It seems like there should be an argument based on applying some kind of power iteration algorithm (https://en.wikipedia.org/wiki/Power_iteration) to vectors with non-negative coefficients.
OK, this is still a little ugly, but at least it all makes sense. ##O## - the place where the perpendicular bisectors meet - is also a the place where the circumcircles meet and there's also a connection between the angle bisectors, and the perpendicular bisectors as marked by the points ##T##...
This smacks of "coffin problem."
Let's say that we start with ##\bigtriangleup ABC##. Then it's easy to find the the perpendicular bisector of ##BC##, the bisector of ##\angle CAB## , the perpendicular to that bisector ##OA## and then the circle centered at ##O## that goes through ##B## (or...
I didn't realize that CQBR really are all on the same circle. Ptolemy's should work for showing that, but ...
For Ptolemy's theorem we want ##\left| QC \right| \left| BR \right| + \left| QB \right| \left| RC \right| = \left| QR \right| \left| CB \right|##.
I imagine it's possible to use the...
So, if ##\delta=\angle QPN##, ##\gamma = \angle CAN ## and ##\alpha=\angle ANP## then I think we have
##\left| \bar{QP }\right| + \left|\bar{PR}\right| =2 \frac{\sin \alpha}{\sin \delta} \left| \bar{AN} \right| + 2 \frac{sin \alpha}{\sin \left(\alpha - \delta \right)} \left| \bar{NP} \right|##...
Ugh. I guess I'm too used to thinking about triangles where there are shared vertices. So there's no shared inner or outer circle, but concentric pairs of circles instead.
We know that there's an "outer circle" that goes through QBRC, and we know that there's an "inner circle" that is tangent to AB, AC, PQ, and PR. The "inner circle" is centered at N, and the outer circle is centered
WOLOG, we can assume that the radius of the outer circle is 1. And, let's...
Looking at this a little more, there should also be a single circle centered on N that's tangent to QP, PR, AB, and AC.
Can you construct an example of this geometry where ABC and PQR are different?
Assuming that CB is not parallel to QR, there has to be a circle that goes through all four points of those points. Maybe drawing in that circle leads somewhere.
Here's a drawing:
N is the centroid of triangle ABC.
If M is the centroid of the rectangle, then triangle ARC has more area than triangle ALC which has half the area of triangle ABC, and triangle IQC has less area than triangle IBC which has half the area of triangle ABC.
So there's some...
It was me thinking that everything can work as a nail when I've got a hammer in my hand kind of thing.
I was in a "compass and straightedge" kind of mindset, so I went looking to draw a tangent to some kind of circle that also went through the centroid of the rectangle. Then I remembered (or...
My capabilities with GeoGebra weren't quite up to making nice illustrations.
Start by constructing the centroid of the rectangle ##C_R## by crossing the diagonals of the rectangle. We know that lines divide the area of the rectangle in two if and only if they go through the centroid. So our...