Book VI · Proposition 28

VI.28

To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.Heath, 1908

The geometric solution of a quadratic. Euclid's proviso is exactly the condition for the discriminant not to be negative.

Every step, checked

• A and B are distinctgiven
• the application is a proper onegiven
• the given area does not exceed that on the half (VI.27)given
• the point falls on AB, between A and the midpointby VI.27
• the rectangle applied equals the given areaby II.5

What it needs, and what needs it

Needs: I.10 II.5 VI.27

Rests on: C.N.1, C.N.2, C.N.3, C.N.4, C.N.5, Def.10, Def.15, Def.22, Def.4, Post.5, V.Def.5

Depth: 17 steps of argument above the first principles. Parallel postulate: needed.

What it takes on trust

• a circle-circle intersection is used, though no postulate asserts that two circles meet (×12)continuity
• a line-circle intersection is used, though no postulate asserts that a line meets a circle (×12)continuity

The proposition as code

@proposition(
    "VI.28",
    CONSTRUCTION,
    sample=_deficient_application,
    note="The geometric solution of a quadratic. Euclid's proviso is exactly the "
    "condition for the discriminant not to be negative.",
)
def prop_VI_28(a: Point, b: Point, part) -> Out:
    """Apply to AB a parallelogram equal to a given area, deficient by a square."""
    hypothesis("A and B are distinct", a != b)
    hypothesis("the application is a proper one", sign(part) > 0 and sign(1 - part) > 0, guard=True)
    line(a, b, "the given line AB")
    middle = posit(prop_I_10(a, b).midpoint, "C")

    whole = length(a, b)
    wanted = whole * whole * part * (1 - part)  # never more than the square on the half
    hypothesis("the given area does not exceed that on the half (VI.27)",
               sign(whole * whole / 4 - wanted) >= 0)

    # x(whole - x) = wanted has the root x = whole/2 - sqrt(whole^2/4 - wanted),
    # which is II.5 read as a formula: the half, less the piece between sections.
    gap = sqrt(whole * whole / 4 - wanted)
    cut = posit(_along(a, b, (whole / 2 - gap) / whole), "S")
    across = (-(b.y - a.y), b.x - a.x)
    applied = _parallelogram_on(a, cut, Point(a.x + across[0], a.y + across[1]))
    outline(*applied)

    because(prop_VI_27, a, b, part)
    because(prop_II_5, a, cut, b)

    claim("the point falls on AB, between A and the midpoint", "VI.27",
          on_line(cut, Line.through(a, b)) and sign(length(a, cut)) >= 0
          and sign(length(a, middle) - length(a, cut)) >= 0)
    claim("the rectangle applied equals the given area", "II.5",
          length(a, cut) * length(cut, b) == wanted)
    return Out(section=cut)