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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.

  Submitted by Bractals    
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Solution: (Hide)


NOTATION:

  PQ denotes the vector from point P to point Q.

   denotes the vector dot product.

PROOF:

WOLOG let the circle have unit radius and O be its center.

Let

    P = (AB&CD) ∩ (AC&BD)

Let M and N be the midpoints of line segments AB and AC
respectively. Therefore,
                   OA + OB       OC + OD
   OP = OM + MP = --------- + λ ---------
                      2             2

      and

                   OA + OC       OB + OD
   OP = ON + NP = --------- + μ ---------
                      2             2

Thus,

   (1-μ)OB + (λ-1)OC + (λ-μ)OD = 0

                1-μ        λ-1
   λ≠μ ⇒ OD = ----- OB + ----- OC
                μ-λ        μ-λ

               1-μ
          τ = ----- ⇒ OD = τ OB - (τ+1)OC
               μ-λ

          1 = OD•OD = τ2 - 2τ(τ+1)OB•OC + (τ+1)2

          0 = τ(τ+1)(1 - OB•OC)

          τ(τ+1) = 0 ⇒ λ = 1 or μ = 1

                    ⇒ OP = (OA+OB+OC+OD)/2

          (1 - OB•OC) = 0 ⇒ B = C which contradicts
                             B and C as distinct

   λ=μ ⇒ (1-λ)(OB - OC) = 0

          (1-λ) = 0 ⇒ OP = (OA+OB+OC+OD)/2

          (OB - OC) = 0 ⇒ B = C which contradicts
                           B and C as distinct

Therefore, OP = (OA+OB+OC+OD)/2
Let

    P = (AB&CD) ∩ (AD&BC)

Then OP = (OA+OB+OC+OD)/2 by the same argument as above
swapping letters C and D.

Therefore, (AB&CD), (AC&BD), and (AD&BC) are concurrent.

QED

HOW CONCYLIC FEET PARTS 1&2 ARE RELATED:

If A, B, C, and D are distinct points on a
circle, then the point of concurrency of Part 2
is the center of the circle on which the feet of
Part 1 lie.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionHarry2012-05-25 23:07:51
ObservationsJer2012-05-24 14:05:14
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