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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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Some Thoughts Sines and secants | Comment 2 of 4 |

Construct the circle. Let B,C be the points as defined. Select some other points B1,C1, near B and C respectively, in similar manner. Line BB1 extended and line CC1 extended meet at a single point outside the circle. This is point A, as will be shown.

The triangles are ABC and A'B'C'. Their sides are a,b,c, and a1,b1,c1 respectively.

By construction, lines BB1 extended and CC1 extended are secants that intersect at A (and only at A). Thus by the secant rule: c*c1=b*b1=x. Also, angle ABC=angle AC1B1. So the triangle are similar, so the proportion (b+c)/a is constant.

So it is true that for any pair of triangles so constructed, point A is unique, and (b+c)/a constant.

However, the broader proposition that all chords BC of the circle through P can share a common A with constant proportion seems not to hold:

Construct a further pair of points B2,C2 on the circle. Fresh sets of secants can be constructed with B2,C2 and either B, C, or B1, C1 respectively (but probably not both) and a new point A' so determined; but the point A' need not coincide with A. While it is true that A' is also unique, that a similar angle equality holds, and that (b+c)/a constant for the new pair of triangles, it is not necessarily the same constant as previously determined.

So on reflection the sine rule is not needed after all.

Edited on March 23, 2015, 8:01 am
  Posted by broll on 2015-03-23 04:54:09

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