In this post I’ll cover three properties of isogonal conjugates which were only recently made known to me. These properties are generalization of some well-known lemmas, such as the incenter/excenter lemma and the nine-point circle.

1. Definitions

Let $ABC$ be a triangle with incenter $I$, and let $P$ be any point in the interior of $ABC$. Then we obtain three lines $AP$, $BP$, $CP$. Then the reflections of these lines across lines $AI$, $BI$, $CI$ always concur at a point $Q$ which is called the isogonal conjugate of $P$. (The proof of this concurrence follows from readily from Trig Ceva.) When $P$ lies inside $ABC$, then $Q$ is the point for which $\angle BAP = \angle CAQ$ and so on.

The isogonal conjugate of $P$ is sometimes denoted $P^\ast$. Note that $(P^\ast)^\ast = P$.

Examples of pairs of isogonal conjugates include the following.

1. The incenter is its own isogonal conjugate. Similarly, each excenter is also its own isogonal conjugate.
2. The isogonal conjugate of the circumcenter is the orthocenter.
3. The isogonal conjugate of the centroid is the symmedian point.
4. The isogonal conjugate of the Nagel point is the point of concurrence of $AT_A$, $BT_B$, $CT_C$, where $T_A$ is the contact point of the $A$-mixtilinear incircle. The proof of this result was essentially given as Problem 5 of the European Girl’s Math Olympiad.

2. Inverses and circumcircles

You may already be aware of the famous result (which I always affectionately call “Fact 5”) that the circumcenter of $BIC$ is the midpoint of arc $BC$ of the circumcircle of $ABC$. Indeed, so is the circumcenter of triangle $BI_AC$, where $I_A$ is the $A$-excenter.

In fact, it turns out that we can generalize this result for arbitrary isogonal conjugates as follows.

Theorem 1. Let $P$ and $Q$ be isogonal conjugates. Then the circumcenters of $\triangle BPC$ and $\triangle BQC$ are inverses with respect to the circumcircle of $\triangle ABC$.

Proof: This is just angle chasing. Let $O_P$ and $O_Q$ be the desired circumcenters. It’s clear that both $O_P$ and $O_Q$ lie on the perpendicular bisector of $\overline{BC}$. Angle chasing allows us to compute that $$\angle BO_PO = \frac 12 \angle BO_PC = 180^{\circ} - \angle BPC.$$ Similarly, $\angle BO_QO = 180^{\circ} - \angle BQC$. But the reader can check that $\angle BPC + \angle BQC = 180^{\circ} + A$. Using this we can show that $\angle OBO_Q = \angle BO_PO$, so $\triangle OBO_P \sim \triangle OO_QB$, as needed. $\Box$

When we take $P$ and $Q$ to be $I$ (or $I_A$), we recover the Fact 5 we mentioned above. When we take $P$ to be the orthocenter and $Q$ to be the circumcenter, we find that the circumcenter of $BHC$ is the inverse of the circumcenter of $BOC$. But the inverse of the circumcenter of $BOC$ is the reflection of $O$ over $\overline{BC}$. Thus we derive that $\triangle BHC$ and $\triangle BOC$ have circumcircles which are just reflections over $\overline{BC}$.

3. Pedal circles

You may already be aware of the nine-point circle, which passes through the midpoints and feet of the altitudes of $ABC$. In fact, we can obtain such a circle for any pair of isogonal conjugates.

Theorem 2. Let $P$ and $Q$ be isogonal conjugates in the interior of $\triangle ABC$. The pedal triangles of $P$ and $Q$ share a circumcircle. Moreover, the center of this circle is the midpoint $M$ of $\overline{PQ}$.

Upon taking $P=H$ and $Q=O$ we recover the nine-point circle. Of course, the incircle is the special case $P=Q=I$!

Proof: Let $\triangle P_AP_BP_C$ and $\triangle Q_AQ_BQ_C$ be the pedal triangles. We leave the reader to check that $$AP_C \cdot AQ_C = AP \cdot AQ \cdot \cos \angle BAP \cdot \cos \angle BAQ = AP_B \cdot AQ_B.$$ Consequently, the points $P_C$, $Q_C$, $P_B$, $Q_B$ are concyclic. The circumcenter of these four points is the intersection of the perpendicular bisectors of segments $\overline{P_CQ_C}$ and $\overline{P_BQ_B}$, which is precisely $M$. Thus $$MP_C = MQ_C = MP_B = MQ_B.$$ Similarly work with the other vertices shows that $M$ is indeed the desired circumcenter. $\Box$

There is a second way to phrase this theorem by taking a homothety at $Q$.

Corollary. If the point $Q$ is reflected about the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$, then the resulting triangle has circumcenter $P$.

4. Ellipses

We can actually derive the following remarkable result from the above theorem.

Theorem 3. An ellipse $\mathcal E$ is inscribed in triangle $ABC$. Then the foci $P$ and $Q$ are isogonal conjugates.

Of course, the incircle is just the special case when the ellipse is a circle.

Proof: We will deduce this from the corollary. Let the ellipse be tangent at points $D$, $E$, $F$. Moreover, let the reflection of $Q$ about the sides of $\triangle ABC$ be points $X$, $Y$, $Z$. By definition, there is a common sum $s$ with $$s = PD + DQ = PE + EQ + PF + FQ.$$ Because of the tangency condition, the points $P$, $D$, $X$ are collinear. But now $$PX = PD+DX = PD+DQ = s$$ and we deduce $$PX = PY = PZ = s.$$ So $P$ is the circumcenter of $\triangle XYZ$. Hence $P$ is the isogonal conjugate of $Q$. $\Box$

The converse of this theorem is also true; given isogonal conjugates $P$ and $Q$ inside $ABC$ we can construct a suitable ellipse. Moreover, it’s worth noting that the lines $AD$, $BE$, $CF$ are also concurrent; one proof is to take a projective transformation which sends the ellipse to a circle.

Using this theorem, we can give a “morally correct” solution to the following problem, which is IMO Shortlist 2000, Problem G3.

Problem. Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that

$$OD + DH = OE + EH = OF + FH$$

and the lines $AD$, $BE$, and $CF$ are concurrent.

Proof: Because $O$ and $H$ are isogonal conjugates we can construct an ellipse tangent to the sides at $D$, $E$, $F$ from which both conditions follow. $\Box$

5. Pascal’s theorem

For more on isogonal conjugates, see e.g. Darij Grinberg. I’ll just leave off with one more nice application of isogonal conjugates, communicated to me by M Kural last August.

Theorem 4 (Pascal). Let $AEBDFC$ by a cyclic hexagon, as shown. Suppose $P = \overline{AB} \cap \overline{DE}$ $Q = \overline{CD} \cap \overline{FA}$, and $X = \overline{BC} \cap \overline{EF}$. Then points $P$, $X$, $Q$ are collinear.

Proof: Notice that $\triangle XEB \sim \triangle XCF$, though the triangles have opposite orientations. Because $\angle BEP = \angle BED = \angle BCD = \angle XCQ$, and so on, the points $P$ and $Q$ correspond to isogonal conjugates. Hence $\angle EXP = \angle QXF$, which gives the collinearity. $\Box$

Thanks to R Alweiss and heron1618 for pointing out a few typos, and Daniel Paleka for noticing a careless application of Brianchon’s theorem.