    Given that there is a specified number of degrees in a section of the circle, I want t ...[text shortened]... ecipher what im attempting to say since i don't know any terminology to simplify this problem
    i'll give some specific numbers to work with for determining the angle

    start with the line y= 4x it contains the point (1, 4)

    perpendicular to y=4x is y = -1/4x it contains the point (4, -1)

    I took those two points and found the mid-point between them.

    I can't prove this next assumption, but graphically it looks as if the mid-point line creates two 45 degree angles, and it seems to make sence because 90/2 is 45 and the formula for finding the mid-point uses division by 2 as well.

    using this logic I'll attempt to find a 40 degree angle

    this time I used this to find my desired point.

    (x1 + x2)/ 9/4 , (y1 + y2)/ 9/4 since 90 divided by 9/4 is 40

    I ended up with the point (20/9, 4/3)

    and graphically it seems to be acurrate, which leads me to believe my assumptions are accurrate......
  3. Subscriber joe shmo On Vacation
    Strange Egg
    24 Dec '07 07:00 / 2 edits
    Originally posted by joe shmo
    i'll give some specific numbers to work with for determining the angle

    start with the line y= 4x it contains the point (1, 4)

    perpendicular to y=4x is y = -1/4x it contains the point (4, -1)

    I took those two points and found the mid-point between them.

    I can't prove this next assumption, but graphically it looks as if the mid-point line hically it seems to be acurrate, which leads me to believe my assumptions are accurrate......
    hmmm.....something doesn't quite make sence....... the equation of the line that i thought was my 45 degree angle appears to have the same slope as the equation of the line that i believed to have a 40 degree angle.......what is going on?.....

    This is telling me that they are in fact the same line... lol

    it seems i havent changed degrees at all, rather i just found a point on that line a bit closer to the origin
  4. Subscriber sonhouse
    Fast and Curious
    24 Dec '07 14:58
    Originally posted by joe shmo
    I have been told by several people that the following problem "NEEDS" to be solved using Trigonometry. I feel I am close to a solution in some other way, well here it is....

    Slap a circle on a graph with a radius (A) centered on the origin

    eqtn: x^2 + y^2 = A^2

    Given that there is a specified number of degrees in a section of the circle, I want t ...[text shortened]... ecipher what im attempting to say since i don't know any terminology to simplify this problem
    Where does the A squared come in? In a circle around an origin, the formula is just X^2+Y^2=1 or R. It doesn't have to be assigned to a power term. R=radius of the circle which we can assign as 1. So in your case, Y^2 would be = to the radius which we can call 1 if we want. I guess we can call it peanut butter if we want but it is still just a unit of 1.
  5. Subscriber joe shmo On Vacation
    Strange Egg
    24 Dec '07 15:26
    Originally posted by sonhouse
    Where does the A squared come in? In a circle around an origin, the formula is just X^2+Y^2=1 or R. It doesn't have to be assigned to a power term. R=radius of the circle which we can assign as 1. So in your case, Y^2 would be = to the radius which we can call 1 if we want. I guess we can call it peanut butter if we want but it is still just a unit of 1.
    sorry, then this is the equation:

    (x+0)^2 + ( y+0 )^2 = A^2 or x^2 + y^2 = A
  6. Standard member PBE6
    Bananarama
    24 Dec '07 15:51 / 1 edit
    Originally posted by joe shmo
    I have been told by several people that the following problem "NEEDS" to be solved using Trigonometry. I feel I am close to a solution in some other way, well here it is....

    Slap a circle on a graph with a radius (A) centered on the origin

    eqtn: x^2 + y^2 = A^2

    Given that there is a specified number of degrees in a section of the circle, I want t ...[text shortened]... ecipher what im attempting to say since i don't know any terminology to simplify this problem
    It sounds like you're trying to find the straight-line distance between Pac-Man's lips.

    The trig solution is straightforward. If the angle of the sector is T, then the straight-line distance is 2*A*sin(T/2). Contrast this with the arc length of T*A.

    The method you're using can be simplified by fixing one point in a useful position and then sweeping out the sector from there. My suggestion is keeping one point at (A,0), so the distance formula becomes:

    D = SQRT((A*cos(T) - A)^2 + (A*sin(T) - 0)^2)

    This simplifies to:

    D = A*SQRT(2-2*cos(T))
  7. Subscriber joe shmo On Vacation
    Strange Egg
    24 Dec '07 16:07 / 1 edit
    Originally posted by PBE6
    It sounds like you're trying to find the straight-line distance between Pac-Man's lips.

    The trig solution is straightforward. If the angle of the sector is T, then the straight-line distance is 2*A*sin(T/2). Contrast this with the arc length of T*A.

    The method you're using can be simplified by fixing one point in a useful position and then sweeping o ...[text shortened]... = SQRT((A*cos(T) - A)^2 + (A*sin(T) - 0)^2)

    This simplifies to:

    D = A*SQRT(2-2*cos(T))
    im actually trying not to use Trig.......because I don't know how.

    i posted a few times with the various steps I ve taken IN AN ATTEMPT to arrive at a solution without using sin or cos...........

    But i've hit a brick wall....
  8. Standard member PBE6
    Bananarama
    24 Dec '07 16:18
    Originally posted by joe shmo
    im actually trying not to use Trig.......because I don't know how.

    i posted a few times with the various steps I ve taken to arrive at a solution without using sin or cos...........

    But i've hit a brick wall....
    OK, lemme try something else.

    You can define a sector as lying between the two lines y = 0 and y = mx, intersecting with the circle x^2 + y^2 = A^2. The points of intersection are then:

    P1 - (A,0)

    P2 - (A/SQRT(m^2+1), A*m/SQRT(m^2+1))

    You can plug these values into the distance formula to get a rather long but straightforward expression for the distance in terms of "m", with nary a trig function to be seen.
  9. Standard member PBE6
    Bananarama
    24 Dec '07 16:19
    Originally posted by joe shmo
    im actually trying not to use Trig.......because I don't know how.

    i posted a few times with the various steps I ve taken to arrive at a solution without using sin or cos...........

    But i've hit a brick wall....
    Wait a second...you don't know how to use trig? It's easy and very useful. If you're wondering about it, just ask.
  10. Subscriber joe shmo On Vacation
    Strange Egg
    24 Dec '07 16:30
    Originally posted by PBE6
    Wait a second...you don't know how to use trig? It's easy and very useful. If you're wondering about it, just ask.
    Im taking Trig this upcomming semester..............I really just wanted to see if i could work something out from scratch...

    I was wondering if you could take a look at my second post and tell me why, what I did, does not put me on the right track to a solution..?
  11. 25 Dec '07 06:58 / 1 edit
    Originally posted by joe shmo
    hmmm.....something doesn't quite make sence....... the equation of the line that i thought was my 45 degree angle appears to have the same slope as the equation of the line that i believed to have a 40 degree angle.......what is going on?.....

    This is telling me that they are in fact the same line... lol

    it seems i havent changed degrees at all, rather i just found a point on that line a bit closer to the origin
    i think what you meant to do, rather than dividing the x and y coordinates of the midpoint by the same number (which would indeed give you a different point on the same line as the one that cuts the angle in half).. you wish to cut the 90 degree angle into a 40 and a 50. Then, you're hoping to use algebra to determine what point would lie along that 40 degree line, and use its y/x value to determine a slope.. and eventually use that line however you want (not sure how you mean to use it in terms of determining chord lengths)

    sadly, because of the way trig functions grow between 0 and 90 degrees, this "cutting" of the angle and its corresponding cutting of the chord do not act in a linear way. EXCEPT at 45 degrees - cutting the angle in half indeed will cut the line in half as well

    for example: if you draw a chord between (0,1) and (1,0) on the unit circle, the midpoint (1/2, 1/2) lies on the 45 degree angle, which cuts the initial 90 degree angle in half.

    however, starting with these same points and looking at cutting the angle into a 30 degree and a 60 degree, it does not cut the line algebraically: (i.e. assuming it cuts it into a 1/3 piece and a 2/3 piece is incorrect).

    now, interestingly you can find a good algebraic APPROXIMATION and get values that are very very close to correct (even closer than a protractor could measure) by using the first few terms of the MacLaurin series for the trig functions sin and cosine.

    this may be a little overboard considering you haven't learned even how to use trig yet, but it could prove useful-ish, if you really want to arrive at working values algebraically. if you are interested in an outline for this method, let me know... it would be very messy in terms of calculation, but not theoretically too hard i don't think (unless i'm being crazy), and would give you points within a few decimal places.
  12. 25 Dec '07 07:10
    and to answer your question about your two lines more clearly:...

    given a point [e.g. (3,4)] the line that connects it with the origin will me y=4/3x. this is because the formula for slope achieves m =(4-0/3-0) = 4/3. now if you divide both the x and y value of the point by the same number c [continuing the example we look at (3/c,4/c)], the slope of that line will be m = [(4/c) - 0]/[(3/c) - 0] = (4/c)/(3/c). notice algebraically, we can factor out a factor of 1/c from both the numerator and denominator... and we realize we have not changed the slope at all.

    in fact, multiplying the numerator and denominator of any fraction by the same number, by definition, will not change the value of that fraction.

    i hope this helped!
  13. Standard member wolfgang59
    Infidel
    25 Dec '07 11:22
    Originally posted by sonhouse
    Where does the A squared come in? In a circle around an origin, the formula is just X^2+Y^2=1 or R. It doesn't have to be assigned to a power term. R=radius of the circle which we can assign as 1. So in your case, Y^2 would be = to the radius which we can call 1 if we want. I guess we can call it peanut butter if we want but it is still just a unit of 1.
    I have to agree with Joe. His 'standard' equation is much better and yours is inaccurate.

    x^2 + y^2 = r^2

    is the formula for a circle of radius r about the origin.
  14. 26 Dec '07 08:46 / 1 edit
    Originally posted by wolfgang59
    I have to agree with Joe. His 'standard' equation is much better and yours is inaccurate.

    x^2 + y^2 = r^2

    is the formula for a circle of radius r about the origin.
    i think what sonhouse was trying to say was that the relative lengths of the chords, with respect to the radius of the circle, do not change no matter what radius you choose for said circle. so why not use the unit circle to make the theory slightly easier to understand, and then multiply by the radius to discover the "true" length of the chord on a given circle?...

    though you are correct in assessing that his equation is inaccurate: the radius of the circle in his equation (x^2 + y^2 = R) would actually be the square root of R.

    just my two cents