 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2021/3/c/i/solution

The four points form a projective frame: no three are collinear. By the allowed uniqueness theorem and cross-ratio invariance, use

P_{0} = [1 : 0 : 0], P_{1} = [0 : 1 : 0], P_{2} = [0 : 0 : 1], P_{3} = [1 : 1 : 1] .

Intersecting the indicated projective lines gives

Q_{0} = [1 : 0 : 1], Q_{1} = [1 : 1 : 0], Q_{2} = [0 : 1 : 1] .

A point of

Q_{0} Q_{2}

has coordinates

[a : b : a + b]

. Intersecting with

P_{0} P_{1}

, whose equation is

z = 0

, gives

Q_{3} = [1 : - 1 : 0]

. On that line use its first two homogeneous coordinates. The cross-ratio from (a)(ii) is

(P_{0} P_{1} : Q_{3} Q_{1}) = \frac{( - 1 ) ( - 1 )}{( 1 ) ( - 1 )} = - 1.

The characteristic of a field being zero ensures that

Q_{3} \neq = Q_{1}

. The construction is the diagonal-point property of a complete quadrangle, producing harmonic conjugate points.

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