The Ex-points and the Queue-points - Part Two

by math_pi_rate, Mar 30, 2019, 8:12 PM

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 :gleam:

). 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 $ABC$ be an acute triangle. Denote by $D$ the foot of the perpendicular line drawn from the point $A$ to the side $BC$ , by $M$ the midpoint of $BC$ , and by $H$ the orthocenter of $ABC$ . Let $E$ be the point of intersection of the circumcircle $\Gamma$ of the triangle $ABC$ and the half line $MH$ , and $F$ be the point of intersection (other than $E$ ) of the line $ED$ and the circle $\Gamma$ . Prove that $\tfrac{BF}{CF} = \tfrac{AB}{AC}$ must hold.

Solution: By Property 1, we know that $E$ is the $A$ -Queue point of $\triangle ABC$ (rename it as $Q_A$ from now on). Seeing this point, we add in the $A$ -Ex point $X_A$ . Also note that we are supposed to show that $ABFC$ is harmonic, for which we will require some sort of perspectivity. As all points are closely linked to the point $Q_A$ , so it is our first and foremost choice for perspector. Using the fact that $CDBX_A$ is harmonic (I guess this has been given in the proof of Property-3), and with the help of Property-1, we get that $-1=(B,C;D,X_A) \overset{Q_A}{=} (B,C;F,A) \Rightarrow ABFC \text{ is harmonic.}$
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 $ABC$ and $(O)$ be its circumcircle, and $H$ is its orthocenter. Let $M, N, P$ be midpoints of $BC, CA, AB$ , respectively. $D, E, F$ are the feet of the altitudes from $A, B$ and $C$ , respectively. Let $K$ symmetry with $H$ through $BC$ . $DE$ intersects $MP$ at $X$ , $DF$ intersects $MN$ at $Y$ .
a) $XY$ intersects smaller arc $BC$ of $(O)$ at $Z$ . Prove that $K, Z, E, F$ are concyclic.
b) $KE, KF$ intersect $(O)$ at $S, T$ ( $S, T\ne K$ ), respectively. Prove that $BS, CT, XY$ 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 $FDENMP$ , to find that $A,X,Y$ 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 $Z$ 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 $\odot (ABC),\odot (BCEF),\odot (KZEF)$ , we get that $KZ$ passes through the $A$ -Ex point $X_A$ of $\triangle ABC$ . Wow! That's interesting. But again we are stuck. At this point of time, I always try to incorporate the fact that $CDBX_A$ is harmonic. Taking perspectivity from $K$ , we get that $-1=(B,C;D,X_A) \overset{K}{=} (B,C;A,Z) \Rightarrow AZ \text{ must be the symmedian}$ So we arrive at the most crucial part of the solution. We just need to show that the line $\overline{AXY}$ is the $A$ -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 $AX$ is the $A$ -symmedian. Because of the weird definition of $X$ , there's not much one can do apart from applying Pascal's Theorem again. Trying to find some nice hexagons involving $X$ , we stumble upon $DEFPMM$ . This is nice, as we know that $ME=MF$ , and so we get that $BX \parallel EF$ . Now, it is some manipulations before finding a way out. What I did was take notice of the fact that $X$ lies on $MP$ , and so the reflection of $B$ in $X$ (say $B'$ ) must lie on $AC$ . This is quite helpful, because we also know that $BB'$ is antiparallel wrt $\angle BAC$ (as $BB' \parallel EF$ ). But, $AX$ bisects $BB'$ , which directly gives that it must be the $A$ -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 $\triangle ABC$ ). This is easy, and with some trial and error, one easily gets the two desired hexagons $KSBACT$ and $AZKSBC$ , after which one can finish using the fact that $KZ,BC,EF$ 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 $ABC$ $(AB < AC)$ let $D$ , $E$ and $F$ be foots of perpedicular from $A$ , $B$ and $C$ to $BC$ , $CA$ and $AB$ , respectively. Let $P$ and $Q$ be points on line $EF$ such that $DP \perp EF$ and $BQ=CQ$ . Prove that $\angle ADP = \angle PBQ$

Solution: Let $M$ be the midpoint of $BC$ (we introduce to get a link to $Q$ ). Then $PQMD$ 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 $X_A$ denotes the $A$ -Ex point of $\triangle ABC$ , then $X_AM \cdot X_AD=X_AB \cdot X_AC$ (as $CDBX_A$ is harmonic). But, $X_A \in PQ$ , so we get $X_AP \cdot X_AQ=X_AD \cdot X_AM$ . Combining these two equalities, we see that $BCQP$ is cyclic. As $QB=QC$ , so we get that in fact $PQ$ bisects $\angle BPC$ 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 $CDBX_A$ is. Always try to use this.

PROBLEM-4 (USA TSTST 2016 Problem 2)
Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$ . Denote by $P$ , $M$ the midpoints of $\overline{AH}$ , $\overline{BC}$ . Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $Q_A \neq A$ , and meets line $AM$ at a point $H_A \neq A$ . The tangent to $\gamma$ at $Q_A$ meets line $OP$ at $K$ . Show that the circumcircles of $\triangle Q_AMH_A$ and $\triangle PBC$ intersect at a point $T$ on $\overline{KM}$ .

Solution: Let's sort this information first (This is really important to all geo problems). By its definition, $Q_A$ is the $A$ -Queue point, and $H_A$ is the $A$ -Humpty point. Also, $K$ lies on $OP$ , which is nothing but the perpendicular bisector of $AQ_A$ . So $KA=KQ_A$ , which means that $KA$ is also tangent to $\gamma$ . So, in quadrilateral $AQ_AEF$ , $K=AA \cap Q_AQ_A$ and $M=EE \cap FF$ (all tangents taken wrt $\gamma$ ). Applying Pascal on $AAEQ_AQ_AF$ and $AEEQ_AFF$ , we get that $AE \cap Q_AF,AF \cap Q_AE,K,M$ 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 $A$ -Ex point $X_A=AQ_A \cap EF$ . 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 $KM$ is the polar of $X_A$ wrt $\gamma$ . So we have removed point $K$ , and now it suffices to show that $T$ lies on the polar of $X_A$ .

Let's look at the two circles now. Using Properties-1,2 and 3, we know that $\angle MQ_AX_A=\angle MH_AX_A=90^{\circ}$ , which means that $X_A \in \odot (Q_AMH_A)$ . Suppose what we are supposed to prove is true, i.e. $TM$ is the polar of $X_A$ wrt $\gamma$ . But, $\angle MTX_A=90^{\circ}$ (as $T \in \odot (Q_AMH_AX_A)$ ). So we get that $T$ must in fact be the inverse of $X$ wrt $\gamma$ . As the original direction is a bit difficult to prove, we will try to use phantom points. Let $T'$ be the inverse of $X_A$ wrt $\gamma$ . As $\angle MT'X_A=90^{\circ}$ , so $T' \in \odot (Q_AMH_AX_A)$ . Thus it suffices to show that $T' \in \odot (PBC)$ , which is equivalent to proving that $X_AB \cdot X_AC=X_AP \cdot X_AT'$ (since $P,T',X_A$ are collinear). However, $X_AP \cdot X_AT'$ is nothing but the power of $X_A$ wrt $\gamma$ (as $T'$ is the inverse of $X_A$ in $\gamma$ ). Thus, $X_AT \cdot X_AP=X_AE \cdot X_AF=X_AB \cdot X_AC$
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 $O$ be the circumcircle of the triangle $ABC$ . Two circles $O_1,O_2$ are tangent to each of the circle $O$ and the rays $\overrightarrow{AB},\overrightarrow{AC}$ , with $O_1$ interior to $O$ , $O_2$ exterior to $O$ . The common tangent of $O,O_1$ and the common tangent of $O,O_2$ intersect at the point $X$ . Let $M$ be the midpoint of the arc $BC$ (not containing the point $A$ ) on the circle $O$ , and the segment $\overline{AA'}$ be the diameter of $O$ . Prove that $X,M$ , and $A'$ are collinear.

Solution: This problem is one of my favourite problems, with a short, yet beautiful solution. Seeing the involvement of both the $A$ -mixtilinear incircle and excircle, we perform $\sqrt{bc}$ inversion, to get the following equivalent problem:-

Inverted problem wrote:

Let $P$ and $Q$ be the $A$ -intouch and $A$ -extouch point in $\triangle ABC$ , $X$ be the foot of internal angle bisector of $\angle BAC$ , and $D$ be the foot of the $A$ -altitude. Let $H_A$ be the $A$ -Humpty point of $\triangle APQ$ . Show that $ADXH_A$ is cyclic.

We are supposed to show that $ADXH_A$ is cyclic; or equivalently that $\angle XH_AA=\angle XDA=90^{\circ}$ . But, applying Property-3 to $\triangle APQ$ (The only major difficulty in this problem), this is only possible if $X$ is the $A$ -Ex point of $\triangle APQ$ . As $D$ is also the foot of $A$ -altitude of $\triangle APQ$ , so it suffices to prove that $DPXQ$ is harmonic (See the Remark given after Problem-3 above). Now, we again turn our attention back to $\triangle ABC$ . 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 :bye:

This post has been edited 2 times. Last edited by math_pi_rate, Dec 4, 2019, 6:56 PM

geometry

altitude

orthocenter

The Ex-points and the Queue-points - Part One

by math_pi_rate, Dec 15, 2018, 2:35 PM

So this post is about six wonderful points related to triangle geometry (sorry for the delay btw

). These points do not have any specific names to the extent I know, but I like to call them the three Queue-points and the three Ex-points, one corresponding to each vertex of a triangle. We start by defining what these points are, followed by some of their basic properties (most of which are well known). Hopefully you all will enjoy this post (I have tried to include motivation to some of the hard problems given below; as promised by me in the shout box). So let's begin:-

Definition: Let $\triangle DEF$ be the orthic triangle (i.e. triangle formed by foot of altitudes) of $\triangle ABC$ , and let $H$ be its orthocenter. Then the point where $\overline{EF}$ and $\overline{BC}$ meet is the $A$ -Ex point (denoted by $X_A$ from now on), and the point where $\odot (AEF)$ and $\odot (ABC)$ meet is the $A$ -Queue point (denoted by $Q_A$ from now on). The points $X_B,Q_B$ and $X_C,Q_C$ are defined in a similar fashion.

PROPERTY-1 The points $A,Q_A,X_A$ are collinear.

Proof 1: By Radical Axis Theorem on $\odot (AQ_ABC),\odot (AQ_AEF),\odot (BCEF)$ , we get that $\overline{AQ_A},\overline{EF},\overline{BC}$ concur; which gives that $A,Q_A,X_A$ are collinear. $\Box$

Proof 2: Invert the figure about $A$ with radius $\sqrt{AH \cdot AD}$ . Then, as $AH \cdot AD=AF \cdot AB=AE \cdot AC$ , we get that this inversion swaps $\{B,F\}$ and $\{C,E\}$ . Thus, this inversion also swaps $\overline{EF} \cap \overline{BC}=X_A$ and $\odot (AEF) \cap \odot (ABC)=Q_A$ ; which gives that $A,Q_A,X_A$ are collinear. $\Box$

PROPERTY-2 Let $M_A$ be this midpoint of $\overline{BC}$ , and $A'$ be the point diametrically opposite to $A$ on $\odot (ABC)$ (i.e. $A'$ is the antipode of $A$ in $\odot (ABC)$ ). Then $Q_A,H,M_A,A'$ are collinear.

Proof: As $\overline{BH} \perp \overline{AC}$ and $\overline{A'C} \perp \overline{AC}$ , we get that $\overline{BH} \parallel \overline{A'C}$ . Similarly, we get that $\overline{CH} \parallel \overline{A'B}$ . This means that $BHCA'$ is a parallelogram, and so $A',M_A,H$ are collinear. Also, as $\measuredangle A'QA=90^{\circ}=\measuredangle HQA$ , we get that $A',H,Q$ are also collinear. Combining these two results, we have that $Q_A,H,M_A,A'$ lie on a line. $\Box$

NOTE: The two properties given above are basically restatement of the fact that $Q_A$ is the Miquel Point of the cyclic quadrilateral $BCEF$ .

PROPERTY-3 Let $H_A$ be the $A$ -Humpty point (See here to find out what Humpty points are). Then $X_A,H,H_A$ lie on a line perpendicular to the $A$ -median $\overline{AM_A}$ . In particular, $AH_ADX_A$ is a cyclic quadrilateral.

Proof: See my proof here. $\Box$

PROPERTY-4 The points $X_A,X_B,X_C$ lie on a line, which is known as the Orthic Axis of $\triangle ABC$ .

Proof: Seeing three collinear points, one is immediately drawn to the idea of radical axis. Note that $X_AE \cdot X_AF=X_AB \cdot X_AC$ , and so $X_A$ lies on the radical axis of $\odot (ABC)$ and $\odot (DEF)$ . By symmetry, we get that $X_B$ and $X_C$ also lie on this line, and so we are done. $\Box$

REMARK: This also shows that the orthic axis is orthogonal to the Euler Line.

Let's begin with some problem solving. We can discover more properties en route.

PROBLEM-1 (Polish MO 2018 Problem 5)
Suppose $E$ and $F$ are the foot of the $B$ and $C$ altitudes of $\triangle ABC$ . The line tangent to $\odot (ABC)$ at point $A$ meets $\overline{BC}$ at $P$ . The line parallel to $\overline{BC}$ that goes through point $A$ meets $\overline{EF}$ at $Q$ . Prove that $\overline{PQ}$ is perpendicular to the $A$ -median of $\triangle ABC$ .

Solution: Seeing the line $\overline{EF}$ and the $A$ -tangent, the first thing that comes to the mind is that both of them are anti-parallel to $\overline{BC}$ , and so they must be in fact parallel to each other. As we are given a line parallel to $\overline{BC}$ meeting a line parallel to $\overline{AP}$ , we are inspired to try completing a parallelogram, and so we define the $A$ -Ex point $X_A$ . Now, we have a parallelogram in front of us, namely $APX_AQ$ . Involvement of the $A$ -median and the orthocenter configuration, along with the perpendicularity to the $A$ -median that needs to be proved, is a big indication to us to introduce the $A$ -Humpty point $H_A$ . Using Property-3, we just need to prove that $\overline{PQ}$ and $\overline{X_AHH_A}$ are parallel to each other.

The parallelism that needs to be shown again motivates us to complete a parallelogram, and so we define the point $\overline{X_AH} \cap \overline{AQ}=T$ . Then it suffices to show that $PX_ATQ$ is a parallelogram; or equivalently that $QT=PX_A$ . As $APX_AQ$ is a parallelogram, we just need to prove that $QT=QA$ . Now note that as $\overline{X_AHH_AT} \perp \overline{AH_A}$ , so in fact $\triangle AH_AT$ is right-angled at vertex $H_A$ . So we only have to show that $Q$ is the circumcenter of $\triangle AH_AT$ , or that $QH_A=QA$ . Noting that $AEH_AF$ is harmonic, and that $\overline{QA}$ is tangent to $\odot (AEHH_AF)$ now easily finishes the problem. For a formal solution, see here. Another solution which I had found (does not involve Ex-points) is this.

REMARK: This problem provides a really nice method to connect tangents to the Ex-points. Also the solution uses an important concept, which is when you see parallel lines, try to find some parallelograms, as they have really nice properties, and also help in proving that some point is the midpoint of a line.

PROBLEM-2 (USA TSTST 2017 Problem 1)
Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$ , circumcenter $O$ , and orthocenter $H$ . Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$ . Let $M$ and $N$ be the midpoints of sides $\overline{AB}$ and $\overline{AC}$ , respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$ , respectively. Let $P$ be the intersection of line $\overline{MN}$ with the tangent line to $\Gamma$ at $A$ . Let $Q$ be the intersection point, other than $A$ , of $\Gamma$ with the circumcircle of $\triangle AEF$ . Let $R$ be the intersection of lines $\overline{AQ}$ and $\overline{EF}$ . Prove that $\overline{PR} \perp \overline{OH}$ .

Solution: Note that, using Property-1, $R$ is in fact the $A$ -Ex point of $\triangle ABC$ . Then, by using the Remark given after Property-4, we just need to show that $P$ lies on the orthic axis; or equivalently that it lies on the radical axis of the circumcircle and the nine-point circle of $\triangle ABC$ . But this easily follows from the equality $PA^2=PM \cdot PN$ . So much for a USA TSTST problem, huh

.

NOTE: There exists another nice solution to this problem, which uses the nice connection we had discovered to connect tangents and the Ex-points in the previous problem. See this solution here (Do read it; it is quite interesting).

PROBLEM-3 (RMM 2013 Problem 3)
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$ . The lines $\overline{AB}$ and $\overline{CD}$ meet at $P$ , the lines $\overline{AD}$ and $\overline{BC}$ meet at $Q$ , and the diagonals $\overline{AC}$ and $\overline{BD}$ meet at $R$ . Let $M$ be the midpoint of the segment $\overline{PQ}$ , and let $K$ be the common point of the segment $\overline{MR}$ and the circle $\omega$ . Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.

Solution: At first glance, you might be surprised to find this problem in this post. But you'll soon see why it is present here. On seeing the problem, the first thing that strikes the mind is Brokard's Theorem (If it doesn't, then just make it a habit to think of Brokard on seeing a complete quadrilateral). According to Brokard's Theorem, $R$ is the orthocenter of $\triangle OPQ$ , and so, in order to make things more familiar, we restate the problem in terms of $\triangle OPQ$ as follows:-

Restated problem wrote:

Let $D$ be the foot of the $A$ -altitude and $H$ be the orthocenter of $\triangle ABC$ . Let $\gamma$ be the circle centered at $A$ with radius $\sqrt{AH \cdot AD}$ . Let $M$ be the midpoint of $\overline{BC}$ , and let $\overline{MH} \cap \gamma=K$ . Show that $\odot (KBC)$ is tangent to $\gamma$ .

On seeing a familiar configuration, I add in all the points which I think might turn out to be useful in the future. So we add the points $E,F,X_A,Q_A$ in the diagram (Inspiration to add $X_A$ and $Q_A$ is that they are inverses wrt $\gamma$ , something we proved in Proof 2 of Property-1). Note that, from Property-2, we get that $\angle KQ_AA=90^{\circ}$ , and so $\odot (KQ_AA)$ is tangent to $\gamma$ . Inverting about $\gamma$ , $\odot (KQ_AA)$ goes to $\overline{X_AK}$ , and so we get that $\overline{X_AK}$ is tangent to both $\gamma$ and $\odot (KQ_AA)$ . This means that $X_AK^2=X_AA \cdot X_AQ_A=X_AB \cdot X_AC$ , which directly gives the result that $\odot (KBC)$ is tangent to $\gamma$ .

REMARK: The fact that $\overline{X_AK}$ is the common tangent of the two circles is a bit unmotivated, and it was by chance that I stumbled upon this fact while solving this problem (basically a nice diagram helped

). The only motivation is that $X_A$ is the only nice point where the common tangent of the two circles might meet $\overline{BC}$ , something which is common to Olympiad geo problems. The solution given above is a refined version of my original solution, which was quite complicated. You can see my original solution here.

I'll post some more harder problems in Part Two, but this is it for now. For a real challenge, try this problem (HINT: Look at the notation used).

This post has been edited 4 times. Last edited by math_pi_rate, Apr 1, 2019, 6:48 AM
Reason: Typo