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KCC's Quizzes AQQ252 about available meadow area for a goat

Dear Brainstormers, 

As promised, here is a new challenge! 

But before to present it, we will start, from now on, with a funny quote (we can be excellent engineers with great sense of humor as well!)

1. Quote of the week: “If someone smiles all the time, he/she ’s probably selling something that doesn’t work” – Woody Allen

2. Now, Our challenge of the week (this one is not an easy one...):

Good luck!

P.S. And don't forget to inform colleagues around you; they might be interested to have brain twisted as well !

Please share your answer to view other submitted answers
  • Question 1:  2*R  = 20m

    Question 2:  around 11.59m

    How to find that last answer (without calculus):

    Draw the following:

    A circle of radius R centered at point C at x=R, y = 0. That is the field circle.
    A circle of radius g (about 1.75 R) centered at the origine, the point O. That is the goat circle. Call G the point of intersection of this circle on the x-axis  (at a distance g to the origine, the point O).
    Call the intersection of the two circles as the point P.
    Draw the lines segments OP  and CP  (their length is R, since O and P lay on the field circle.
    Call t the angles POC  (and OPC)    (t is in radians). Call u the angle OCP. From the triangle OCP we get:
                                  u = π  – 2t

    The area of the sector OCP (portion of the field circle with an angle t ) =  (u /(2 π)) πR2.  (Or, if you prefer the proportion of the full area of circle of radius R in the ratio of the sustained angle to a full turn.)

    The area of the sector GOP = ( t/(2 π)) πg2  since the radius of that circle is g, not R, and the sustaining angle is t.

    The area of the triangle OCP is 2 * (R sint(t)) (R cos(t))/2  =  R2 sin t cos t.

    Adding the area of two sectors and subtracting the area of the triangle (included twice in the sectors) should, from the problem, equal half of half the circle of radius R  (one of the half comes from the symmetry, the other from the question itself) which is ¼ πR2.

    We are simply missing a relation between t and g (R is given in the question).

    Consider the point P. It is part of the two circles of (in cartesian coordinates):
    x2 + y2 = g2        --  the goat circle
    x2 + y2 = 2Rx      -- the field circle

    Giving:  2Rxp = g2.

    Furthermore, the field circle, in polar coordinates, is  ρ = 2R cos θ. For the point P, that would be ρ = OP  and θ = t.  So, segment length OP = 2R cos(t).
    Projecting OP on the x axis, that is equal to xp. So we get: g = 2R cos(t) after elementary algebraic manipulations.

    So, using Excel, from a value of t, we compute g, u, sum the area of the two sectors, subtract the area of the triangle and compare the result to a quarter of a full circle of radius R.
    When they are equal, we read t around 54.6 degree, and g as about 11.59m.

  • The first sector is sustained by the angle u, not by the angle t, as I typed in the first message.

  • 1. 20m

    2. 11.587m

    Be the fix stake the origin of a cartesian plane. You then have a circle with center = (R, 0). From the origin of the cartesian plan, draw a circle with radius r (r>0, <=2R).The two circle will intersect in two point simmetric with respect to x axis. With these two point you can then evaluate the area of two circle sectors: first one relative two the circle with radius r and starting at the origin and second one relative to the circle with radius R and starting in point (R, 0). Now, the area we're looking for is the sum of these two circle sectors minus the area of the triangle isosceles with sides R, R and r (which is counted twice when you sum the two circle sector areas). You then set this area be equal to half the area of the meadow area. So, with geometric considerations you get to the following equation:

    r^2*arccos(r/2R)+2R^2*arcsin(r/2R)-r*sqrt(R^2-r^2/4)-PI*R^2/2=0

    Set R=10 and solve numerically, you get r=11.587m 

  • 1. Minimum rope for full meadow is 20m (2*R)

    2. Goat's Rope = 11.5872m approximately.

    I don't have an exact solution. But the area should be 2 segments (sector minus triangle) by the goat's rope and the radius R. 

    Sector = 0.5*radius^2* theta. 

    Triangle by sector = 0.5* radius^2 * sin(theta)

    Let A1 be the segment of the goat's rope, and A2 be the segment of the Field's Radius.

    A1 = 0.5 * g^2  * (theta - sin (theta))

    A2 = 0.5 * R^2 * ( (2*pi - 2*theta) - sin(2*pi - 2*theta))

    (the reason why it's 2pi - 2theta is because it's the angle that faces the angle of the goat's area.

    A1 + A2 = 100 * pi / 2

    Using law of cosines because the segments can share the same line that bifurcates them...

    g^2 + g^2 - 2g^2 cos (theta) = R^2 + R^2 - 2R^2 * cos(2* pi - 2*theta)

    we can get g^2 = 200 * (1 + cos(theta)).

    Substituting everything back to that A1 + A2 and removing g^2, it should result to: - pi / 2 = theta * cos(theta) - sin(theta)

    I have no idea how to solve this exactly without a calculator.  Results are theta = 1.9056957 radians. Plug that back in to the result of our law of cosines and we get goat's rope = 11.5872m. Approximately.

  • 1) If the ropes length (L) is 20 meters then 100% of the meadow is covered.

    2) Dust off the geometry and trig books.  If the rope’s length is 11.58728473.. meters it will create two meadow areas that can be accessed.  These are the Pie and Segment areas which should sum to 1/4 of the meadow area.  The lower left area can also be accessed equally so 50% of the meadow is covered.  The approach taken is to express these two areas as functions of alpha and then use a numerical solver (MathCAD, MATLAB, etc.) to compute the angle alpha from which L is calculated.  All angles are in radians.

    Rotate L from a horizontal position upward to the perimeter of the meadow through angle beta.  This creates a pie slice shaped area, a segment area with L as its cord length, and an isosceles triangle whose base is L with equal sides R and vertex angle alpha.

    Now from the web and the triangle we get these functions.

    Segment Area=(R2/2)[alpha-sin(alpha)]=50[alpha-sin(alpha)]

    L2=2R2[1-cos(alpha)]=200[1-cos(alpha)]

    beta=(Pi-alpha)/2

    Pie Area=50[1-cos(alpha)](Pi-alpha)

    Solve:    Segment Area + Pie Area – (Meadow Area)/4 =0

                   50[alpha-sin(alpha)] + 50[1-cos(alpha)](Pi-alpha) – 25Pi = 0

    The above gives alpha equal to 1.235896924 from which L can be calculated.

    Using the calculator here: https://www.123calculus.com/en/two-circles-calculator-page-7-60-400.html, I matched the value of L as calculated by the above technique.

  • BINGO  ! You get it, congratulations since this quiz is not an easy one! And you are the first!

  • Big applause to you Gaetano! You get it; this one was not easy to solve!

  • Fantastic, you find it, congratulations  !

  • yes! you solved it, congratulation