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 :P). 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 :D.

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 :P). 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
geometry
altitude
orthocenter
humpty point
Inversion

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8 Comments

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Dual Desgarues gives an one line proof of Property 4. ISL 2005 G5 gives another way to define the Q points.

I don't know much about involution (never really used it on a problem), but I guess maybe!

Also hau r u so gud

Shouldn't I be asking this question to you considering you are so much better.
This post has been edited 3 times. Last edited by math_pi_rate, Dec 3, 2019, 10:55 AM

by Kayak, Dec 16, 2018, 4:41 PM

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Sorry by Dual Desgarues I mean that two triangles are perspective axially iff they are perspective radially (apply it to the main triangle and the orthic triangle) ; not anything related to involution since idk involution.

by Kayak, Dec 17, 2018, 11:42 AM

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oh yes, this is the classic orthocenter configuration. Two more problems in this area are TSTST 2016/2 and USAMO 2017/3

by yayups, Jan 2, 2019, 8:26 AM

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Yeah I'll probably add those in part 2 of this topic. I have got some tst problems lined up :P

by math_pi_rate, Jan 4, 2019, 4:41 PM

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Hi!! This is a really wonderful blog! I learnt a lot from this !! :thumbup:
Just wanted to ask:
Isn't Property 3 just Brokard's Thm ..... or Am I wrong?

No you are correct. Property 3 is just Brocard's Theorem on $BEFC$
This post has been edited 1 time. Last edited by math_pi_rate, Jan 16, 2019, 10:21 AM

by AlastorMoody, Jan 15, 2019, 4:22 PM

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Property 3 seemed to be radical axis on $\odot (MEF)$(nine pt circle) , $\odot (HM)$ , $\odot (AH)$

Yeah, they are. Just look at the solution given in the link.
This post has been edited 1 time. Last edited by math_pi_rate, Apr 6, 2019, 3:42 PM

by Aryan-23, Apr 5, 2019, 5:05 PM

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Another problem which can be yrivialized
USATSTST 2012#4
Follows directly from property 1 and property 2

by HoRI_DA_GRe8, Dec 9, 2021, 5:10 PM

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Property 3 is from India TST 2011

by Project_Donkey_into_M4, Mar 3, 2022, 10:33 AM

Geo Geo everywhere, nor a point to see.

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