# Yet Another Concurrence on the 9-Point Circle

### Problem

Lines $m_a,$ $m_b,$ $m_c,$ are perpendicular to the three parallel lines $l_a,$ $l_b,$ $l_c$. Denote the intersections $P_{ab}=l_{a}\cap m_{b},$ $P_{ba}=l_{b}\cap m_{a},$ $P_{ac}=l_{a}\cap m_{c},$ $P_{ca}=l_{c}\cap m_{a},$ $P_{bc}=l_{b}\cap m_{c},$ $P_{cb}=l_{c}\cap m_{b},$ $A=l_{a}\cap m_{a},$ $B=l_{b}\cap m_{b},$ $C=l_{c}\cap m_{c}.$ Prove that

1. Lines $P_{ab}P_{ba},$ $P_{ac}P_{ca},$ $P_{cb}P_{bc}$ are concurrent;

2. The point of concurrency lies on the 9-point circle of $\Delta ABC.$

### Hint

The problem may be solved with relative ease by choosing a convenient coordinate system.

### Solution

Assume the given lines are parallel to the coordinate axes, so that the lines are given by the equations:

$l_a:\space y=2n,$ $l_b:\space y=2q,$ $l_c:\space y=2v,$
$m_a:\space y=2m,$ $m_b:\space y=2p,$ $m_c:\space y=2u.$

These determine the coordinates of nine points: $P_{ab}(2p,2n),$ $P_{ba}(2m,2q),$ $P_{bc}(2u,2q),$ $P_{cb}(2p,2v),$ $P_{ca}(2m,2v),$ $P_{ac}(2u,2n),$ $A(2m,2n),$ $B(2p,2q),$ $C(2u,2v).$ The equations of the lines through two given points are found to be

\begin{align} P_{ab}P_{ba} &:\space x(q-n)+y(p-m)=2(pq-mn),\\ P_{bc}P_{cb} &:\space x(v-q)+y(u-p)=2(uv-pq),\\ P_{ca}P_{ac} &:\space x(n-v)+y(m-u)=2(mn-uv). \end{align}

The condition for the concurrency of the three lines,

$\left|\begin{array}{ccc} q-n & p-m & pq-mn \\ b-q & u-p & uv-pq \\ n-v & m-u & mn-uv \end{array}\right|=0,$

is easily verified, as, say, the rows of the determinant are clearly linearly dependent (the sum of all three is identically zero.)

The three lines $P_{ab}P_{ba},$ $P_{bc}P_{cb},$ and $P_{ca}P_{ac}$ cross the sides of $\Delta ABC$ at the midpoints, so that $P_{ab}P_{ba}\cap AB=M_c,$ $P_{bc}P_{cb}\cap BC=M_a,$ and $P_{ca}P_{ac}\cap AC=M_b.$ The three midpoints lie on the 9-point circle of $\Delta ABC.$ This allows finding their coordinates, e.g., $M_a(p+u,q+v),$ etc.

In order to prove that the point of concurrency $P$ lies on the 9-point circle suffice it to prove that the four points $P,$ $M_a,$ $M_b,$ $M_c$ are concyclic. One way to do that is to consider the points as complex numbers and establish that the harmonic ratio,

$\displaystyle\frac{M_{a}-P}{M_{b}-P}:\frac{M_{a}-M_{c}}{M_{b}-M_{c}}$

of the four points is real. This is not a difficult exercise in the algebra of complex numbers, as the explicit formula for the harmonic ratio permits significant simplifications.

### Acknowledgment

The problem has been pointed to at the CutTheKnotMath facebook page by Antreas Hatzipolakis (Greece); the proof has been supplied by Leonard Giugiuc (Romania).

### References

1. I. Panakis: 2500 Problems of Geometric Loci With Their Solutions [in Greek]. Athens, ca 1965, p. 654, #582. 