# 9-point Circle as a Locus of Concurrency

What Is This About?

A Mathematical Droodle

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 $\Delta A_{1}A_{2}A_{2}.$ Then the pedal circles of points $P$ and $Q$ meet on the $9$-point circle of $\Delta A_{1}A_{2}A_{2},$ 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 © 1996-2018 Alexander Bogomolny

Let the line joining two points $P$ and $Q$ pass through the circumcenter of $\Delta A_{1}A_{2}A_{2}.$ Then the pedal circles of points $P$ and $Q$ meet on the $9$-point circle of $\Delta A_{1}A_{2}A_{2},$ 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

Assume the circumcenter of $\Delta A_{1}A_{2}A_{3}$ 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 $a_{i}$ corresponds to $A_{i},$ $i = 1, 2, 3.$ We have

$|a_{1}| = |a_{2}| = |a_{3}| = 1.$

Let $P$ lie on a straight line $l$ through the origin and $C_{1}.$ We choose $l$ to coincide with the real axis which in particular implies $p = p\space '.$ Let $C_{2},$ $C_{3}$ be the feet of perpendiculars from $P$ to the side lines $A_{2}A_{3},$ $A_{1}A_{3},$ and $A_{1}A_{2}.$

Since $C_{1}$ is collinear with $A_{2}$ and $A_{3}$

$(c_{1} - a_{2}) / (a_{3} - a_{2}) = (c\space ^{'}_{1} - a\space ^{'}_{2}) / (a\space ^{'}_{3} - a\space ^{'}_{2})$

From $|a_{2}| = 1,$ $a\space ^{'}_{2} = 1/a_{2}$ and $a\space ^{'}_{3} = 1/a_{3}.$ So we have

$\begin{align} (c_{1} - a_{2}) / (a_{3} - a_{2})&= (c\space ^{'}_{1} - a\space ^{'}_{2}) / (a\space ^{'}_{3} - a\space ^{'}_{2})\\ &= (c\space ^{'}_{1} - 1/a_{2}) / (1/a_{3} - 1/a_{2})\\ &= (c\space ^{'}_{1}a_{2} - 1) a_{3} / (a_{2} - a_{3}) \end{align}$

from which

$c_{1} - a_{2} = a_{3} - a_{2}a_{3}c\space ^{'}_{1}$

or

$a_{2} + a_{3} = c_{1} + a_{2}a_{3}c\space ^{'}_{1}$

If the indices $i, j, k$ are distinct and come from the set $\{1, 2, 3\}$ then, more generally,

(1)

$c\space ^{'}_{k} a_{i} a_{j} = a_{i} + a_{j} - c_{k}.$

On the other hand, $PC_{k}$ is perpendicular to $A_{i}A_{j},$ so that

$(p - c_{k})/(a_{i} - a_{j}) + (p\space ' - c\space ^{'}_{k})/(a\space ^{'}_{i} - a\space ^{'}_{j}) = 0,$

which, using $|a_{i}| = |a_{j}| = 1,$ we modify as follows:

$(p - c_{k})/(a_{i} - a_{j}) + (p\space ' - c\space ^{'}_{k}) a_{i} a_{j} /(a_{j} - a_{i}) = 0,$

or, with $P = p\space ',$

$p - c_{k} - (p - c\space ^{'}_{k}) a_{i} a_{j} = 0.$

Finally, we obtain

(2)

$c\space ^{'}_{k} a_{i} a_{j} = p a_{i} a_{j} - p + c_{k}.$

A comparison of (2) and (1) gives

(3)

$c_{k} = (a_{i} + a_{j} + p - p a_{i} a_{j}) / 2.$

Introduce point

(4)

$w = (a_{1} + a_{2} + a_{3} - a_{1} a_{2} a_{3}) / 2.$

We shall show that four points $C_{1},$ $C_{2},$ $C_{3},$ $W$ are concyclic, i.e. $W$ lies on the circumcircle of $\Delta C_{1}C_{2}C_{3}.$ 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 $\Delta C_{1}C_{2}C_{3}$ is the nine point circle of $\Delta A_{1}A_{2}A_{3}.$ Thus we shall have proved Griffiths' theorem and showed that $W$ is exactly the *Griffiths point*.

For the concyclicity of $C_{1},$ $C_{2},$ $C_{3},$ $W$ we have to check that

$(w - c_{2})/(c_{1} - c_{2}) : (w - c_{3})/(c_{1} - c_{3}) = (w' - c\space ^{'}_{2})/(c\space ^{'}_{1} - c\space ^{'}_{2}) : (w' - c\space ^{'}_{3})/(c\space ^{'}_{1} - c\space ^{'}_{3}).$

which we shall prove in another form

(5)

$(w - c_{2})/(c_{1} - c_{2}) : (w' - c\space ^{'}_{2})/(c\space ^{'}_{1} - c\space ^{'}_{2}) = (w - c_{3})/(c_{1} - c_{3}) : (w' - c\space ^{'}_{3})/(c\space ^{'}_{1} - c\space ^{'}_{3}).$

First of all, from (3) and (4),

$\begin{align} w - c_{2}&= (a_{1} + a_{2} + a_{3} - a_{1} a_{2} a_{3}) / 2 - (a_{1} + a_{3} + P - P a_{1} a_{3}) / 2\\ &= (p - a_{2})(a_{1} a_{3} - 1) / 2. \end{align}$

From here, using $|a_{i}| = 1,$

$\begin{align} w' - c\space ^{'}_{2}&= (p - a\space ^{'}_{2})(a\space ^{'}_{1} a\space ^{'}_{3} - 1) / 2\\ &= (p - 1/a_{2})(1/a_{1} 1/a_{3} - 1) / 2\\ &= -(p a_{2} - 1)(a_{1} a_{3} - 1) / 2a, \end{align}$

where $a = a_{1}a_{2}a_{3}.$ Also

$c_{1} - c_{2} = (a_{1} - a_{2})(p a_{3} - 1) / 2$

and

$c\space ^{'}_{1} - c\space ^{'}_{2} = -(a_{1} - a_{2})(p - a_{3}) / 2a.$

Similar expressions are obtained for the differences in (5) that include $c_{3}.$ The left hand side in (5) becomes

$(p - a_{2})(a_{1} a_{3} - 1)(a_{1} - a_{2})(p - a_{3}) / (a_{1} - a_{2})(p a_{3} - 1)(p a_{2} - 1)(a_{1} a_{3} - 1)$

or, after reduction,

$(p - a_{2})(p - a_{3}) / (p a_{3} - 1)(p a_{2} - 1),$

which is symmetric in $a_{2}$ and $a_{3}$ 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 $C_{1},$ $C_{2},$ $C_{3}$ are collinear and lie on the simsons of $\Delta A_{1}A_{2}A_{3}.$ It follows that the two simsons meet at the Griffiths point associated with $l$.)

Assume now there are four concyclic points $A_{1},$ $A_{2},$ $A_{3},$ and $A_{4}.$ Taken by three, they form four triangles $T_{m} = \Delta A_{i}A_{j}A_{k},$ where $i, j, k, m$ are distinct elements from $\{1, 2, 3, 4\}.$ To each $T_{m}$ and the given line $l$ there correspond Griffiths point $W_{m}.$ Using the same technique as above we can show (after J. Tabov) that the four Griffiths points $W_{1},$ $W_{2},$ $W_{3},$ $W_{4}$ are collinear. It is sufficient to show that

$(w_{1} - w_{2}) / (w\space ^{'}_{1} - w\space ^{'}_{2}) = (w_{1} - w_{3}) / (w\space ^{'}_{1} - w\space ^{'}_{3}) = (w_{1} - w_{4}) / (w\space ^{'}_{1} - w\space ^{'}_{4}).$

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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Copyright © 1996-2018 Alexander Bogomolny