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

Comment

6 Comments

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
You forgot about this problem!!!
USA Winter Team Selection Test #2 for IMO 2018, Problem 2 wrote:
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
  • ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
  • $\angle BEN = \angle MED$.

Proposed by Evan Chen

Thanks!! I am sure readers would be interested in solving this problem too with the help of the above configuration :)
This post has been edited 2 times. Last edited by math_pi_rate, Mar 13, 2020, 8:21 PM

by Aryan-23, Jan 30, 2020, 6:31 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Sir I think there is a typo in Property 1. Thanks.

Fixed. Thanks :)

Also you don't have to call me "Sir" each time you mention me :P
This post has been edited 2 times. Last edited by math_pi_rate, Jan 30, 2020, 6:48 PM

by amar_04, Jan 30, 2020, 6:32 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
@above OK. :P, but where did you change? I meant that part $T$ is the foot of perpendicular from $D_1$ to $BC$. $T$ and $D$ becomes same isn't it?

Sorry! I corrected some other typos :oops:. Anyway, edited now.
This post has been edited 1 time. Last edited by math_pi_rate, Jan 31, 2020, 6:05 AM

by amar_04, Jan 30, 2020, 6:55 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hau so Pro. This is a really nice configuration.

Thanks :)
This post has been edited 1 time. Last edited by math_pi_rate, Mar 13, 2020, 8:22 PM

by GeoMetrix, Mar 1, 2020, 3:31 PM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Here's another trivial property
Let $DX_a$ intersect $EF$ in $Y_a$.Define $Y_b$ and $Y_c$ similarly.Then $AY_a$,$BY_b$ and $CY_c$ concur on the trillinear pole of radical axis of incircle and circumcircle of $\triangle ABC$

by mmathss, Mar 3, 2020, 7:18 AM

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
This is also quite nice:-

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incircle $\gamma$. Let $A'$ be the mixtilinear excircle touch point, and let $A'B'$ and $A'C'$ be tangent to $\gamma$ with $B',C'\in\Gamma$. Let $X$ be the tangency point of $B'C'$ with $\gamma$ which exists by Poncelet's Porism. Show that $(XBC)$ is tangent to $\gamma$.

by math_pi_rate, Mar 13, 2020, 8:23 PM

Geo Geo everywhere, nor a point to see.

avatar

math_pi_rate
Shouts
Submit

  • First comment of January 26, 2025!

    by Yiyj1, Jan 27, 2025, 1:02 AM

  • And here's the first of 2025!

    by CrimsonBlade273, Jan 9, 2025, 6:16 AM

  • First comment of 2024!

    by mannshah1211, Jan 14, 2024, 3:01 PM

  • Wowowooo my man came backkkk

    by HoRI_DA_GRe8, Nov 11, 2023, 4:21 PM

  • Aah the thrill of coming back to a dead blog every year once!

    by math_pi_rate, Jul 28, 2023, 8:43 PM

  • kukuku 1st comment of 2023

    by kamatadu, Jan 3, 2023, 7:32 PM

  • 1st comment of 2022

    by HoRI_DA_GRe8, May 25, 2022, 8:29 PM

  • 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 :oops:

    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

68 shouts
Tags
geometry
Inversion
altitude
ex-point
humpty point
orthocenter
APMO
arc midpoint
Casey theorem
Circumcenter
collinearity
external angle bisector
Functional Equations
IMO
IMOSL 2002 G7
IMOSL 2011 G4
isogonal
length bash
overlay
perpendicular diagonals
Pointwise Value Trap
queue-point
ratio
reflection
Steiner
tangency of circles
About Owner
  • Posts: 1218
  • Joined: May 25, 2018
Blog Stats
  • Blog created: Aug 19, 2018
  • Total entries: 9
  • Total visits: 25935
  • Total comments: 30
Search Blog
a