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    Constructing a Lattice from a Heron Triangle

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    The triangle (15,34, 35; 252) is the smallest acute Heron triangle indecomposable into two Pythagorean triangles. Realize it as a lattice triangle, with one vertex at the origin.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:37 pm ad1c9bdddf
    https://brainmass.com/math/triangles/constructing-lattice-heron-triangle-493879

    SOLUTION This solution is FREE courtesy of BrainMass!

    Let the vertices of the triangle be given by (0, 0), (a, b), and (c, d).

    We divide the work into cases, looking for a lattice triangle with sides of 15, 34, or 35, or proving that no such lattice triangle exists.

    Case 1: the side length of 15 touches the origin.
    Case 2: the side length of 15 does not touch the origin.

    To analyze case 1, first assume that the side of length 15 lies between the origin and (a, b). Also, without loss of generality, assume that a and b are at least 0 (the vertex (a, b) could be flipped into any quadrant while retaining its status as a lattice point or non-lattice point).

    Then we have 15 = a^2 + b^2, so b = sqrt(225 - a^2). The only integer possibilities for (a, b) are then (15, 0) and (9, 12), so we divide into cases again.

    Case 1.a: (a, b) = (0, 15)
    Case 1.b: (a, b) = (9, 12)

    Case 1.a.1: (0, 0) to (c, d) is 34, and (15, 0) to (c,d) is 35

    c^2 + d^2 = 1156
    (15 - c)^2 + d^2 = 1225
    therefore
    225 - 30c = 69
    c = 156/30 = not integer

    Case 1.a.2: (0, 0) to (c, d) is 35, and (15, 0) to (c,d) is 34

    c^2 + d^2 = 1225
    (15 - c)^2 + d^2 = 1156
    therefore
    225 - 30c = -69
    c = 294/30 = not integer

    Case 1.b.1
    (0, 0) to (c, d) is 34, and (9, 12) to (c,d) is 35

    c^2 + d^2 = 1156
    (9 - c)^2 + (12 - d)^2 = 1225

    therefore

    81 - 18c + 144 - 24d = 69
    156 - 18c - 24d = 0
    26 - 3c - 4d = 0
    3c + 4d = 26

    solve this Diophantine equation to get

    c = 2 + 4m, d = 5 - 3m, for any integer m

    substitute into c^2 + d^2 = 1156:

    (2 + 4m)^2 + (5 - 3m)^2 = 1156
    4 + 16m + 16m^2 + 25 - 30m + 9m^2 = 1156
    0 = 25m^2 - 14m - 1127
    m = [14 +- sqrt(196 + 112700)]/50
    m = [14 +- 336]/50
    m = 7
    (the other solution to the quadratic is -161/25, which is not an integer)

    Substitute m = 7 into c = 2 + 4m, d = 5 - 3m to get:

    c = 30, d = -16

    We have now found a solution:

    Vertex 1: (0,0)
    Vertex 2: (9, 12)
    Vertex 3: (30, -16)

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:37 pm ad1c9bdddf>
    https://brainmass.com/math/triangles/constructing-lattice-heron-triangle-493879

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