Steve Sigur posted recently posted to the geometry.college newsgroup a conjucture by one of his students, Adam Bliss:
"Draw an arbitrary line in the plane of the triangle. At each of the midpoints of the triangle, draw a line parallel to the defined line. Reflect each side of the triangle across the line that goes through its midpoint. The reflected lines are concurrent. As the defined line sweeps through any quarter-arc, the point of concurrency sweeps out the nine-point circle.
Here I discuss the proof
Lou Talman presented a proof of the concurrency and the fact that the point of concurrency lies on the ninepointcircle. This proof is accessable from his homepage.
I will give an alternative proof.
Proof
Let us consider a triangle ABC (labelled counterclockwise), and a line l, which I choose to pass through A. The midpoints of AB, AC and BC are denoted by C', B' and A' respectively.
Let l' be the line through C' parallel to l, then we can construct the reflection of AB through l' in the following way:
figure 1 |
Construct a rectangle of which l is one of the sidelines, and AB is a diagonal. The other diagonal is the reflection we wish to construct. This can be seen because H is the image of B after reflection through l' and I the image of A after reflection through l'. Note that the image line (the blue line in the figure) would be the same when not AH would be chosen to be l, but a line perpendicular to AH, for instance AI. This shows why the last statement of the theorem holds, when the concurrency of the reflected lines on the ninepointcircle is proven (after a check, which I omit, that all points of the circle will be reached). |
The use of a rectangle in the construction, points us to the use of a rectangular grid with cartesian coordinates to do some calculations. We can choose the grid in such a way that A lies on the y-axis, B on the x-axis and that one of the axes is parallel to line l. We find a picture like the following:
figure 2 |
The red lines are the reflected lines of which the concurrence is to be proven. They have the following equations: y = a/b x y = 2d - (d-a)/c x y = 2cd/(c-b) - d/(c-b) x It is easy to check that they all intersect for x = 2bcd/(bd+ac-ab). |
So the concurrency of the three lines in the theorem, the red lines in figure 2, is proven. Let us name the point of concurrency X.
When X coincides with A', B' or C' it clearly is on the circumcircle of A'B'C', the ninepointcircle ABC.
So let us suppose that X is not one of the points A', B' or C'. Note that the three red lines from figure 3 are antiparallel to the triangle sides. Without losing generality we can assume that X is opposite to A w.r.t. the line B'C', as it is in figure 2 (this can be fixed by choosing the two right vertices to be on the axes). Since the red lines are anti-parallel to the triangle sides we see that angle C'XB' = angle CAB = angle C'A'B'. Since A' and X are both opposite to A w.r.t. B'C', we can conclude that A'B'C'X is a cyclic quadrilateral, and X lies on the ninepointcircle of ABC.
And the theorem is proven.
This theorem has been expanded to find a relation to Morley's trisector theorem. See:
Floor van Lamoen, Morley related triangles on the Nine Point Circle, American Mathematical Monthly, 107 (2000) 941–945.
Back to Floors wiskunde pagina (Dutch).
Home (Dutch).