9-point Circle as a Locus of Concurrency
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The Simson lines of two diametrically opposite points on a circumcircle of a triangle intersect on the 9-point circle of that triangle. A generalization of this fact is known as Griffiths' theorem:
Let the line joining two points P and Q pass through the circumcenter of ΔA1A2A2. Then the pedal circles of points P and Q meet on the 9-point circle of ΔA1A2A2, i.e. the three circles are concurrent. This is known as Griffiths' theorem. In fact the point of concurrency only depends on the line the two points lie on. In other words, the pedal circles of the points on a line through the circumcenter of a triangle all concur at the same point on its 9-point circle.
The point is known as Griffiths' point.
(It is worthwhile to note that the 9-point circle is the pedal circle of the circumcenter as well as that of the orthocenter.)
Griffiths' theorem admits a proof based on complex numbers.
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Copyright © 1996-2012 Alexander Bogomolny
Let the line joining two points P and Q pass through the circumcenter of ΔA1A2A2. Then the pedal circles of points P and Q meet on the 9-point circle of ΔA1A2A2, i.e. the three circles are concurrent. This is known as Griffiths' theorem. In fact the point of concurrency only depends on the line the two points lie on. In other words, the pedal circles of the points on a line through the circumcenter of a triangle all concur at the same point on its 9-point circle.
Proof
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Assume the circumcenter of ΔA1A2A3 is at the origin and the circumradius equals 1. Lower case letters denote complex numbers associated with the points denoted by the corresponding upper case so that, for example ai corresponds to Ai, i = 1, 2, 3. We have
|a1| = |a2| = |a3| = 1.
Let P lie on a straight line l through the origin and C1. We choose l to coincide with the real axis which in particular implies
Since C1 is collinear A2 and A3
(c1 - a2) / (a3 - a2) = (c'1 - a'2) / (a'3 - a'2)
From |a2| = 1, a'2 = 1/a2 and a'3 = 1/a3. So we have
| (c1 - a2) / (a3 - a2) | = (c'1 - a'2) / (a'3 - a'2) |
| = (c'1 - 1/a2) / (1/a3 - 1/a2) | |
| = (c'1a2 - 1) a3 / (a2 - a3) |
from which
c1 - a2 = a3 - a2a3c'1
or
a2 + a3 = c1 + a2a3c'1
If the indices i, j, k are distinct and come from the set {1, 2, 3} then, more generally,
| (1) | c'k ai aj = ai + aj - ck. |
On the other hand, PCk is perpendicular to AiAj, so that
(p - ck)/(ai - aj) + (p' - c'k)/(a'i - a'j) = 0,
which, using |ai| = |aj| = 1, we modify as follows:
(p - ck)/(ai - aj) + (p' - c'k) ai aj /(aj - ai) = 0,
or, with p = p',
p - ck - (p - c'k) ai aj = 0.
Finally, we obtain
| (2) | c'k ai aj = p ai aj - p + ck. |
A comparison of (2) and (1) gives
| (3) | ck = (ai + aj + p - p ai aj) / 2. |
Introduce point
| (4) | w = (a1 + a2 + a3 - a1 a2 a3) / 2. |
We shall show that four points C1, C2, C3, W are concyclic, i.e. W lies on the circumcircle of ΔC1C2C3. Note W is independent of P. In particular, W is the same when P is at the origin, i.e. when P coincides with the circumcenter in which case the circumcircle of ΔC1C2C3 is the nine point circle of ΔA1A2A3. Thus we shall have proved Griffiths' theorem and showed that W is exactly the Griffiths point.
For the concyclicity of C1, C2, C3, W we have to check that
(w - c2)/(c1 - c2) : (w - c3)/(c1 - c3) = (w' - c'2)/(c'1 - c'2) : (w' - c'3)/(c'1 - c'3).
which we shall prove in another form
| (5) | (w - c2)/(c1 - c2) : (w' - c'2)/(c'1 - c'2) = (w - c3)/(c1 - c3) : (w' - c'3)/(c'1 - c'3). |
First of all, from (3) and (4),
| w - c2 | = (a1 + a2 + a3 - a1 a2 a3) / 2 - (a1 + a3 + p - p a1 a3) / 2 |
| = (p - a2)(a1 a3 - 1) / 2. |
From here, using |ai| = 1,
| w' - c'2 | = (p - a'2)(a'1 a'3 - 1) / 2 |
| = (p - 1/a2)(1/a1 1/a3 - 1) / 2 | |
| = -(p a2 - 1)(a1 a3 - 1) / 2a, |
where a = a1a2a3. Also
c1 - c2 = (a1 - a2)(p a3 - 1) / 2
and
c'1 - c'2 = -(a1 - a2)(p - a3) / 2a.
Similar expressions are obtained for the differences in (5) that include c3. The left hand side in (5) becomes
(p - a2)(a1 a3 - 1)(a1 - a2)(p - a3) / (a1 - a2)(p a3 - 1)(p a2 - 1)(a1 a3 - 1)
or, after reduction,
(p - a2)(p - a3) / (p a3 - 1)(p a2 - 1),
which is symmetric in a2 and a3 showing that the right hand side in (5) equals the same expression, thus proving (5) and, with it, Griffiths' theorem.
(Line l passing through the circumcenter intersects the circumcircle at two points. For each of these, the three projections C1, C2, C3 are collinear and lie on the simsons of ΔA1A2A3. It follows that the two simsons meet at the Griffiths point associated with l.)
Assume now there are four concyclic points A1, A2, A3, and A4. Taken by three, they form four triangles Tm = ΔAiAjAk, where i, j, k, m are distinct elements from
(w1 - w2) / (w'1 - w'2) = (w1 - w3) / (w'1 - w'3) = (w1 - w4) / (w'1 - w'4).
This and an additional feature of the configuration is considered on a separate page.
A purely geometric proof can be found in [Advanced Euclidean Geometry, pp. 245-246].
References
- R. A. Johnson, Advanced Euclidean Geometry, Dover, 2007
- J. Tabov, Four Collinear Griffiths Points, Mathematics Magazine, v. 68, n 1, February 1995, pp. 61-64
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