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Chord Triangle Constant (Posted on 2015-03-21) Difficulty: 4 of 5

  
Let Γ be a circle with center O and radius r. Let P be a point inside Γ
( different from O ) with |OP| = p.

a) Prove there exists a point A outside Γ such that for all chords BC
of Γ through P the quantity (b+c)/a is constant ( where a, b, and c
are the side lengths of ΔABC ).

b) What is the constant in terms of p and r?

c) Prove that the point A is unique.
  

See The Solution Submitted by Bractals    
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Solution Possible Solution | Comment 3 of 4 |

Let A denote the point where a second circle, through B,
O and C, cuts the line OP produced.
BC is a common chord, and the intersecting chord theorem
applied in both circles gives:
            |BP||CP| = (r + p)(r – p) = p(|OA| – p)

from which:       |OA| = r2/p

Since r and p are independent of the position of B and C, it
follows that A is fixed relative to O.

/OAB = /OAC (since they are subtended by equal arcs in the

2nd circle) so AO bisects /BAC, and therefore b/|PC| = c/|PB|.

Also, since triangles OPB and CPA are similar (angles subtended

by common arcs),          b/|PC| = c/|PB| = r/p,

from which        b + c = (r/p)(|PC| + |PB|) = r*a/p.

So   (b + c)/a = r/p,  and is a constant satisfying parts (a) and (b).

Uniqueness?
Since b + c is a constant (for a given value of a), A must lie on an
ellipse with focal points at B and C. In fact, since AB and AC make
equal angles with AO, it follows that AO is normal to the ellipse at A.
Since A is fixed, as B and C ‘move’ on the circle the point on the
ellipse at A is moving perpendicular to OA, so the instantaneous
centre of rotation of the ellipse must lie on OA. However the other
point of intersection of the ellipse with OA is clearly not fixed since
the ellipse is changing in size and position as it rotates; so A is
unique.
(There must be a neater explanation..  but this ellipse has given
me a lot of fun recently).



  Posted by Harry on 2015-03-24 17:51:33
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