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Strawberry field, my bed is lost!

Discussion in 'Mathematics' started by AVDBA, Sep 25, 2019.

  1. AVDBA

    AVDBA New commenter

    An ancient manuscript describing the location of a buried treasure has been found:

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    A plane terrain with a bed for strawberries and having just two trees (T1 and T2). From the center C of the bed, measuring the straight line to reach T1. Turning ninety degrees to the left, and then this same distance is needed to straightly reach a point P. Back to the bed to measure the straight distance to reach T2. Turning ninety degrees to the right, and then this same distance is needed to straightly reach the point Q. The treasure is buried down the midpoint of PQ.

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    The plane terrain has been found - but there is no bed! Find the treasure as a function depending on the coordinates of T1 and T2.
     
  2. AVDBA

    AVDBA New commenter

    This problem seems interesting, since, even with a nonexistent information (the strawberry bed), one can use it as an ancillary information, since the result does not depend on this information.

    Let T1(s,t) = (s, t), T2(u, v) = (u, v) and C(b, c) = (b, c) be, respectively, the Cartesian coordinates of the trees, T1 and T2, and of the strawberry bed. Hence, the oriented segment (vector) from the center of the bed to the tree 1 reads:

    CT1 = T1 - C = (s, t) - (b, c) = (s - b, t - c)

    and the oriented segment (vector) from the center of the bed to the tree 2 reads:

    CT2 = T2 - C = (u, v) - (b, c) = (u - b, v - c)

    An arbitrary point (x, y) may be written as (rcos(a), rsin(a)) [r = (x^2 + y^2)^(1/2)], being a the measured angle (counterclockwise) from the 0x axis. A rotation of 90 degrees to the left (+90 degrees) leads to (rcos(a + 90), rsin(a + 90)) = (-rsin(a), rcos(a)) = (-y, x); and a rotation of 90 degrees to the right (-90 degrees) leads to (rcos(a - 90), rsin(a - 90)) = (rsin(a), -rcos(a)) = (y, -x). Henceforth, turning CT1 90 degrees to the left leads to (c - t, s - b), and turning CT2 90 degrees to the right leads to (v - c, b - u).

    Thus, the points P and Q read:

    P = T1 + (c - t, s - b) = (s, t) + (c - t, s - b) = (s + c - t, t + s - b)
    Q = T2 + (v - c, b - u) = (u, v) + (v - c, b - u) = (u + v - c, v + b - u)

    The midpoint M of the segment PQ is given by (P + Q)/2:

    M = [(s + c - t, t + s - b) + (u + v - c, v + b - u)]/2 = (s - t + u + v, t + s + v - u)/2, that does not depend on the [supposed] position of the nonexistent bed.

    And the treasure is buried down the point M = (s - t + u + v, t + s + v - u)/2.
     
  3. adamcreen

    adamcreen Occasional commenter

    We. Don't. Care.
     
  4. AVDBA

    AVDBA New commenter

    Thanks a lot adamcreen.
     

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