Bill Millar I have seen several three-circle theorems on CTK, but not this one. I came across it by generalising an unpublished result of Brian Stonebridge. PQR is any triangle, and A,B,C are any points on the (possibly extended) respective sides QR, RP, PQ. Then the three circles on ARB,BPC, and CQA are concurrent. My proof is very simple, though it needs six cases. I give here just the case in which all the points A,B,C lie within their segments QR, RP, PQ. Let the circles on ARB and on BPC meet again in J, which in this case lies inside the triangle. We have to show that J lies on the third circle. The angle-value CJA IS 360 degrees - (AJB + BJC). Since the points J,R are on opposite sides of the line AB, the angles AJB and ARB are supplementary. So also are BJC and BPC. Thus from the sum of the angles of triangle PQR, we find CJA and CQA are supplementary, giving the required result. The proof is similar for the other cases. Surely the theorem has cropped up before, but I would be glad to hear from anyone who can give a reference.
2. "RE: Three Concurrent Circles"
In response to message #1
Bill Millar Yes - many thanks for introducing me to the name. I tried searching on Google as well as on CTK for "three circles", but found nothing. Now with "Pivot Theorem" I have found all sorts of references. The special case you asked about was that of an equilateral triangle, with the three extra points on non-adjacent extensions of its sides.
Printer-friendly copy
Email this topic to a friend
