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Concylic Feet Part 2 (Posted on 2012-05-24) Difficulty: 3 of 5


DEFINITION

If P, Q, R, and S are distinct points, then
let (PQ&RS) denote the line through the
midpoint of line segment PQ and
perpendicular to line RS.

PROVE THE FOLLOWING

If A, B, C, and D are distinct points on a
circle, then (AB&CD), (AC&BD), and
(AD&CB) are concurrent.

See The Solution Submitted by Bractals    
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Solution Solution Comment 2 of 2 |
Let a, b, c, d, be the position vectors of A, B, C, D, relative to the centre, O, of the circle, and let P be a point with position vector p =  0.5(a + b + c + d).

Let M be the mid-point of AB, so that the vector MP can be written as:

                 MP   =  p – OM
                        =  0.5(a + b + c + d) – 0.5(a + b)
                        =  0.5(c + d)

Then     MP.CD =  0.5(c + d).(d – c)  =  0.5(d2 – c2)  =  0  since |c| = |d|.

So MP and CD are perpendicular and MP must be the line (AB&CD).

Since p is symmetrical in a, b, c and d, it follows that P will lie on all of the following lines, and therefore that they are all concurrent:

            (AB&CD), (AC&BD), (AD&BC), (BC&AD), (BD&AC), (CD&AB).

It also confirms Jer’s discovery, since  p – 0.5(a + b + c) = 0.5d, so that as D moves on the circle, with A, B and C fixed, P describes a circle with half that radius, and centre at 0.5(a + b + c). This will be concentric with the original circle only when a + b + c = 0; i.e. when triangle ABC is equilateral.



  Posted by Harry on 2012-05-25 23:07:51
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