Intersection of a circle centered on a cevian

Posted on by Lasitha Jayasinghe

Today we shall have a look at a generalized configuration, with certain noteworthy special cases. Problems from this configuration seem to not yet have appeared in olympiads, with the only problem I am aware of being due to Ethan Wang. Also, shoutouts to Yanqing Yang for several key ideas and claims.

Since most points are unfamiliar, we shall first take a moment to define all points used in the above diagram (the following title is a drop-down list).

Definitions

Let \omega be a fixed arbitrary circle passing through B and C with center O.
Let T denote the point on segment AO and \omega.
Let D denote the intersection of the internal \angle BAC-bisector with BC.
Let X and Y denote the intersections of the internal \angle TAB-bisector and \angle TAC-bisector with segments BT and CT respectively.
Let R denote the intersection of lines \overline{XY} and \overline{BC}.
Let E and F denote the intersections of lines \overline{BT} and \overline{DY} and lines \overline{CT} and \overline{DX} respectively.
Let P be the foot of the altitude from T to BC and N is the midpoint of segment DR.

We shall follow the point definitions in the above diagram, to which a clickable GeoGebra link is attached.

A summary of the key results pertaining to this configuration,

  1. First, an easy lengths calculation (for example Menelaus’s theorem) shows that lines \overline{BY} , \overline{CX} and \overline{DT} concur. From the Ceva/Menelaus picture it then follows that (RD;BC)=-1. This also implies that \angle RAD = 90^\circ.
  2. Projecting through X and Y it follows that (CT;YF)=-1 and (BT;XE)=-1 respectively.
    • Now, applying Prism lemma we have that lines \overline{BC}, \overline{XY} and \overline{EF} concur at R.
  3. An easy angle chase shows that lines \overline{AT} and \overline{AD} are isogonal with respect to \angle XAY.
    • Then, Desargues Dual Involution theorem at A on quadrilateral TXDY implies that the pairs (AT,AD); (AX,AY) and (AE,AF) form an involution. The former two imply that this involution is reflection across the internal \angle XAY-bisector, so it follows that AE and AF are isogonal with respect to \angle XAY.
    • An alternate proof of this isogonality is to note that since (BT;XE)=-1 and \overline{BX} is the internal \angle BAT-bisector, we must have \angle EAX = 90^\circ and similarly, \angle FAY= 90^\circ.
  4. Note that since (BT;XE)=-1, it follows that X lies on the polar of E with respect to \omega. The self-polar orthogonality lemma now implies that the circle with diameter EX is orthogonal to \omega, and similarly the circle with diameter FY is orthogonal to \omega.
    • As a result, inversion about \omega keeps both circles (EX) and (FY) fixed, implying that their radical axis contains O.
    • Moreover, we know from the previous bullet point that A lies on this radical axis as well, so \overline{AO} is the radical axis of circles (EX) and (FY).
    • Since T also lies on this line, it follows that TE\cdot TX = TF \cdot TY and thus EFXY must be cyclic. In particular, this means that \angle TXO' = \angle TYO'.
  5. With the claim that EFXY is cyclic, we can invoke well-known results on the complete quadrilateral EFXYDR.
    • For example, this means that \overline{AO} is the Aubert line of complete quadrilateral EFXYDR, so the orthocenters of \triangle DXY and \triangle DEF must both lie on the line \overline{AO}.
    • Further, the Miquel point of quadrilateral EFXY is the foot P from T to DR, which implies that P lies on all of the circles (DXY), (DEF), (RXF) and (RYE).
    • Also, it is well known that \text{Pow}_{(EFXY)}(N)=ND^2 = \text{Pow}_{(ABC)}(N), so N lies on the radical axis of circles (ABC) and (EFXY).
  6. The special cases when O' coincides with the midpoint of segment BC and minor/major BC arc midpoints of (ABC) are noteworthy.
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Published by Lasitha Jayasinghe

3 thoughts on “Intersection of a circle centered on a cevian”

  3. I’m surprised that math oly is like 60 years old already and still has a config that has not appeared lol! How do u even find those so frequently!

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