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A seemingly impossible problem from my teacher!

  1. Oct 25, 2005 #1
    No its not homework, but my teacher (pre-cal) couldn't figure it out and neither can I, so see if you can!

    I can't draw it so i'll explain it, pretty simple.

    You have a square (ABCD) with a random point closest to the bottom-left corner of the square (it doesn't really matter which corner). The point is 3 units from A (the closest corner), 7 units from B (the point directly above A), and 5 units from D (the point directly right of A)

    Keep in mind you don't no any angles except of course the 90 degrees of each corner.

    Solve for - any side length!
     
  2. jcsd
  3. Oct 26, 2005 #2

    Tide

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    The point lies at the intersection of 3 circles centered on the given vertices with their respective radii. You have 3 equations and 3 unknowns (x, y, a) where a is the length of each side of the circle. You're only interested in a so you want to eliminate x and y.

    After a quick run through I get [itex]a = 4[/itex] or [itex]a = 3 \sqrt {11}[/itex].
     
  4. Oct 26, 2005 #3

    AKG

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    Let X be your point. Let x be the angle BXA, y is the angle DXA, and z is the angle DXB. Let s be the side length of your square. Using the cosine law, you know:

    1) 2a² = 25 + 49 - 2(5)(7)cos(z)
    2) a² = 9 + 25 - 2(3)(5)cos(y)
    3) a² = 9 + 49 - 2(3)(7)cos(x)

    Treat a², cos(x), cos(y), and cos(z) as your unknowns, then you have three linear equations in terms of 4 unknowns. You can then solve for all of them in terms of one of them, say cos(x).

    But you also know that x + y + z = 2π, and that π/2 < x < π. With a bunch of messy algebra (it will boil down to solving a cubic polynomial with integer coefficients), or using a computer program, you can find a solution for cos(x), and hence a², and hence a.
     
    Last edited: Oct 26, 2005
  5. Oct 26, 2005 #4
    well, one triangle has sides 7, 3, a
    another has sides 3, 5, a
    a third has sides 7,5,a* sqrt {2}

    you can use a lot of law of cosines, a big ugly heron formula for the whole square or some plain geometry. If i were to gwt this problem i'd go for the Cosines.
     
    Last edited: Oct 26, 2005
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