Equilateral ΔABC inscribed in a circle. 2nd equilateral ΔPQR such that P is on segment AB and Q & R are on arc AB. Locus of center of ΔPQR?— Matt Enlow (@CmonMattTHINK) November 17, 2016
My scribblesI will admit that, initially, I had no good ideas about how to approach this puzzle. Intuitively, I was drawn to the idea of specifying the point P on the chord AB, building the equilateral triangle PQR, and then finding the center of that triangle. In the spirit of doing something, I decided to set up a coordinate system and think about the resulting equations:
Now, these are a mess and didn't really go anywhere, but, I had two thoughts as I was writing these out:
- Since we are just intersecting lines and circles, we are probably looking for a conic section
- The fact that triangle ABC is equilateral gives us a relationship between the side length and the radius of the largest circle, but I doubt whether it is critical for this problem. My suspicion is that we will get something similar for any chord.
Now, I'd had the suspicion that the coordinates for the center of an equilateral triangle was the average of the coordinates of the vertices, so I did a little work to confirm that:
At this point, I felt that I wasn't really making progress. In particular, I felt that I didn't really have the right ideas about how to find Q and R, given P.
I turned to geogebra to build the picture. Given Matt's request not to send screenshots, I felt that this was a bit of a cheat, but anyway...
I still didn't have a good idea about how to find Q and R, so I first built a worksheet where Q was a free point, from which R would be constructed so that the triangle PQR is equilateral. Then, I played around with Q until it and R sat on the original circle.
After doing this, I had two observations/conjectures. Taking C as the center of the original circle:
- we know the angle CPQ is 150 degrees
- we know that the line CP is the perpendicular bisector of segment QR
Neither is really tricky, so I was a bit chagrined to have missed them earlier, particularly the second point.
With that I now had a method to (a) draw more realistic diagrams and (b) reduce my Geogebra sheet so that only P's location on AB was an open degree of freedom. Here's a link to that Geogebra work.