The Fables of a G4 and a G7

by math_pi_rate, Jan 30, 2020, 2:27 PM

Hey guys! I know this is many months late :oops:, but I finally got something worthy to post on (after the massive success of the Queue and Ex points post :)). The main content of the discussion ahead is based on two extremely well-known problems, namely IMOSL 2002 G7 and IMOSL 2011 G4. The basic agenda is to unravel the relation between these two seemingly unrelated configurations, and then to pick up some problems around this. Hopefully you guys will enjoy this one too. Anyway, let's begin by first showing the two problems about which this whole post is (with solutions included for completeness' sake):

IMOSL 2002 G7: The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.

Solution: Suppose the tangent to $\Omega$ at $N$ meets $BC$ (which is the tangent to $\Omega$ at $K$) at some point $T$. Let $I,I_A$ be the incenter,$A$-excenter of $\triangle ABC$, and $Z$ be the midpoint of $KN$. It is well known that $M,K,I_A$ are collinear, so we get $$\measuredangle I_AZI=\measuredangle KZI=90^{\circ} \Rightarrow Z \in \odot (BICI_A)$$This gives $TN^2=TZ \cdot TP=TB \cdot TC$ (since $Z$ is the inverse of $T$ in $\Omega$), and hence the result.

IMOSL 2011 G4: Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Solution: Let $A_0$ be the midpoint of $BC$, and $D_0$ be the foot of perpendicular from $A_0$ to $B_0C_0$. Notice that the negative homothety centered at $G$ that takes $\triangle ABC$ to $\triangle A_0B_0C_0$, also takes $D$ to $D_0$, and so $D,G,D_0$ are collinear. Thus it suffices to show that $X$ lies on $DD_0$. Let $O$ be the circumcenter of $\triangle ABC$, and suppose the tangent to $\Omega$ at $A$ meets $B_0C_0$ at $K$. As $\odot (AB_0C_0)$ is also tangent to $\Omega$, from the Radical Axes Theorem on $\odot (AB_0C_0), \Omega, \omega$ we get that $KX$ is tangent to $\Omega$ at $X$. Now, $\measuredangle OD_0K=\measuredangle OAK= \measuredangle OXK=90^{\circ} \Rightarrow KAD_0OX$ is cyclic. As $KX=KA$, we have that $\measuredangle XD_0K=\measuredangle KD_0A=\measuredangle DD_0K$, where the last equality follows from the fact that $D_0$ lies on $B_0C_0$, which is the perpendicular bisector of $AD$. Thus, $X$ lies on $DD_0$, hence proving what was required.

Now that we are done with the formalities, let's move onto the more important part, i.e. the relation between these two problems.

The Whole Configuration: Let $\triangle ABC$ have incircle $\omega$, incenter $I$ and intouch triangle $\triangle DEF$. Let $\triangle A_0B_0C_0$ be the medial triangle of $\triangle DEF$. Let $M$ be the midpoint of the $A$-altitude, and $N$ be the point where $DM$ meets $\omega$ again. By IMOSL 2002 G7, $\odot (BCN)$ is tangent to $\omega$. Let $X \neq D$ be the point on $\omega$ such that $\odot (B_0C_0X)$ is tangent to $\omega$. Then $N=X$.

Proof: Invert about $\omega$. Then $\{B,B_0\}$ and $\{C,C_0\}$ are swapped.. Thus, $\odot (BCN)$ and $\odot (B_0C_0N)$ get swapped. But the inverse of $\odot (BCN)$ (i.e. $\odot (B_0C_0N)$) must also be tangent to $\omega$. This gives $N=X$ directly.

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.86, xmax = 9.86, ymin = -4.91, ymax = 6.31; /* image dimensions */ pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */ draw((-2.12,5.43)--(-5.42,-1.57), linewidth(1.2) + wvvxds); draw((-5.42,-1.57)--(4.48,-1.53), linewidth(1.2) + wvvxds); draw((4.48,-1.53)--(-2.12,5.43), linewidth(1.2) + wvvxds); draw(circle((-1.4066970274945345,0.9863112648557781), 2.540075166203071), linewidth(1.2) + rvwvcq); draw((-3.7042601504632997,2.06944816568391)--(0.43644209687573987,2.734115606931038), linewidth(1.2) + wrwrwr); draw((0.43644209687573987,2.734115606931038)--(-1.3964340996325002,-1.5537431680793232), linewidth(1.2) + wrwrwr); draw((-1.3964340996325002,-1.5537431680793232)--(-3.7042601504632997,2.06944816568391), linewidth(1.2) + wrwrwr); draw(circle((-0.47123281122031063,-1.2448792229730814), 4.959435472752621), linewidth(1.2) + dtsfsf); draw(circle((-1.7287210006171598,1.7543758359074733), 1.7072350707642758), linewidth(1.2) + sexdts); /* dots and labels */ dot((-2.12,5.43),linewidth(4pt) + dotstyle); label("A", (-2.08,5.67), NE * labelscalefactor); dot((-5.42,-1.57),linewidth(4pt) + dotstyle); label("B", (-6.02,-1.85), NE * labelscalefactor); dot((4.48,-1.53),linewidth(4pt) + dotstyle); label("C", (4.86,-1.77), NE * labelscalefactor); dot((-1.3964340996325002,-1.5537431680793232),linewidth(4pt) + dotstyle); label("$D$", (-1.42,-2.11), NE * labelscalefactor); dot((0.43644209687573987,2.734115606931038),linewidth(4pt) + dotstyle); label("$E$", (0.52,2.89), NE * labelscalefactor); dot((-3.7042601504632997,2.06944816568391),linewidth(4pt) + dotstyle); label("$F$", (-4.12,1.69), NE * labelscalefactor); dot((-0.47999600137838017,0.5901862194258574),linewidth(4pt) + dotstyle); label("$C_0$", (-0.2,0.41), NE * labelscalefactor); dot((-2.5503471250479,0.25785249880229344),linewidth(4pt) + dotstyle); label("$B_0$", (-2.98,-0.21), NE * labelscalefactor); dot((-1.63390902679378,2.4017818863074742),linewidth(4pt) + dotstyle); label("$A_0$", (-1.7,2.78), NE * labelscalefactor); dot((-1.4066970274945345,0.9863112648557781),linewidth(4pt) + dotstyle); label("$I$", (-1.32,1.25), NE * labelscalefactor); dot((-2.3888366000379633,3.3288277984165924),linewidth(4pt) + dotstyle); label("N=X", (-2.68,3.67), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Well having seen the relation, I guess it's time for some properties (From now on we will work with the above configuration only, i.e. same notations).

PROPERTY-1 Let $D_1$ be the antipode of $D$ in $\omega$, and $T$ be the foot of perpendicular from $D_1$ to $EF$ (i.e. $T$ is the isotomic point of the foot of the $D$-altitude). Then $DX,DT$ are isogonal in $\angle EDF$. Also lines $D_1X,EF,BC$ are concurrent.

Proof: Seeing isogonal lines with one foot on the circumcircle and the other on the corresponding side inspires us to consider $\sqrt{bc}$ inversion (If it doesn't, then make sure it does the next time :D). So we invert about $D$ with radius $\sqrt{DE \cdot DF}$, followed by reflection in $\angle EDF$ (denoting images under this transformation by $'$). Then $B_0',C_0'$ are the reflections of $D$ in $F,E$ respectively, while $X' \in EF$ is a point such that $\odot (B_0'C_0'X')$ is tangent to $EF$. Thus, we have $$\angle B_0'C_0'X'=\angle EX'C_0'=\angle X'B_0'C_0' \Rightarrow X' \text{ lies on the perpendicular bisector of } B_0'C_0'$$Also, as $D_1$ is the center of $\odot (DB_0'C_0')$, so we have that $D_1X'$ is the perpendicular bisector of $B_0'C_0'$. Combining this with $B_0'C_0' \parallel EF$, we get that $D_1X' \perp EF$, as desired.

For the second part, let $S$ be the reflection of $D$ in line $AI$ (i.e. $S$ lies on $\omega$ such that $DS \parallel EF$). Note that $\odot (SDT)$ passes through the foot of the $D$-altitude. Then the same inversion gives that $S'T'$ passes through $D_1$. Combined with $S'=EF \cap BC$ and $T'=X$ (which has been shown above), we have the desired result $\Box$

PROPERTY-2 (Probably my favourite property related to this configuration): Replace $X$ by $X_a$, and define $X_b,X_c$ similarly (We'll continue this notation from now on). Then $DX_a,EX_b,FX_c$ concur at the homothety center $P$ of $\triangle DEF$ and the excentral triangle of $\triangle ABC$. Finally, $P$ also lies on the Euler line of $\triangle DEF$. (NOTE: $P$ is known as the Isogonal Mittenpunkt Point of $\triangle ABC$, and is also the homothety center of $\triangle ABC$ and the orthic triangle of $\triangle DEF$.)

Proof: Using Property-1, lines $DX_a,EX_b,FX_c$ concur at the isogonal conjugate of the isotomic conjugate of the orthocenter $H$ of $\triangle DEF$. Now, let $\triangle I_AI_BI_C$ be the excentral triangle of $\triangle ABC$, and $V$ be the Bevan Point of $\triangle ABC$ (i.e. the center of $\odot (I_AI_BI_C)$). By IMOSL 2002 G7's proof, we have $X_a \in DI_A$. This means that the concurrency point of $DX_a,EX_b,FX_c$ is the homothety center $P$ of $\triangle DEF$ and $\triangle I_AI_BI_C$.

This gives that $P$ also lies on the line joining the circumcenters of $\triangle DEF$ and $\triangle I_AI_BI_C$, i.e. $P \in IV$. As $I$ is the orthocenter of $\triangle I_AI_BI_C$, so line $IV$ is the Euler line of $\triangle I_AI_BI_C$. But, it is well known that the Euler lines of $\triangle DEF$ and $\triangle I_AI_BI_C$ are the same. Thus, we get that $P$ lies on the Euler line of $\triangle DEF$ also, as desired. $\Box$

PROPERTY-3 Lines $AX_a,BX_b,CX_c$ are concurrent.

Proof: We prove the following stronger result (which is known as the Steinbart theorem).
Steinbart wrote:
Let $P,Q,R$ be points on $\omega$ such that $DP,EQ,FR$ are concurrent at a point $K$. Then $AP,BQ,CR$ are also concurrent.

This can easily be proved using Trig Ceva, but I'll be modest, and just use moving points :D. Fix $P \in \omega$, and animate $K$ on line $DP$. Then $$K \mapsto EK \cap \omega=Q \mapsto BQ \cap AP \text{ and } K \mapsto FK \cap \omega=R \mapsto CR \cap AP$$are projective. Thus, it suffices to prove that $BQ \cap AP=CR \cap AP$ for three distinct positions of $K$ on line $AP$. The result trivially follows on taking $K=D,P,DP \cap EF$, as desired.

REMARK (IMPORTANT): Unlike most of the famous theorems, the converse of Steinbart Theorem is not completely true. The extended Steinbart theorem deals with this anomaly. You can find it here (written by Darij Grinberg).

Well, it's time to move on to some problems (wait, what's the distinction between problems and properties :oops:). Anyway, let's hope I add enough motivational content for each one :P.

PROBLEM-1 (Generalization of IMOSL 2011 G4; given by babu2001 here)
Let $X\in BC$. Let $Y$ be the reflection of $X$ in the midpoint of $BC$. Let the isogonal ray to $AY$ intersect $\odot (ABC)$ at $Z$. Let $A'$ be the reflection of $A$ in the perpendicular bisector of $BC$. Then $Z,X,A'$ are collinear.

Solution 1: Let $M$ be the midpoint of $BC$, and $T \in \odot (ABC)$ such that $AT$ is the $A$-symmedian . We will do a cross ratio chase (don't get afraid if you haven't done one before; it is a wonderful method definitely worth knowing about). The basic inspiration is to note that isogonal rays and isotomic points are related to each other projectively (i.e. they preserve cross ratios). We start with the ratio $\zeta=(B,C;M,X)$ and try to prove it is equal to $(B,C;M,A'Z \cap BC)$, after which we are obviously done. First, by isotomic conjugation, $\zeta=(C,B;M,Y)$. Then we have $$\zeta=(C,B;M,Y)=(AC,AB;AM,AY) \overset{\text{Isogonality}}{=} (AB,AC;AT,AZ)=(B,C;T,Z) \overset{A'}{=} (B,C;A'T \cap BC,A'Z \cap BC)$$Thus, to show the original problem it suffices to prove that $A',M,T$ are collinear. This can equivalently be formulated as $$A',X,Z \text{ are collinear} \Leftrightarrow A',M,T \text{ are collinear}$$This equivalent statement gives that if we show the problem for even one finite position of $X (\neq B,C)$ (which is not necessarily $M$), we can deduce that the problem is true for all positions of $X$. Taking $X$ as the isotomic point of the foot of the $A$-altitude (i.e. the foot of perpendicular from the $A$-antipode on $BC$) quite easily works. Hence, done. $\blacksquare$

NOTE (A possible trap): We cannot just show the problem to be true for $X=B$ or $X=C$, and say it is always true, after we have deduced the equivalence relation. This is because, when $X=B/C$, the cross ratio $\zeta$ becomes invariant of point $M$. A similar thing happens when $X$ is the point at infinity, which is why we must ensure that $X$ is a finite point distinct from $B,C$.

Solution 2: There is an easy alternative moving points solution (those who don't know moving points may skip this). Animate $X$ on $BC$. Then, since isogonal and isotomic conjugation are projective maps, so the map $$X \mapsto Y \mapsto Z \mapsto A'Z \cap BC$$must also be projective. Thus, it suffices to prove the problem for three distinct positions of $X$. Taking $X=B,C$ and the point at infinity on $BC$ trivially give the conclusion. $\blacksquare$

PROBLEM-2 (EMMO 2016 Junior P5)
Let $\triangle ABC$ be a triangle with circumcenter $O$ and circumcircle $\Gamma.$ The point $X$ lies on $\Gamma$ such that $AX$ is the $A$- symmedian of triangle $\triangle ABC.$ The line through $X$ perpendicular to $AX$ intersects $AB,AC$ in $F,E,$ respectively. Denote by $\gamma$ the nine-point circle of triangle $\triangle AEF,$ and let $\Gamma$ and $\gamma$ intersect again in $P \neq X.$ Further, let the tangent to $\Gamma$ at $A$ meet the line $BC$ in $Y,$ and let $Z$ be the antipode of $A$ with respect to circle $\Gamma.$ Prove that the points $Y,P,Z$ are collinear.

Solution: Redefine $P$ as the point where $YZ$ meets $\Gamma$ again, in which case we have to show that $P$ lies on the nine-point circle of $\triangle AEF$. Note that, by Property-1 (second part), $P$ is in fact the 2011 G4 point of $\triangle ABC$. Suppose the perpendicular bisector of $AX$ (i.e. line $OY$) meets $AB,AC$ at $M,N$. Then we wish to show that $P \in \odot (MNX)$. Consider $\sqrt{bc}$ inversion. Then, using Property-1, and with the help of the inversive distance formula, we get the following problem:-
Inverted problem wrote:
Let $M,D$ be the midpoint and the isotomic point of the $A$-altitude in $BC$. Suppose $\odot (M,MA)$ meets $AB,AC$ at $X,Y$. Then show that $DMXY$ is cyclic.

As we do with almost all problems, we will assume that the problem is true, and try to deduce whatever we can. Note that, if we denote the reflection of $A$ in $M$ by $S$, then $SX \perp AB$ and $SY \perp AC$. The reason we consider $S$ is that it also lets us characterize point $D$; it is simply the foot of perpendicular from $S$ to $BC$. Thus, $\odot (DXY)$ is nothing but the pedal circle of $S$ wrt $\triangle ABC$. Now, whenever pedal circles are involved, I have a tendency to go towards the isogonal conjugate (and so should you, if you don't have such an urge already). Anyway, the main point here is that isogonal conjugates share their pedal circles (which is also called the six point circle theorem). So we consider the isogonal conjugate of $S$, which is nothing but the point $T$ where the tangent to $\odot (ABC)$ at $B$ and $C$ meet. Note that $TM \perp BC$, and so using the fact given above, we get that $M \in \odot (DXY)$, as desired. $\blacksquare$

REMARK: The trick of considering the common pedal circle of isogonal conjugates is a wonderful tool for tackling problems involving these. Here's another example of this trick (the geometry one, not the combi one :P).

PROBLEM-3 (Proposed by AlastorMoody here)
In an acute triangle $\Delta ABC$, Let $H'$ be the isotomic conjugate of the orthocenter $H$ of $\Delta ABC$. Let $AH' \cap BC=X$ and $M$ be the midpoint of $AX$. Let $\Delta DEF$ be the orthic triangle WRT $\Delta ABC$ and $G$ be the centroid of $\Delta ABC$. Define $AG \cap \odot (ABC) = P$ and Let $M_A$ is the midpoint of $BC$

(i) Let $N$ be the midpoint of $AC$. Show that the intersection of $\odot (EFP)$ and $\odot (MNC)$ lies on $\odot (ABC)$. Moreover, show that this intersection point lies on $\odot (DM_AP)$
(ii) Suppose that the intersection point in (i) is $Y$, then prove, $Y$ lies on $DM$
(iii) Let $MY \cap \odot (ABC)$ at $T$, then prove, $AT \parallel BC$

Solution: Redefine $Y$ as the 2011 G4 point of $\triangle ABC$, since Part (ii) and (iii) kind of give away that this should be $Y$. In that case, (ii) and (iii) are true by solution of IMOSL 2011 G4. So we'll just show that $Y \in \odot (EFP),\odot (MNC)$ and $\odot (DM_AP)$. Using $T$ as given in (iii), we notice that $AT \parallel DM_A$ directly gives $Y \in \odot (DM_AP)$ by converse of Reim's Theorem. Again using converse of Reim's Theorem, we get $Y \in \odot (MNC)$ from $MN \parallel AT$ (since $M \in TY$ and $N \in AC$).

For the final part, i.e. showing $Y \in \odot (EFP)$, we bring our old friend, the $A$-Ex point $X_A$, into picture (You'll see soon why). Seeing so many circles, for obvious reasons, we try to use radical axes theorem. Applying it to $\odot (DM_AYP),\odot (ABC),\odot (DM_AEF)$, we see that $YP \cap BC$ lies on radical axis of $\odot (ABC)$ and $\odot (DEF)$. But, by Property-4 here, we know that point is nothing but $X_A$, which in turn means that $EF,BC,PY$ are concurrent at $X_A$. Then $$X_AE \cdot X_AF=X_AB \cdot X_AC=X_AY \cdot X_AP$$gives the result. $\blacksquare$

REMARK: I believe the problem would have been much harder (especially in exam conditions), if the last 2 parts weren't given. They are sort of the giveaways. Apart from that, the part of getting $PY,BC,EF$ concurrent could be observed even by a good eye. Like in my diagram, I had a good hunch that it might be true (and these hunches go a long way with geometry).

PROBLEM-4 (STEMS 2020 Category B Subjective P4)
In triangle $\triangle{ABC}$ with incenter $I$, the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$, respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$. The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$. Prove that $FQ$ is parallel to $BI$.

Solution: Let $D$ be the $A$-intouch point of $\triangle ABC$, and note that $K$ is the 2011 G4 point of $\triangle DEF$. Redefine $Q$ as the point such that $FQ \parallel BI$. Then $FQ \perp DF$ combined with $BF=BD$ gives that $B$ is the center of $\odot (DFQ)$, i.e. $Q$ is the reflection of $D$ in $B$. Now, we consider $\sqrt{DE \cdot DF}$ inversion followed by reflection in angle bisector of $\angle EDF$ (denoting inverse of point $Z$ by $Z'$). Let $\ell$ be the line through $D$ parallel to $EF$. Then $B',C' \in \ell$ such that $EC'=ED$ and $FB'=FD$. Since, $Q'$ is the midpoint of $DB'$, so we get that $Q'$ is the foot of perpendicular from $F$ to $\ell$. Note that $DQ'$ is equal to the distance of $F$ from the $D$-altitude, and so we have $DQ'=EK'$. Combined with $DQ' \parallel EK'$, we get that $DQ'K'E$ is a parallelogram. Thus, we have $Q'K'=DE=EC'$, and so $Q'K'EC'$ must be an isosceles trapezoid, or equivalently, $Q'K'EC'$ is cyclic, as desired. $\blacksquare$

REMARKS: Wonderfully easy if you know the above theory :D.

With that, we come to an end of this section. For a bonus problem, try this one using what we have done till now (Don't see my solution :P). After this I will probably be posting on this blog after May only. So enjoy :)

P.S. Do write in the comments section if you wish to share something (like a problem or another property). Also, I am extremely sorry for any typos. You can notify me about those too in the comments section. Thanks :)
This post has been edited 4 times. Last edited by math_pi_rate, Jan 31, 2020, 6:04 AM
IMOSL 2011 G4
IMOSL 2002 G7
ex-point
geometry
Inversion

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 :P). 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 :P). 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 :D). 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 :D). 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 :D. 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
humpty point
Inversion
ex-point
queue-point

Happy New Year!!

by math_pi_rate, Dec 31, 2018, 11:13 AM

Sorry for an off-topic post, but Happy New Year everyone. And for this New Year, I'd suggest everyone to look at this wonderful handout (attached below). It contains some well known properties of mixtilinear incircles, and some other not so well known ones. I'd like to thank Jafet98 for releasing such a nice handout, though I think it wasn't publicised that much. So enjoy! :-D
Attachments:
On mixtilinear incircles.pdf (470kb)

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

A Powerful Ratio Theorem

by math_pi_rate, Sep 27, 2018, 12:43 PM

So this topic is for a wonderful theorem about isogonal lines, namely Steiner's Ratio Theorem. Here's the theorem:

THEOREM (Steiner) Let $D$ and $E$ be points on $BC$, so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$. Then we have the following equality:- $$\left (\frac{AB}{AC} \right)^2=\frac{BD \cdot BE}{CD \cdot CE}$$
REMARK: The converse of Steiner's Theorem is also true.

PROOF 1 (Trig Bash :D) Let $\angle BAD=\angle CAE=x$ and $\angle BAE=\angle CAD=y$. Apply sine law in $\triangle ADB, \triangle ADC, \triangle AEB, \triangle AEC$. Then we have $$\frac{BD}{AB}=\frac{\sin x}{\sin \angle BDA},\frac{BE}{AB}=\frac{\sin y}{\sin \angle BEA},\frac{CD}{AC}=\frac{\sin y}{\sin \angle CDA},\frac{CE}{AC}=\frac{\sin x}{\sin \angle CEA}$$Multiplying the first two fractions, dividing them by the next two fractions, and using the fact that $\sin(180^{\circ}-\theta)=\sin \theta$, one gets the required equality. The converse can be proved in a similar way.

PROOF 2 (Inversion) Let $\omega$ be the circumcircle of $\triangle ABC$, and let $AD \cap \omega=P, AE \cap \omega =Q$. Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the internal angle bisector of $\angle BAC$. Then $D \rightarrow Q$ and $E \rightarrow P$. Thus it suffices to show that $$\frac{BP}{CQ}=\frac{CP}{BQ} \Leftrightarrow \frac{BP}{CP}=\frac{CQ}{BQ}$$As $\angle BPC=\angle BQC$, this is equivalent to proving that $\triangle BPC \sim \triangle CQB$. But this is obviously true, cause we have $PQ \parallel BC$, i.e. $BPQC$ is an isosceles trapezoid. The converse can be proved in a similar fashion.

So now let's move on to some questions which actually use this theorem. Remember that this theorem is just a part of these solutions (albeit an important one), and so most solutions using this theorem might require other lemmas and observations too. Having said that, let's move forward.

PROBLEM 1 (Source=Balamatda)
Let $ABC$ be a triangle inscribed in a circle $(O)$. Suppose that $AH$ is the altitude and the line $AO$ intersects $BC$ at $D$. The circumcircle of $ADC$ intersects the circumcircle of $AHB$ at $E$. The tangent at $A$ of $(O)$ meets $BC$ at $I$. Prove that $IA=IE$.
https://scontent.fdad3-3.fna.fbcdn.net/v/t1.0-9/42614647_2256121971285643_3789226474465132544_n.jpg?_nc_cat=100&oh=dddf8d8f46971fe51fc3e49a6899879e&oe=5C61361D

SOLUTION: We have $\angle BEH=\angle BAH=\angle CAO=\angle CED \Rightarrow ED$ and $EH$ are isogonal wrt $\angle BEC$. By Steiner's Ratio Theorem, we get that $$\left (\frac{AB}{AC} \right)^2=\frac{BH \cdot BD}{CH \cdot CD}=\left (\frac{BE}{CE} \right)^2 \Rightarrow \frac{AB}{AC}=\frac{BE}{CE}$$This means that $E$ lies on the $A$-Apollonius circle, giving that $IA=IE$ (as $I$ is the center of the $A$-Apollonius circle). $\blacksquare$

REMARK: The above solution also shows that the problem is true for any two isogonal lines.


PROBLEM 2 (ELMO 2016 P6)
Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$.

(a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$.

(b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$.

James Lin

SOLUTION: WLOG assume $AB \leq AC$. Let $I$ be the incenter of $\triangle ABC$.

Let $\omega$ be the circumcircle of $XSYT$, and $Z$ be the midpoint of $XY$, i.e. center of $\omega$. Also, Let $AI \cap BC = P$.

Let $DA$ meet the incircle again at $K$. Then $DEKF$ is harmonic $\Rightarrow -1=(K,D;E,F) \overset{D}{=} (A,P;X,Y)$.

This means that $A$ and $P$ are inverses w.r.t. $\omega$, i.e. $ZX^2=ZP \cdot ZA = PZ(PZ+PA) =PZ^2+PZ \cdot PA$

$\Rightarrow PZ \cdot PA = ZX^2-ZP^2 = (ZX-ZP)(ZX+ZP) = (ZX-ZP)(ZY+ZP) = PX \cdot PY = PS \cdot PT \Rightarrow ASZT$ is cyclic.

As $ZS = ZT$, $AZ$ is the internal angle bisector of $AS$ and $AT$ $\Rightarrow AS$ and $AT$ are isogonal.

(a) By Steiner's Ratio Theorem, $\frac{BT}{CT} \cdot \frac{BS}{CS} = \left(\frac{AB}{AC} \right)^2 \Rightarrow AC \cdot \sqrt{BS \cdot BT} - AB \cdot \sqrt{CS \cdot CT} = 0$. By the Converse of Casey's Theorem on point circles $\odot (A), \odot (B), \odot (C)$ and $\odot (AST)$, we get that $\odot (AST)$ and $\odot (ABC)$ are tangent to each other at $A$. $\blacksquare$

(b) Let $AI \cap EF = T$, then $\angle FTY = 90^{\circ} \Rightarrow \angle DYX = \angle FYT = 90^{\circ}-\angle EFD = \angle IDE = \angle IDX$

$\Rightarrow ID$ is tangent to $\odot (DXY) \Rightarrow ID^2 = IX \cdot IY \Rightarrow X$ and $Y$ are inverses w.r.t. the incircle.

Thus, $\omega$ and the incircle are orthogonal $\Rightarrow$ Length of tangent from $Z$ to $\omega$, i.e. $\ell$, is equal to $ZS$.

Now, $ZS \cdot ST = ZS(SD+TD) = ZS \cdot TD+ZT \cdot SD \Rightarrow ZS \cdot TD +ZT \cdot SD - ST \cdot \ell = 0$. By the Converse of Casey's Theorem on point circles $\odot (Z), \odot (S), \odot (T)$ and the incircle, and using the fact that $Z$ lies on $\odot (AST)$, we get that $\odot (AST)$ and the incircle are tangent to each other. $\blacksquare$

REMARK: The first part gives an important result, which has been stated more clearly below.


RESULT Let $D$ and $E$ be points on $BC$, so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$. Then $\odot (ADE)$ is tangent to $\odot (ABC)$ at $A$. The converse is also true.

PROOF: See the solution to the first part of ELMO 2016 P6 given above.


PROBLEM 3 (Sharygin 2018 Correspondence Round Problem 16)
Let $ABC$ be a triangle with $AB < BC$. The bisector of angle $C$ meets the line parallel to $AC$ and passing through $B$, at point $P$. The tangent at $B$ to the circumcircle of $ABC$ meets this bisector at point $R$. Let $R'$ be the reflection of $R$ with respect to $AB$. Prove that $\angle R'PB = \angle RPA$.

SOLUTION: (Bary bash :D) The attached solution is the one that I submitted during the actual contest.
Attachments:
Solution_Q 16.docx (72kb)
This post has been edited 3 times. Last edited by math_pi_rate, Feb 13, 2019, 4:18 PM
Steiner
ratio
isogonal
reflection
Inversion
overlay

Functional Equations: A Beauty in Themselves

by math_pi_rate, Sep 17, 2018, 2:56 PM

Even though the name of the blog suggests only geo (and I support it by heart as usual :P), but I'll post some of the FE's which I found out to be really interesting (Another reason for posting something apart from geo is to keep the interests of guys like kayak going #seemyshoutbox)

I'll start with IMO 2017 Problem 2, which is the most beautiful FE I have ever seen (Before doing this problem, remember that looks can be deceiving). First of all notice that the equation is symmetric. Also on seeing an FE on reals, one obvious thing is to plug in $x=y=0$, although I personally prefer plugging just $y=0$, as that helps build a condition on $f(x)$. This same trick can be seen in IMO 2010 Problem 1 also, but we'll return to that later. Anyway, so on plugging $y=0$, we get $f(f(x) \cdot f(0))+f(x) = f(0)$, which is not as helpful as we thought. So to make things easier we make cases on $f(0)$, i.e. when it is zero and when it is not (making these cases is one of the most important part of this question, and I remember using this trick in some other problem, although I can't remember which). Anyway, this trick of making cases is very important. (For eg. See here). Now $f(0)=0$ trivially gives $f(x)=0$. So now we focus on $f(0) \neq 0$. After some playing around with $f$, I couldn't get much, and so I decided to make another function $g$, defining $g(x)=f(x)-x$. This is another major trick (you can try comparing this trick with phantom points in geometry, oh sorry, forgot about no geo :D). It helps using this trick when you have guessed the functions (which is mostly the case in all FE's). Like, by now, I had guessed that the solutions are most probably $f(x)=x-1$ and $f(x)=1-x$. Then I noted that the second solution follows from the first, as if $f$ was a solution, then $-f$ will also be a solution. This brought me to a nice revelation, i.e. $f(0)=1$ can be assumed and $f(0)=-1$ will immediately give the other solution (as I had by now proved that $f(1)=0$ and $f(0)= \pm 1$). Now some manipulations with $g$ brought me to the necessary solution. Finally while posting the FE, I completely removed the mention of any such function $g$, thus making the solution in $f$ purely.

Before proceeding to the next FE, I'd like to mention another trick which I couldn't mention before. This is basically used to show that if $f(c)=k$, then $c=p$, where $k,c,p$ are constants. It has two main parts: First of all show that there exists some $x \in \mathbb{R}$ such that $f(x)=k$. Then assume that $f(c)=k$, and try to arrive at some sort of contradiction to prove that $c=p$. Remember that the first step is really important, although it is oft forgotten. You can see some examples which use this trick here and here.

Now, Let's go on to APMO 2016 Problem 5. This is an FE on positive reals, meaning that we no longer have zeroes at our service. I consider it to be one of the most difficult FE's to do on an exam, because of how time consuming it can be (If I remember correctly, it took me 4 hours to solve this one :oops:). We start with some obvious things, like proving injectivity and surjectivity, which any person having some basic knowledge in FE's should be able to do. Although there is a small room for mistake. One can easily prove that $f$ is surjective $\forall x> \frac{1}{2}$. But from this concluding that $f$ is surjective for all $x$ is a big mistake, something which I did while solving the problem, only to rectify it later. After this the problem actually flows on, slowing moving us to $f$ being involutive, multiplicative and finally additive, proving and using $f(1)=1$ and $f(2)=2$ in the way. In the end, all one needs to do is use the famous identity that if $f$ is multiplicative as well as additive, then either $f \equiv 0$ or $f(x)=x$. Plugging this into the original equation, we get the required solution.

Let's move on to the last FE of this topic: IMO 2008 Problem 4. This problem has a very short solution, but if not taken care of then one can easily make the huge mistake of ignoring the so-called Pointwise Value Trap (To learn what they are see this handout by v_Enhance). After getting that either $f(a)=a$ or $f(a)=\frac{1}{a} \text{ } \forall a \in \mathbb{R^+}$, which is quite easy to discover, one needs to check on the aforementioned Trap. I know of two methods to check: First is assuming that it is possible for the two values of $f(x)$ to hold for distinct numbers simultaneously, and then arriving at a contradiction. The second is to assume that if one of the solution holds for some number $t$, then it holds for all $x$ which lie in the domain of $f$. I personally prefer the first one, but the choice can depend from person to person and question to question. To understand it better, see the discussion in the beginning here. Also see my solution to IMOSL 2007 A4. It gives a nice insight on FE's in general.

Before concluding with this topic, I'll just post some tips.

1. Try making your solution presentable. Using lemmas while writing is a beautiful way to do this. See my solution to APMO 2016 Problem 5 for an example. (This can be used in geometry problems also)

2. After solving an FE, always remember to plug in the solution and check it, as this always carries some marks.

3. If an FE has more than one non-symmetric solutions, then always check for the Trap mentioned above.

4. Before starting an FE, always try to discover the solutions to it first. This helps a lot later on in how to move forward with the FE, and motivates what to plug into the FE. Basically check whether $f$ can be constant or not, and then try putting $f(x)=ax+b$ or $f(x)=ax^2+bx+c$ and try solving for $a,b,c$. Remember that mostly all Olympiad FE's have usually a max of 2 or 3 solutions, and which generally are of the above given forms only.

In the end, remember that FE's have room for a lot of creativity. So don't think too much before trying something out of the way, as it always helps while solving them.
This post has been edited 6 times. Last edited by math_pi_rate, Oct 16, 2018, 7:27 AM
Functional Equations
IMO
APMO
Pointwise Value Trap

Collinearity in a Quadrilateral with Perpendicular Diagonals

by math_pi_rate, Sep 10, 2018, 10:04 AM

PROBLEM (Source=ayan.nmath)
In a convex quadrilateral $ABCD,$ the diagonals are perpendicular to each other and they intersect at $E.$ Let $P\ne A$ be a point on the side $AD$ such that $PE=EC.$ $\odot(BCD)\cap AD=Q\ne A.$ The circle passing through $A$ and tangent to line $EP$ at $P,$ intersects $\overline{AC}$ at $R.$ Prove synthetically that if $\angle BCD=90^{\circ}$ then $B,R,Q$ are collinear.

SOLUTION 1 (My solution): Let $\odot (BCD) \cap \odot (BER) = T$ and $CA \cap \odot (BCD) = X$. Note that $\angle BTD = 90^{\circ}$ and $\angle BTR = \angle BER = 90^{\circ}$ $\Rightarrow D,T,R$ are collinear.

Also, $BD$ is the perpendicular bisector of $CX$ $\Rightarrow -1=(B,D;C,X) \overset{R}{=} (Q,T;X,C)$.

And, By Pascal's Theorem on $BTTDQQ$, we get $BT \cap DQ$ lies on $CX$ $\Rightarrow B,A,T$ are collinear.

Thus, As $AQBE$ is cyclic, so $DA \cdot DQ = DE \cdot DB = DT \cdot DR \Rightarrow AQRT$ is cyclic $\Rightarrow \angle AQR = \angle ATD = 90^{\circ} \Rightarrow B,Q,R$ are collinear.

SOLUTION 2 (Synthetic_Potato): From power of point, we see that $EA\cdot ER = EP^2= EC^2$. Thus if $C'$ is the reflection of $C$ over $E$, we see that $EC^2=EA\cdot ER$ implies $(C,C';A,R)=-1$.
Let $R'$ be $BQ\cap CC'$. It suffices to now show that $(C,C';A,R')=-1$. Let $BA\cap \odot (CBQC'D) = A'$. Now from radical axis theorem on cyclic quadrilaterals $EAQB, EAA'D, BQA'D$, we see that $R'\equiv BQ\cap DA'\cap CC'$. Thus from brokard's theorem on $BQA'D$, the polar of $R'$ passes through $A$. From the defination of Polar, $(R',A;C',C)=-1$. Thus $R=R'$ and so, $B,Q,R$ are collinear.
This post has been edited 2 times. Last edited by math_pi_rate, Dec 31, 2018, 11:01 AM
collinearity
geometry
perpendicular diagonals

Circumcenter lies on line parallel to BC

by math_pi_rate, Aug 20, 2018, 12:14 PM

Here is a problem I made a few days ago (which I'll later be using as a lemma).
LEMMA Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. Let $D$ be the midpoint of arc $\overarc{BC}$ (not containing $A$), and let $K$ be the antipode of $D$ in $\omega$. Let $AK \cap CD = T$. Finally let $O$ be the circumcenter of $\triangle ABT$. Then $OD$ is tangent to $\omega$.

Solution 1 (My solution, along with some motivation)

Solution 2 (Solution by TheDarkPrince)

Anyway, here's a problem that uses this lemma.
Problem (Polish MO 2018 P5): Point $O$ is a center of circumcircle of acute triangle $ABC$, bisector of angle $BAC$ cuts side $BC$ in point $D$. Let $M$ be a point such that, $MC \perp BC$ and $MA \perp AD$. Lines $BM$ and $OA$ intersect in point $P$. Show that circle of center in point $P$ passing through a point $A$ is tangent to line $BC$.

Solution
This post has been edited 6 times. Last edited by math_pi_rate, Sep 27, 2018, 12:58 PM
geometry
Circumcenter
external angle bisector
arc midpoint

A beautiful theorem about tangency of circles

by math_pi_rate, Aug 19, 2018, 2:22 PM

So a week before I went for the Sharygin Finals, I striked upon this beautiful theorem, called Casey's Theorem (also known as Generalised Ptolemy's Theorem). So here is the theorem, in all its might:

THEOREM (Casey) Given four circles $\omega_i, i=1,2,3,4$, let $t_{ij}$ denote the length of a common tangent (either internal or external) between $\omega_i$ and $\omega_j$. Then the four circles are tangent to a fifth circle $\Gamma$ (or line) if and only if for appropriate choice of signs, $t_{12} \cdot t_{34} \pm t_{13} \cdot t_{42} \pm t_{14} \cdot t_{23}=0$.

For a proof, see this handout by Luis Gonzales or have a look at Problem 239 and 240 of Problems in Plane Geometry by I. F. Sharygin.

Anyway, I'll just post a few problems which are reduced to mere computations by this theorem (mainly the if part).

PROBLEM 1 (Sharygin Finals 2017 Problem 9.4) Points $M$ and $K$ are chosen on lateral sides $AB,AC$ of an isosceles triangle $ABC$ and point $D$ is chosen on $BC$ such that $AMDK$ is a parallelogram. Let the lines $MK$ and $BC$ meet at point $L$, and let $X,Y$ be the intersection points of $AB,AC$ with the perpendicular line from $D$ to $BC$. Prove that the circle with center $L$ and radius $LD$ and the circumcircle of triangle $AXY$ are tangent.

SOLUTION

PROBLEM 2 (Tuymaada 2018 Senior/Junior League Problem 8) Quadrilateral $ABCD$ with perpendicular diagonals is inscribed in a circle with centre $O$. The tangents to this circle at $A$ and $C$ together with line $BD$ form the triangle $\Delta$. Prove that the circumcircles of $BOD$ and $\Delta$ are tangent.

SOLUTION

PROBLEM 3 (Source=buratinogigle) Let $ABC$ be a triangle inscribed in circle $(O).$ The circle $(X)$ passes through $A,O$ with $X$ is on perpendicular bisector of $BC.$ Similarly, we have the circles $(Y)$ and $(Z). $ Prove that the circle is tangent to $(X),$ $(Y)$ and $(Z)$ internally then is tangent to $(O)$.

SOLUTION

PROBLEM 4 (Source=buratinogigle) Let $ABC$ be a triangle inscribed in circle $(O)$ with altitude $AH$. Incircle $(I)$ touches $BC$ at $D$. $K$ is midpoint of $AH$. $L$ is projection of $K$ on $ID$. Prove that circle $(L,LD)$ is tangent to $(O)$.

SOLUTION

PROBLEM 5 (Source=buratinogigle) Let $ABC$ be a triangle with circumcenter $O$ and altitude $AH.$ $AO$ meets $BC$ at $M$ and meets the circle $(BOC)$ again at $N.$ $P$ is the midpoint of $MN.$ $K$ is the projection of $P$ on line $AH.$ Prove that the circle $(K,KH)$ is tangent to the circle $(BOC)$.

SOLUTION
This post has been edited 3 times. Last edited by math_pi_rate, Sep 27, 2018, 12:57 PM
Casey theorem
tangency of circles
length bash

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