I know some friends who are fantastic at synthetic geometry. I can give them any problem and they’ll come up with an incredibly impressive synthetic solution. I also have some friends who are very bad at synthetic geometry, but have such good fortitude at computations that they can get away with using Cartesian coordinates for everything.

I don’t consider myself either of these types; I don’t have much ingenuity when it comes to my solutions, and I’m actually quite clumsy when it comes to long calculations. But nonetheless I have a high success rate with olympiad geometry problems. Not only that, but my solutions are often very algorithmic, in the sense that any well-trained student should be able to come up with this solution.

In this article I try to describe how I come up which such solutions.

## 1. The Three Reductions

Very roughly, there are three different ways I try to make progress on a geometry problem.

• (I) The standard synthetic techniques; angle chasing, cyclic quadrilaterals, homothety, radical axis / power of a point, etc. My own personal arsenal contains some weapons not known to many contestants as well, most notably inversion, harmonic bundles and quadrilaterals, and spiral similarity / Miquel points.For this part, it’s highly advantageous to be well-versed with “standard” configurations and tricks. To give an extreme example: to solve Iran TST 2009, Problem 9 one essentially needs only recognize two configurations: a lemma about the midpoint of an altitude (2002 G7) and another lemma about the line ${EF}$ (USAJMO 2014/6). Not knowing either of these makes it more difficult to solve the problem synthetically in the time limit. As a reference, Yufei Zhao’s lemmas handout contains a fairly comprehensive list of these configurations.

Easier problems don’t require as much in this way of configuration recognition.

• (II) Standard computational techniques (aka bashing). Personally, I prefer complex numbers and barycentric coordinates but I know other students who will use Cartesian coordinates and trigonometry to great success. The advantage of such methods is that they are straightforward and reliable, albeit tedious and time-consuming. It is mostly a matter of experience to understand whether a calculation can be carried out within the time limit — I can basically tell just by looking at a setup whether it can be solved in this time.
• (III) Most surprisingly: simply finding crucial claims. Especially for harder problems like IMO 3/6 much of the time the key to solving a problem is making some key observation. Said another way: a difficult IMO 3/6 problem which asks you to prove ${A \implies B}$ might have a solution which goes like,

$\displaystyle A \implies X \implies Y \implies B.$

Each of the individual implications might be no harder than an IMO 1/4 but the difficulty rests in finding what to prove. The most reliable way to do such things is to draw large, in-scale diagrams. If you are good at recognizing cyclic quadrilaterals, collinear points, etc. then the correct claims will naturally suggest themselves; conversely, good diagrams will prevent you from wasting time trying to prove things that aren’t true (effectively letting you test your claims “experimentally” before trying to prove them).

Type (III) deserves some comment here. There is more to making progress on a problem than simply trying things you think will solve the problem: there is some “scouting” involved that you will need to do for any difficult problems. As a terrible analogy, in StarCraft you have to scout an experienced opponent to understand what they’re doing before you try to attack them. The situation with IMO 3/6 is no different: you have to have some understanding of the problem before you stand a chance of being able to solve it.

Easy problems can often succumb to just one class of attacks, but the interesting and difficult problems can require two or all three classes in order to solve. How much you use each type of strategy is in my opinion a matter of personal taste — some people don’t use (II) at all and rely on (I) to prove everything, and even vice versa! I like to think I balance (I) and (II) evenly. But (III) is indispensable, and in any case I think part of the reason I have been so successful with geometry problems is precisely that I can draw on all three strategies in tandem, rather than being limited to one or two.

In fact, a good rule of thumb that I use for judging the difficulty of a problem is how many of the above methods I had to use: the ${n}$th problem on an IMO paper should require me to resort to about ${n}$ of these strategies.

## 2. Concrete Examples

I’ll now give some concrete examples of the things I said above. Warning: spoilers follow, and hyperlinks lead to my solutions on Art of Problem Solving. You are encouraged to try the problems yourself before reading the comments.

Example [EGMO 2012/1] Let ${ABC}$ be a triangle with circumcenter ${O}$. The points ${D}$, ${E}$, ${F}$ lie in the interiors of the sides ${BC}$, ${CA}$, ${AB}$ respectively, such that ${\overline{DE}}$ is perpendicular to ${\overline{CO}}$ and ${\overline{DF}}$ is perpendicular to ${\overline{BO}}$. Let ${K}$ be the circumcenter of triangle ${AFE}$. Prove that the lines ${\overline{DK}}$ and ${\overline{BC}}$ are perpendicular.

This is a pretty typical entry-level geometry problem. Do some angle chasing (I) to find one cyclic quad (III), and then follow through to solve the problem (I). If you are good enough, you don’t even need to find the cyclic quad in advance; just play around with the angles until you notice it.

Example [IMO 2014, Problem 4] Let ${P}$ and ${Q}$ be on segment ${BC}$ of an acute triangle ${ABC}$ such that ${\angle PAB=\angle BCA}$ and ${\angle CAQ=\angle ABC}$. Let ${M}$ and ${N}$ be the points on ${AP}$ and ${AQ}$, respectively, such that ${P}$ is the midpoint of ${AM}$ and ${Q}$ is the midpoint of ${AN}$. Prove that the intersection of ${BM}$ and ${CN}$ is on the circumference of triangle ${ABC}$.

You can solve this problem by barycentric coordinates (II) instantly (textbook example). Also similar triangles (I) solves the problem pretty quickly as well. Again, this problem is “easy” in the sense that one can directly approach it with either (I) or (II), not needing (III) at all.

Example [USAMO 2015/2] Quadrilateral ${APBQ}$ is inscribed in circle ${\omega}$ with ${\angle P = \angle Q = 90^{\circ}}$ and ${AP = AQ < BP}$. Let ${X}$ be a variable point on segment ${\overline{PQ}}$. Line ${AX}$ meets ${\omega}$ again at ${S}$ (other than ${A}$). Point ${T}$ lies on arc ${AQB}$ of ${\omega}$ such that ${\overline{XT}}$ is perpendicular to ${\overline{AX}}$. Let ${M}$ denote the midpoint of chord ${\overline{ST}}$. As ${X}$ varies on segment ${\overline{PQ}}$, show that ${M}$ moves along a circle.

This was not supposed to be a very difficult problem, but it seems to have nearly swept the JMO group. Essentially, the key to this problem is to notice that the center of the desired circle is in fact the midpoint of ${AO}$ (with ${O}$ the center of the circle). This is a huge example of (III) — after this observation, one can solve the problem very quickly using complex numbers (II). It is much harder (though not impossible) to solve the problem without knowing the desired center.

Example [USAMO 2014/5] Let ${ABC}$ be a triangle with orthocenter ${H}$ and let ${P}$ be the second intersection of the circumcircle of triangle ${AHC}$ with the internal bisector of the angle ${\angle BAC}$. Let ${X}$ be the circumcenter of triangle ${APB}$ and ${Y}$ the orthocenter of triangle ${APC}$. Prove that the length of segment ${XY}$ is equal to the circumradius of triangle ${ABC}$.

Personally I think the most straightforward solution is to use (I) to eliminate the orthocenter condition, and then finish with complex numbers (II). Normally, you won’t see a medium-level problem that dies immediately to (II), and the only reason a problem like this could end up as a problem 5 is that there is a tiny bit of (I) that needs to happen before the complex numbers becomes feasible.

Example [IMO 2014/3] Convex quadrilateral ${ABCD}$ has ${\angle ABC = \angle CDA = 90^{\circ}}$. Point ${H}$ is the foot of the perpendicular from ${A}$ to ${BD}$. Points ${S}$ and ${T}$ lie on sides ${AB}$ and ${AD}$, respectively, such that ${H}$ lies inside triangle ${SCT}$ and

$\displaystyle \angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}.$

Prove that line ${BD}$ is tangent to the circumcircle of triangle ${TSH}$.

Like most IMO 3/6’s I had to resort to using all three methods in order to solve this problem. The first important step was finding out what to do with the angle condition. It turns out that in fact, it’s equivalent to the circumcenter of triangle ${TCH}$ lying on side ${AD}$ of the triangle (III); proving this is then a matter of angle chasing (I). Afterwards, one has to recognize a tricky usage of the angle bisector theorem (I) to reduce it to something that can be computed with trigonometry (II). This leads to a direct solution that, while not elegant, also requires much less ingenuity then most of the solutions found by friends I know.

I really want to stress that being proficient in all three strategies is key to getting “straightforward” solutions like this to IMO 3/6 caliber problems. If you miss any of these components, you are not going to solve the problem.

Example [IMO 2011/6] Let ${ABC}$ be an acute triangle with circumcircle ${\Gamma}$. Let ${\ell}$ be a tangent line to ${\Gamma}$, and let ${\ell_a}$, ${\ell_b}$, ${\ell_c}$ be the lines obtained by reflecting ${\ell}$ in the lines ${BC}$, ${CA}$, and ${AB}$, respectively. Show that the circumcircle of the triangle determined by the lines ${\ell_a}$, ${\ell_b}$, and ${\ell_c}$ is tangent to the circle ${\Gamma}$.

The ultimate example of these three principles. Using a trick that showed up on APMO 2014/5 and RMM 2013/3, one constructs the tangency point ${T}$ and connects the points ${A_1}$, ${B_1}$, ${C_1}$, as I explain in this post, yielding points ${A_2}$, ${B_2}$, ${C_2}$. After that, a very careful examination of the diagram (possibly several diagrams) leads to a conjecture that ${A_1A=AP}$, et cetera. This is the key observation (III), and leads to highly direct solution via (II). But the point of this problem is that you need to have the guts to construct those auxiliary points and then boldly claim they are the desired “squared” points.

## 3. Comparison with Other Subjects

The approaches I’ve described highlight some of the features of olympiad geometry which distinguish it from other subjects.

• Unlike other olympiad subjects, you can actually obtain a big advantage by just knowing lots of theory. Experienced contestants simply “recognize” a large body of common configurations that those without access to training materials have never seen before. Similarly, there are a lot of fancy techniques that can make a big difference. This is much less true of other subjects (for example combinatorics is the opposite extreme).
• There’s less variance in the subject: lots of Euclidean geometry problems feel the same, and all of them use the same body of techniques. It reminds me of chess: it’s very “narrow” in the sense that at the end of the day, there are only so many possible moves. (Olympiad inequalities also has this kind of behavior.) Again combinatorics is the opposite of this.
• You have a reliable backup in case you can’t find the official solution: bash. Moreover, in general there are often many different ways to solve a problem; not true of other subjects.
• If you want to make some “critical claim” you can quickly test it empirically (by drawing a good diagram).

You can use a wide range of wild, cultivated or supermarket greens in this recipe. Consider nettles, beet tops, turnip tops, spinach, or watercress in place of chard. The combination is also up to you so choose the ones you like most.

— Y. Ottolenghi. Plenty More

In this post I’ll describe how I come up with geometry proposals for olympiad-style contests. In particular, I’ll go into detail about how I created the following two problems, which were the first olympiad problems which I got onto a contest. Note that I don’t claim this is the only way to write such problems, it just happens to be the approach I use, and has consistently gotten me reasonably good results.

[USA December TST for 56th IMO] Let ${ABC}$ be a triangle with incenter ${I}$ whose incircle is tangent to ${\overline{BC}}$, ${\overline{CA}}$, ${\overline{AB}}$ at ${D}$, ${E}$, ${F}$, respectively. Denote by ${M}$ the midpoint of ${\overline{BC}}$ and let ${P}$ be a point in the interior of ${\triangle ABC}$ so that ${MD = MP}$ and ${\angle PAB = \angle PAC}$. Let ${Q}$ be a point on the incircle such that ${\angle AQD = 90^{\circ}}$. Prove that either ${\angle PQE = 90^{\circ}}$ or ${\angle PQF = 90^{\circ}}$.

[Taiwan TST Quiz for 56th IMO] In scalene triangle ${ABC}$ with incenter ${I}$, the incircle is tangent to sides ${CA}$ and ${AB}$ at points ${E}$ and ${F}$. The tangents to the circumcircle of ${\triangle AEF}$ at ${E}$ and ${F}$ meet at ${S}$. Lines ${EF}$ and ${BC}$ intersect at ${T}$. Prove that the circle with diameter ${ST}$ is orthogonal to the nine-point circle of ${\triangle BIC}$.

## 1. General Procedure

Here are the main ingredients you’ll need.

• The ability to consistently solve medium to hard olympiad geometry problems. The intuition you have from being a contestant proves valuable when you go about looking for things.
• In particular, a good eye: in an accurate diagram, you should be able to notice if three points look collinear or if four points are concyclic, and so on. Fortunately, this is something you’ll hopefully have just from having done enough olympiad problems.
• Geogebra, or some other software that will let you quickly draw and edit diagrams.

With that in mind, here’s the gist of what you do.

1. Start with a configuration of your choice; something that has a bit of nontrivial structure in it, and add something more to it. For example, you might draw a triangle with its incircle and then add in the excircle tangency point, and the circle centered at ${BC}$ passing through both points (taking advantage of the fact that the two tangency points are equidistant from ${B}$ and ${C}$).
2. Start playing around, adding in points and so on to see if anything interesting happens. You might be guided by some actual geometry constructions: for example, if you know that the starting configuration has a harmonic bundle in it, you might project this bundle to obtain the new points to play with.
3. Keep going with this until you find something unexpected: three points are collinear, four points are cyclic, or so on. Perturb the diagram to make sure your conjecture looks like it’s true in all cases.
4. Figure out why this coincidence happened. This will probably add more points to you figure, since you often need to construct more auxiliary points to prove the conjecture that you have found.
5. Repeat the previous two steps to your satisfaction.
6. Once you are happy with what you have, you have a nontrivial statement and probably several things that are equivalent to it. Pick the one that is most elegant (or hardest), and erase auxiliary points you added that are not needed for the problem statement.
7. Look for other ways to reduce the number of points even further, by finding other equivalent formulations that have fewer points.

Or shorter yet: build up, then tear down.

None of this makes sense written this abstractly, so now let me walk you through the two problems I wrote.

## 2. The December TST Problem

In this narrative, the point names might be a little strange at first, because (to make the story follow-able) I used the point names that ended up in the final problem, rather than ones I initially gave. Please bear with me!

I began by drawing a triangle ${ABC}$ (always a good start\dots) and its incircle, tangent to side ${BC}$ at ${D}$. Then, I added in the excircle touch point ${T}$, and drew in the circle with diameter ${DT}$, which was centered at the midpoint ${M}$. This was a coy way of using the fact that ${MD = MT}$; I wanted to see whether it would give me anything interesting.

So, I now had the following picture.

Now I had two circles intersecting at a single point ${D}$, so I added in ${Q}$, the second intersection. But really, this point ${Q}$ can be thought of another way. If we let ${DS}$ be the diameter of the incircle, then as ${DT}$ is the other diameter, ${Q}$ is actually just the foot of the altitude from ${D}$ to line ${ST}$.

But recall that ${A}$, ${S}$, ${T}$ are collinear! (Again, this is why it’s helpful to be familiar with “standard” contest configurations; you see these kind of things immediately.) So ${Q}$ in fact lies on line ${AT}$.

This was pretty cool, though not yet interesting enough to be a contest problem. So I looked for most things that might be true.

I don’t remember what I tried next; it didn’t do anything interesting. But I do remember the thing I tried after that: I drew in the angle bisector, line ${AI}$. And then, I noticed a big coincidence: the first intersection of ${AI}$ with the circle with diameter ${DT}$ seemed to lie on line ${DE}$! I was initially confused by this; it didn’t seem like it could possibly be true due to symmetry reasons. But in my diagram, it was indeed correct. A moment later, I realized the reason why this was plausible: in fact, the second intersection of line ${AI}$ with the circle was on line ${DF}$.

Now, I could not see quickly at all why this was true. So I started trying to prove it, but initially failed: however, I managed to show (via angle chasing) that

$\displaystyle D, P, E \text{ collinear} \iff \angle PQE = 90^\circ.$

So, at least I had an interesting equivalent statement.

After another half hour of trying to prove my conjecture, I finally realized what was happening. The point ${P}$ was the one attached to a particular lemma: the ${A}$-bisector, ${B}$-midline, and ${C}$ touch-chord are concurrent, and from this ${MD = MP}$ just follows by some similar triangles. So, drawing in the point ${N}$ (the midpoint of ${AB}$), I had the full configuration which gave the answer to my conjecture.

Finally, I had to clean up the mess that I had made. How could I do this? Well, the points ${N}$, ${S}$ could be eliminated easily enough. And we could re-define ${Q}$ to be a point on the incircle such that ${\angle AQD = 90^\circ}$. This actually eliminated the green circle and point ${T}$ altogether, provided we defined ${P}$ by just saying that it was on the angle bisector, and that ${MD = MP}$. (So while the circle was still implicit in the condition ${MD = MP}$, it was no longer explicitly part of the problem.)

Finally, we could even remove the line through ${D}$, ${P}$ and ${E}$; we ask the contestant to prove ${\angle PQE = 90^\circ}$.

And that was it!

## 3. The Taiwan TST Problem

In fact, the starting point of this problem was the same lemma which provided the key to the previous solution: the circle with diameter ${BC}$ intersects the ${B}$ and ${C}$ bisectors on the ${A}$ touch chord. Thus, we had the following diagram.

The main idea I had was to look at the points ${D}$, ${X}$, ${Y}$ in conjunction with each other. Specifically, this was the orthic triangle of ${\triangle BIC}$, a situation which I had remembered from working on Iran TST 2009, Problem 9. So, I decided to see what would happen if I drew in the nine-point circle of ${\triangle BIC}$. Naturally, this induces the midpoint ${M}$ of ${BC}$.

At this point, notice (or recall!) that line ${AM}$ is concurrent with lines ${DI}$ and ${EF}$.

So the nine-point circle of the problem is very tied down to the triangle ${BIC}$. Now, since I was in the mood for something projective, I constructed the point ${T}$, the intersection of lines ${EF}$ and ${BC}$. In fact, what I was trying to do was take perspectivity through ${I}$. From this we actually deduce that ${(T,K;X,Y)}$ is a harmonic bundle.

Now, what could I do with this picture? I played around looking for some coincidences, but none immediately presented themselves. But I was enticed by the point ${T}$, which was somehow related to the cyclic complete quadrilateral ${XYMD}$. So, I went ahead and constructed the pole of ${T}$ to the nine-point circle, letting it hit line ${BC}$ at ${L}$. This was aimed at “completing” the picture of a cyclic quadrilateral and the pole of an intersection of two sides. In particular, ${(T,L;D,M)}$ was harmonic too.

I spent a long time thinking about how I could make this into a problem. I unfortunately don’t remember exactly what things I tried, other than the fact that I was taking a lot of perspectivity. In particular, the “busiest” point in the picture is ${K}$, so it makes sense to try and take perspectives through it. Especially enticing was the harmonic bundle

$\displaystyle \left( \overline{KT}, \overline{KL}; \overline{KD}, \overline{KM} \right) = -1.$

How could I use this to get a nice result?

Finally about half an hour I got the right idea. We could take this bundle and intersect it with the ray ${AI}$! Now, letting ${N}$ be the midpoint ${EF}$, we find that three of the points in the harmonic bundle we obtain are ${A}$, ${I}$, and ${N}$; let ${S}$ be the fourth point, which is the intersection of line ${KL}$ with ${AI}$. Then by hypothesis, we ought to have ${(A,I;N,S) = -1}$. But from this we know exactly what the point ${S}$. Just look at the circumcircle of triangle ${AEF}$: as this has diameter ${AI}$, we see that ${S}$ is the intersection of the tangents at ${E}$ and ${F}$.

Consequently, we know that the point ${S}$, defined very naturally in terms of the original picture, lies on the polar of ${T}$ to the nine-point circle. By simply asking the contestant to prove this, we thus eliminate all the points ${K}$, ${M}$, ${D}$, ${N}$, ${I}$, ${X}$, and ${Y}$ completely from the picture, leaving only the nine-point circle. Finally, instead of directly asking the contestant to show that ${T}$ lies on the polar of ${S}$, one can rephrase the problem as saying “the circle with diameter ${ST}$ is orthogonal to the nine-point circle of ${\triangle BIC}$”, concealing all the work that went into the creation of the problem.

Fantastic.

# The Mixtilinear Incircle

This blog post corresponds to my newest olympiad handout on mixtilinear incircles.

My favorite circle associated to a triangle is the ${A}$-mixtilinear incircle. While it rarely shows up on olympiads, it is one of the richest configurations I have seen, with many unexpected coincidences showing up, and I would be overjoyed if they become fashionable within the coming years.

Here’s the picture:

The points ${D}$ and ${E}$ are the contact points of the incircle and ${A}$-excircle on the side ${BC}$. Points ${M_A}$, ${M_B}$, ${M_C}$ are the midpoints of the arcs.

As a challenge to my recent USAMO class (I taught at A* Summer Camp this year), I asked them to find as many “coincidences” in the picture as I could (just to illustrate the richness of the configuration). I invite you to do the same with the picture above.

The results of this exercise were somewhat surprising. Firstly, I found out that students without significant olympiad experience can’t “see” cyclic quadrilaterals in a picture. Through lots of training I’ve gained the ability to notice, with some accuracy, when four points in a diagram are concyclic. This has taken me a long way both in setting problems and solving them. (Aside: I wonder if it might be possible to train this skill by e.g. designing an “eyeballing” game with real olympiad problems. I would totally like to make this happen.)

The other two things that happened: one, I discovered one new property while preparing the handout, and two, a student found yet another property which I hadn’t known to be true before. In any case, I ended up covering the board in plenty of ink.

Here’s the list of properties I have.

1. First, the classic: by Pascal’s Theorem on ${TM_CCABM_B}$, we find that points ${B_1}$, ${I}$, ${C}$ are collinear; hence the contact chord of the ${A}$-mixtilinear incircle passes through the incenter. The special case of this problem with ${AB = AC}$ appeared in IMO 1978.
• Then, by Pascal on ${BCM_CTM_AA}$, we discover that lines ${BC}$, ${B_1C_1}$, and ${TM_A}$ are also concurrent.
• This also lets us establish (by angle chasing) that ${BB_1IT}$ and ${CC_1IT}$ are concyclic. In addition, lines ${BM_B}$ and ${CM_C}$ are tangents to these circumcircles at ${I}$ (again by angle chasing).
2. An Iran 2002 problem asks to show that ray ${TI}$ passes through the point diametrically opposite ${M_A}$ on the circumcircle. This is solved by noticing that ${TA}$ is a symmedian of the triangle ${TB_1C_1}$ and (by the previous fact) that ${TI}$ is a median. This is the key lemma in Taiwan TST 2014, Problem 3, which is one of my favorite problems (a nice result by Cosmin Pohoatza).
3. Lines ${AT}$ and ${AE}$ are isogonal. This was essentially EGMO 2012, Problem 5, and the “morally correct” solution is to do an inversion at ${A}$ followed by a reflection along the ${\angle A}$-bisector (sometimes we call this a “${\sqrt{bc}}$ inversion”).
• As a consequence of this, one can also show that lines ${TA}$ and ${TD}$ are isogonal (with respect to ${\angle BTC}$).
• One can also deduce from this that the circumcircle of ${\triangle TDM_A}$ passes through the intersection of ${BC}$ and ${AM_A}$.
4. Lines ${AD}$ and ${TM_A}$ meet on the mixtilinear incircle. (Homothety!)
5. Moreover, line ${AT}$ passes through the exsimilicenter of the incircle and circumcircle, by, say Monge d’Alembert. Said another way, the mentioned exsimilicenter is the isogonal conjugate of the Nagel point.

To put that all into one picture:

# Three Properties of Isogonal Conjugates

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

$\displaystyle \angle BO_PO = \frac 12 \angle BO_PC = 180^{\circ} - \angle BPC.$

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

$\displaystyle AP_C \cdot AQ_C = AP \cdot AQ \cdot \cos \angle BAP \cdot \cos \angle BAQ = AP_B \cdot AQ_B.$

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

$\displaystyle MP_C = MQ_C = MP_B = MQ_B.$

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

$\displaystyle s = PD + DQ = PE + EQ + PF + FQ.$

Because of the tangency condition, the points ${P}$, ${D}$, ${X}$ are collinear. But now

$\displaystyle PX = PD+DX = PD+DQ = s$

and we deduce

$\displaystyle PX = PY = PZ = s.$

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

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

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.