Miquel Point of the Angle Bisectors

This article will explore a relatively new configuration which I have talked about extensively in my handout Neat Config which was crafted from a series of notes made by Om Kutte for lectures conducted by him at the (Unofficial) Indian National Mathematical Olympiad Training Camp.

This is the master diagram. All the point definitions (and most of the key claims) are shown in this picture. However, in the case of an unclear or confusing point simply refer the attached handout.

Here is a summary of the key properties that we discovered.

First, lines $AL$ , $EF$ and $BC$ concur at $Z$ (radical center). Further, Miquel properties tell us that $ZSFB$ and $ZSEC$ are cyclic. Also a straightforward angle chase then shows that $R$ lies on $(AZS)$ and the tangent to $(AEF)$ at $A$ .
Next, by Pascal’s Theorem on the intersection of the tangent to at and , lies on line . Since is harmonic (project through I!) it follows that also lies on the tangent to at . This point plays a great vouge in mixtillinear incircle problems. Particularly,
- Lines $BC$ and $QI$ are parallel. This is essentially 2006 Russia Regional Grade 11/4 and this parallelism has been explored in mixtillinear contexts in many other sources.
- Point $Q$ lies on circle $(ASR)$ while points $Q$ , $L'$ and $S$ are collinear.
- Line $RQ$ is the perpendicular bisector of segment $AI$ , which implies Brazil Olympic Revenge 2017/2.
Line $AW$ and circle $(ARS)$ intersect a second time on the line $QI$ from which it follows that $RH'$ is the second tangent from $R$ to $(AEF)$ . This statement is equivalent to Israel TST 8 2022/3 which is one of the few truly non-trivial problems that have been composed in relation to this configuration.
Lines $WI$ and $LD'$ are parallel. This property generalizes when $I$ is replaced by an arbitrary point inside the triangle and can be easily proven via homothety arguments.
Using the converse of Pascal’s Theorem on allows us to show that points , , , and lie on a conic which encapsulates a lot of weird concurrency claims that can be seen. Applying a more general result on quadrilateral circumconics passing through the Miquel Point (by yours truly!) indicates that point also lies on this conic.
- Now we can apply Pascal’s Theorem on coconic hexagon $CFEWW'S$ to observe that points $W'$ , $W$ and $I$ are collinear (this may require the previous claim).
Another set of coconic points surface as $A$ , $B$ , $C$ , $I$ , $S$ and $Q'$ turn out to lie on the same conic.
The intersection of ray with circle seems to be the most promising point in this picture. Rushing during the last days of the handout we left this point relatively unexplored.
- Both quadrilaterals $AGTI$ and $SRTI$ are cyclic. Utilizing this a simple angle chase leads to the claim that points $G$ , $T$ and $M$ lie on the same line.
- Also ray $SI$ intersects $(ABC)$ on the line through $T$ parallel to side $BC$ .
Line also pops up in other more well-known configurations. It is most effective to apply Radical Center revolving around these points.
- Points $I_b$ , $I_c$ , $I$ , $E_1$ and $F_1$ lie on the same circle since $EF$ is the radical axis of circles $(I_bII_c)$ and $(ABC)$ .
- Theorem 2.7.2 on i3435’s Muricaaa states that lines $I_aO$ and $EF$ are perpendicular to each other.
The angle bisector Miquel point also has connections with the symmedian and median. With some simple angle chasing and a sprinkle of harmonic bundles one can show several claims.
- Since we know that $ABYC$ is harmonic it turns out that ray $SP$ hits $(AEF)$ again at $M'$ .
- Points $A$ , $S$ , $P$ and $K$ are concyclic.
Points $P$ , $I$ and $M_a$ are collinear. More generally, for any pair of points $P$ and $P'$ on $EF$ and $BC$ such that $AP$ and $AP'$ are isogonal, points $P$ , $I$ and $P'$ are collinear. The “morally correct” way to do this is animating $P$ on side $BC$ , although rather involved solutions via DDIT and lengths exist.
The reflection of $P$ across the $\angle A-$ bisector lies on the circle $(LIX)$ which is the pith of 2020 Iran TST 2/3 which disguises this result underneath a pile of reflections cleverly.