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

5 Comments

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Best of Luck!

Thanks

This post has been edited 1 time. Last edited by math_pi_rate, Mar 31, 2019, 2:37 PM

by a_simple_guy, Mar 31, 2019, 2:00 PM

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Another property similar to P1 (Taken from one of my old posts):

Let $ABC$ be a triangle with orthocenter $H$ and let $M$ be the midpoint of $BC$ . Let $AH$ intersect the circumcircle of $ABC$ at $H'$ and let ray $MH$ intersect the circumcircle at points $M'$ . Then $M'BH'C$ is harmonic.

Actually the problem given in the remarks after Problem-1 asks to prove this only

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by Synthetic_Potato, Mar 31, 2019, 4:36 PM

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Quote:

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 $\angle ADP = \angle PBQ$

Another Solution:

$EF$ intersect $BC$ at $D'$ . Since $(B,C;D,D')=-1$ and $DP\perp EF$ , $PD, PD'$ bisect $\angle BPC$ . Combining with $QB=QC$ , $B,C,P,Q$ are concyclic. Let $Q'$ be the antipode of $Q$ in $\odot(BPQC)$ . Then $P,D,Q'$ are collinear since $PD$ bisects $\angle BPC$ . So $\angle PBQ = \angle PQ'Q \stackrel{QQ'\parallel AD}{=} \angle ADP$ . $\blacksquare$ .

Oops!! Seems like my extra long angle chase was unnecessary

This post has been edited 1 time. Last edited by math_pi_rate, Mar 31, 2019, 5:00 PM

by Synthetic_Potato, Mar 31, 2019, 4:45 PM

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All problems from contests at TST level :huuh:

Not at all fair :bruce:

You should have included moderate/easy examples for noobs like us :bruce:

Anyways Great Job!! :thumbup:

I'll include some problems that I recall (also again best of luck for IMOTC)

Serbia NMO 2010 P2 wrote:

In an acute-angled triangle $ABC$ , $M$ is the midpoint of side $BC$ , and $D, E$ and $F$ the feet of the altitudes from $A, B$ and $C$ , respectively. Let $H$ be the orthocenter of $\Delta ABC$ , $S$ the midpoint of $AH$ , and $G$ the intersection of $FE$ and $AH$ . If $N$ is the intersection of the median $AM$ and the circumcircle of $\Delta BCH$ , prove that $\angle HMA = \angle GNS$ .
Proposed by Marko Djikic

Mexico NMO 2012 P6 wrote:

Consider an acute triangle $ABC$ with circumcircle $\mathcal{C}$ . Let $H$ be the orthocenter of $ABC$ and $M$ the midpoint of $BC$ . Lines $AH$ , $BH$ and $CH$ cut $\mathcal{C}$ again at points $D$ , $E$ , and $F$ respectively; line $MH$ cuts $\mathcal{C}$ at $J$ such that $H$ lies between $J$ and $M$ . Let $K$ and $L$ be the incenters of triangles $DEJ$ and $DFJ$ respectively. Prove $KL$ is parallel to $BC$ .

by AlastorMoody, Mar 31, 2019, 6:52 PM

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Wow! Amazing!! :thumbup:

by L567, Feb 25, 2021, 5:04 PM

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And here's the first of 2025!
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First comment of 2024!
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Wowowooo my man came backkkk
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Aah the thrill of coming back to a dead blog every year once!
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kukuku 1st comment of 2023
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1st comment of 2022
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duh i found this masterpiece after the owner went inactive noooooooooooo :sadge:
by Project_Donkey_into_M4, Nov 23, 2021, 2:56 PM
Hi everyone! Sorry but I doubt if I'll be reviving this anytime soon

I might post something if I find anything interesting, but they'll mostly be short posts, unlike the previous ones!
by math_pi_rate, Nov 7, 2020, 7:39 PM
@below be patient. He must be busy with some other work.
by amar_04, Oct 22, 2020, 10:23 AM
Advanced is over now, please revive this please!
by Geronimo_1501, Oct 12, 2020, 5:10 AM
REVIVE please
by Gaussian_cyber, Aug 28, 2020, 2:19 PM
re-vi-ve
by nprime06, Jun 12, 2020, 2:17 AM
re-vi-ve
by fukano_2, May 30, 2020, 2:09 AM
Try this: Prove that a two colored cow has an odd number of Agis Phesis
by Synthetic_Potato, Apr 18, 2020, 5:29 PM
Nice blog!!
by enthusiast101, Nov 9, 2018, 10:44 AM
Updates required
by TheDarkPrince, Oct 29, 2018, 10:07 AM
All hail to geo god math_pi_rate -- who solved 2011 G8 in kindergarten
by Kayak, Oct 24, 2018, 6:46 AM
Thanks. Will probably post after RMO only now.
by math_pi_rate, Sep 30, 2018, 7:22 AM
Nice blog!!! Keep posting!
by ayan.nmath, Sep 28, 2018, 9:01 PM
Thanks. I'll make sure that I include my thought process in the next post.
by math_pi_rate, Sep 27, 2018, 1:01 PM
Can you post some advice on how to get better at synthetic geometry on your blog or say your thought process while solving geo problems (like your post on FE)? Nice post on FE though !
by Kayak, Sep 27, 2018, 11:38 AM
I am sorry, but what do u mean by tangents?
by math_pi_rate, Sep 27, 2018, 9:23 AM
Can you post something on Tangents?
by PhysicsMonster_01, Sep 27, 2018, 6:42 AM
Thanks @below and @2below.
@below Posted smth apart from geo just for you. Jai Mahismati.
by math_pi_rate, Sep 17, 2018, 4:47 PM
How are you so good in geo...

#geoishard
by Kayak, Sep 16, 2018, 4:45 PM
Nice work Genius :O
by TheDarkPrince, Sep 13, 2018, 1:08 PM
Thanks a lot!
by math_pi_rate, Aug 20, 2018, 12:17 PM
First shout! Lovely tangency theorem
by silverivy, Aug 19, 2018, 4:08 PM

68 shouts

Well! All I can do is Geo, so...

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

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

5 Comments