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As promised by me in the shout box, here's the second part to the intro of these beautiful points (I guess a little late ). I'll try to cover some nice and hidden applications of this configuration, touching upon a variety of problems (hopefully ). So, without further ado, let's start this 'safar' (I have tried to arrange the problems in order of difficulty):-

PROBLEM-1 (APMO 2012 Problem 4)
Let be an acute triangle. Denote by the foot of the perpendicular line drawn from the point to the side , by the midpoint of , and by the orthocenter of . Let be the point of intersection of the circumcircle of the triangle and the half line , and be the point of intersection (other than ) of the line and the circle . Prove that must hold.

Solution: By Property 1, we know that is the -Queue point of (rename it as from now on). Seeing this point, we add in the -Ex point . Also note that we are supposed to show that is harmonic, for which we will require some sort of perspectivity. As all points are closely linked to the point , so it is our first and foremost choice for perspector. Using the fact that is harmonic (I guess this has been given in the proof of Property-3), and with the help of Property-1, we get that
REMARK: Remember this problem as a property itself. Another similar problem is Brazil MO 2011 Problem 5. The details of the proof are left to the reader as an exercise (If someone wants, they can post their proof in the comments section later).

PROBLEM-2 (Vietnam MO 2019 Problem 6)
Given an acute triangle and be its circumcircle, and is its orthocenter. Let be midpoints of , respectively. are the feet of the altitudes from and , respectively. Let symmetry with through . intersects at , intersects at .
a) intersects smaller arc of at . Prove that are concyclic.
b) intersect at (), respectively. Prove that are concurent.

Solution: As there are too many points on a single circle (namely the nine-point circle), and the intersections of some lines formed by them is concerned, so we resort to Pascal. Trying to form the desired intersections, we apply Pascal to , to find that are in fact collinear. At first glance, the given conditions do not give any more significant information (atleast not to me ). So we look at what we are supposed to prove. As the point must be fixed, we get interested in finding that point first. Suppose we assume that what we are supposed to prove is true (Remember this trick). Then by Radical Axes Theorem on , we get that passes through the -Ex point of . Wow! That's interesting. But again we are stuck. At this point of time, I always try to incorporate the fact that is harmonic. Taking perspectivity from , we get that So we arrive at the most crucial part of the solution. We just need to show that the line is the -symmedian, in which case the above harmonic bundles (and some use of Power of Point in place of Radical Axes Theorem) will do the job.

Now, the main question is how to show that is the -symmedian. Because of the weird definition of , there's not much one can do apart from applying Pascal's Theorem again. Trying to find some nice hexagons involving , we stumble upon . This is nice, as we know that , and so we get that . Now, it is some manipulations before finding a way out. What I did was take notice of the fact that lies on , and so the reflection of in (say ) must lie on . This is quite helpful, because we also know that is antiparallel wrt (as ). But, bisects , which directly gives that it must be the -symmedian (this is quite well known I guess). Thus, we are done with Part (a).

For Part (b), we again use our friend this evening, Pascal (the motivation comes from the number of points involved on the circumcircle of ). This is easy, and with some trial and error, one easily gets the two desired hexagons and , after which one can finish using the fact that are concurrent.

REMARK: This solution illustrates the power of Pascal. Most of the students are aware of it, but still its true nature is never used to the fullest. You can find my original solution here.

So let's move on to some TST problems (due to time constraint, I'll only be able to take up 3 problems; readers can include more problems in the comments section if they wish to ).

PROBLEM-3 (Bosnia and Herzegovina 2018 TST Day 1 Problem 1)
In acute triangle let , and be foots of perpedicular from , and to , and , respectively. Let and be points on line such that and . Prove that

Solution: Let be the midpoint of (we introduce to get a link to ). Then is cyclic. Our solution is motivated by this well-known lemma and its converse (Lemma 9.17 in EGMO). Without knowing that lemma, it might be difficult to produce this proof (Basically I am just telling you to learn it ASAP ). According to this lemma, if denotes the -Ex point of , then (as is harmonic). But, , so we get . Combining these two equalities, we see that is cyclic. As , so we get that in fact bisects externally. This is the crucial part of the solution. After this it is just some angle chasing (unfortunately there's not much motivation for that; you just try to find nice angles). The full proof can be found here.

REMARK: All the above solutions show how important the harmonic bundle is. Always try to use this.

PROBLEM-4 (USA TSTST 2016 Problem 2)
Let be a scalene triangle with orthocenter and circumcenter . Denote by , the midpoints of , . Suppose the circle with diameter meets the circumcircle of at , and meets line at a point . The tangent to at meets line at . Show that the circumcircles of and intersect at a point on .

Solution: Let's sort this information first (This is really important to all geo problems). By its definition, is the -Queue point, and is the -Humpty point. Also, lies on , which is nothing but the perpendicular bisector of . So , which means that is also tangent to . So, in quadrilateral , and (all tangents taken wrt ). Applying Pascal on and , we get that are collinear (this is true for all cyclic quadrilaterals; learn this result as a fact as it is quite useful). Now, these two new intersection points move us closer towards a complete quadrilateral. So let's "complete" this complete quadrilateral configuration (pun intended ). We add in the -Ex point . As discussed in Problem-3 of Part One, this should motivate us to look for Brokard's Theorem. Then using Brokard (and the collinearity discussed earlier), we get that is the polar of wrt . So we have removed point , and now it suffices to show that lies on the polar of .

Let's look at the two circles now. Using Properties-1,2 and 3, we know that , which means that . Suppose what we are supposed to prove is true, i.e. is the polar of wrt . But, (as ). So we get that must in fact be the inverse of wrt . As the original direction is a bit difficult to prove, we will try to use phantom points. Let be the inverse of wrt . As , so . Thus it suffices to show that , which is equivalent to proving that (since are collinear). However, is nothing but the power of wrt (as is the inverse of in ). Thus,
REMARK: To be honest, this problem doesn't require any remarks . A "complete" solution can be seen here.

PROBLEM-5 (2015 Taiwan TST Round 3 Quiz 1 Problem 2)
Let be the circumcircle of the triangle . Two circles are tangent to each of the circle and the rays , with interior to , exterior to . The common tangent of and the common tangent of intersect at the point . Let be the midpoint of the arc (not containing the point ) on the circle , and the segment be the diameter of . Prove that , and are collinear.

Solution: This problem is one of my favourite problems, with a short, yet beautiful solution. Seeing the involvement of both the -mixtilinear incircle and excircle, we perform inversion, to get the following equivalent problem:- We are supposed to show that is cyclic; or equivalently that . But, applying Property-3 to (The only major difficulty in this problem), this is only possible if is the -Ex point of . As is also the foot of -altitude of , so it suffices to prove that is harmonic (See the Remark given after Problem-3 above). Now, we again turn our attention back to . Then one can easily finish as given in my solution here.

REMARK: After reading the solution above, one might be compelled to think that this problem is not that hard. However, according to me, the main difficulty one faces in this solution is to quickly turn one's attention from one triangle to another. This is really important, and so I'll again bring the reader's attention to this point. Basically, do not be constrained by the reference triangle given in the problem. Try exploring other triangles. Often hard problems are simply restatements of well known facts wrt some other triangle. Two triangles which easily interchange amongst themselves are the excentral triangle and the orthic triangle. I prefer working with the orthocenter configuration, but your preference might be something else. A nice example of this is USAMO 2017 P3.

With this, I come to an end to this section. Hopefully, you guys enjoyed this post (If that is so, feel free to comment). My next post will most probably come after IMOTC only now, so with that I bid adieu to all. Sayonara