In $\triangle ABC$, a point $D$ lies on line $BC$. The circumcircle of $ABD$ meets $AC$ at $F$ (other than $A$), and the circumcircle of $ADC$ meets $AB$ at $E$ (other than $A$). Prove that as $D$ varies, the circumcircle of $AEF$ always passes through a fixed point other than $A$, and that this point lies on the median from $A$ to $BC$. Proposed by Allen Liu
Problem
Source: ELMO Shortlist 2013: Problem G3, by Allen Liu
Tags: geometry, circumcircle, trigonometry
23.07.2013 07:28
Let $P \equiv BF \cap CE.$ $\angle (FA,FB)=\angle (DA,DB)=\angle (EA,EC)$ $\Longrightarrow$ $A,E,P,F$ are concyclic. From the complete cyclic $AEPF,$ it follows that the polar of $B$ WRT $\odot(AEF)$ goes through $C$ and vice versa $\Longrightarrow$ $B,C$ are conjugate points WRT $\odot(AEF)$ $\Longrightarrow$ $\odot(AEF)$ is orthogonal to the circle $(M)$ with diameter $\overline{BC}.$ Now all circles through a fixed point $A$ and orthogonal to a fixed circle $(M)$ forms a pencil, the second common point of this pencil is the intersection of $AM$ with the polar of $A$ WRT $(M),$ which in fact is the projection of the orthocenter of $\triangle ABC$ on the median $AM.$
31.07.2013 16:14
We know that if $BF$ cuts $CE$ at $P$ then $P$ lies on the circle $(AEF)$. Let $H$ be the orthocenter of $\Delta ABC$ with altitudes $AI$ and $BK$, then $B, P, H, C$ are concyclic. Easy to see that the reflection of $A$ over $BC$ lies on this circle $(BHC)$. Notice that both $BDPE$ and $DCFP$ are cyclic. Let $(AEF)$ cuts $(BHC)$ at another point $M$ and the line through $A$ parallel to $BC$ cuts $(AEF)$ at $G$. As $\angle GPF = 180^{\circ} - \angle GAF = \angle BAC = \angle BED = \angle BPD$, $G, P, D$ are collinear. Let $GD, AM$ cut $(BHC)$ at $N, Q$, resp., then $QN\parallel GA\parallel BC$. As $\angle BCN = \angle BPD = \angle ACB$, $N$ is the reflection of $A$ over $BC$ and $N$ lies on the altitude $AI$. Let $AQ$ meets $BC$ at $S$; $I$ is the midpoint of $AN$, $S$ is the midpoint of $AQ$, buts as $BQ = CN = AC$, $S$ is the midpoint of $BC$. As $M$ also lies on $(BHC), M$ is a fixed point on the $A$-median of $\Delta ABC$. (Note: $M$ also lies on the circle $(AHK)$).
Attachments:
24.04.2014 00:27
24.04.2014 02:14
Here is a solution that uses $0$ creativity xD $b=AC,c=AB,a=BC$ Let $K$ be on $AC$ such that $\angle KBC=\angle BAC$ and let $X$ be on circle $ABK$ on arc $BK$ without $A$ such that $\frac {BX}{XK}=\frac {b}{c}$. Now by power of point $c\cdot BE=BD\cdot a,CK\cdot b=a^2,CF\cdot b=CD\cdot a$ now $BE=\frac {BD\cdot a}{c}$ and $KF=CK-CF=\frac{BD\cdot a}{b}$ so we get $\frac {BE}{KF}=\frac {b}{c}=\frac {BX}{XK}$ now along with $\angle XBE=\angle XKF$ we have that $BXE\sim XKE$ yeilding $\angle BEX=\angle XFA$ means $AEXF$ is cyclic. The only thing you need to do is notice $K$ when $D=B$ and just check that $\frac {BE}{KF}$ is fixed.
23.06.2014 06:30
Barycentrics WRT $\triangle ABC$ gives a 5 minute solution. The equation of $(AEF)$ is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$ where $u,v,w$ are the powers of $A,B,C$ WRT $(AEF)$, respectively. So $u=0, v=aBD, w=aCD$, by Power of a Point and the cyclic quadrilaterals $AEDB,AFDC$. Now let $y=z=1$ in the equation, which gives $-a^2-b^2x-c^2x+(x+2)a^2=0$ since $a(BD)+a(CD)=a^2$. We get a fixed value for $x$ so we're done.
28.07.2014 19:41
I will also provide a barycentric coordinate solution. Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ a = BC, b = CA, c = AB $. Let $ D = (0, r, 1 - r) $ for some real $ r $. We find that the equation of the circumcircle of $ \triangle{ABD} $ is given by: $ a^2yz + b^2xz + c^2xy = (x + y + z)a^2rz $ and it is easy to see that this intersects line $ AC $ at the point $ E = (a^2r : 0 : b^2 - a^2r) $. Similarly we find that $ F = (a^2(1 - r) : c^2 - a^2(1 - r) : 0) $. Now it is easy to find that the circumcircle of $ \triangle{AEF} $ has equation $ a^2yz + b^2xz + c^2xy = (x + y + z)(a^2(1 - r)y + a^2rz) $. It is easy to compute that regardless of the value of $ r $, the point $ (a^2 : b^2 + c^2 - a^2 : b^2 + c^2 - a^2) $ lies on this circumcircle, and moreover lies on the $ A $-median of $ \triangle{ABC} $ so we are done. The motivation for using barycentric coordinates is that despite the large number of circles in this diagram, all are defined by vertices or points on sides of $ \triangle{ABC} $ and so their equations are easy to deal with. Moreover, the condition that our fixed point lies on the relevant median implies that its barycentric coordinates will be nice.
28.07.2014 21:21
Let $DE,DF\cap (AEF)=E',F'$ again, since $\angle E'DC=\angle A=\angle EF'D$, therefore $EE'\parallel BC$ and $EE'\parallel FF'\parallel BC$ analogously. Let $BE'\cap (AEF)=G$, since $BE*AB=BG*BE'=BD*BC$, $G,E',C,D$ are concyclic $\Rightarrow \angle E'GC=\angle E'DC=\angle 180-\angle F'GE'\Rightarrow G\in CF'$. Now let $AG\cap BC=M$, since $\angle GCM=\angle FF'C=\angle FAG, \angle GBC=\angle BAG$, therefore by similarity $CM^2=GM*AM=BM^2\Rightarrow$ M is the midpoint of $BC$. The length of $GM$ dictates that $G$ is a fixed point and we are done.
22.04.2015 19:59
My solution: apply an inversion with center $A$ with radius $\sqrt{AB\cdot AC}$ and reflection WRT bisector of $\angle BAC$(assume that $ X'$ is inverse of $X$) then $B\longleftrightarrow C$ under the inversion and $D'$ lies on the arc $ BC$ and $(C, D',E')$ , $(B, D', F')$ are collinear. let $AD'\cap BC=S$ and the tangents from $B,C$ to $\odot (\triangle ABC)$ meet at $R$ because $ E'F'$ is a polar of $S$ ;$E'F'$ passes throw point $R$ which is fixed. So $\odot (\triangle AEF)$ passes throw the inverse of $R$ which lies on the median $M_a$ DONE
16.05.2015 02:10
Let $\mathcal{I} : X \mapsto X'$ be the composition of an inversion with center $A$ and radius $\sqrt{bc}$, combined with a reflection in the $A$-angle bisector. It is easy to see that $B' \equiv C$ and $C' \equiv B.$ Hence, $\mathcal{I}$ takes line $BC$ to $\omega \equiv \odot (ABC).$ It follows that $D' \in \omega$, and from basic inversive properties (circles through the center of inversion are sent to lines not passing through the center of inversion), we find that $E' \equiv A'B' \cap C'D'$ and $F' \equiv A'C' \cap B'D'.$ Now, note that the image of $\odot (AEF)$ is the line $E'F'$, and the image of the $A$-median is the $A$-symmedian. Therefore, to prove that $\odot (AEF)$ passes through a fixed point on the $A$-median, it suffices to prove that $E'F'$ passes through a fixed point on the $A$-symmedian. To see this, let the tangents to $\omega$ at $X$ (note that $X$ is a fixed point, regardless of where $D'$ lies on $\omega$). We will prove that $X$ is the desired point that we seek. First, recall that $X$ lies on the $A$-symmedian (Lemma 1), as needed. Second, by applying Pascal's Theorem to cyclic "hexagon" $AC'C'D'B'B'$, it follows that $F', X, E'$ are collinear, i.e. $E'F'$ passes through $X.$ This completes the proof. $\square$
14.09.2016 07:30
Define $\mathbb{P}(P, \Gamma_1, \Gamma_2) = \mathbb{P}(P, \Gamma_1) - \mathbb{P}(P, \Gamma_2)$, where $\mathbb{P}$ denotes the power of a point. It is not hard to check that this function is linear. Let $\omega$ and $\omega_1$ denote the circumcircles of $\triangle ABC$ and $\triangle AEF$. It suffices to show that $\mathbb{P}(M, \omega_1)$ is constant. We compute \begin{align*} \mathbb{P}(M, \omega_1, \omega) &= \frac{1}{2}(\mathbb{P}(B, \omega_1, \omega) + \mathbb{P}(C, \omega_1, \omega)) \\ &= \frac{1}{2}(BE\cdot BA - 0 + CF\cdot CA - 0) \\ &= \frac{1}{2}(BD\cdot BC + CD\cdot BC)\\ &= \frac{BC^2}{2}. \end{align*}So $$\frac{BC^2}{4}-\mathbb{P}(M, \omega_1) = \mathbb{P}(M, \omega_1, \omega)=\frac{BC^2}{2}.$$Therefore, $\mathbb{P}(M, \omega_1)$ is constant as desired.
10.03.2017 16:09
Here is my solution. I will do bary bash. Actually ELMO 2013 G3 wrote: and that this point lies on the median from $A$ to $BC$. this sentence is just a hint for barycentric. If problem asks to prove that there is fixed point on that circle. The problem would be more difficulter for bary bash. Let $A=(1,0,0)$,$B=(0,1,0)$,$C=(0,0,1)$ Since $D$ is on $BC$ we have $D=(0,t,1-t)$.Intersecting $(ABD)$ and $AC$ we get $F=(a^2t,0,b^2-a^2t)$ Again Intersecting $(ADC)$ and $AB$ we get $E=(a^2(1-t),c^2-a^2(1-t),0)$.Now we have that $(AEF)$ has equation $a^2yz+b^2zx+c^2xy=(x+y+z)(a^2(1-t)y+a^2tz)$ Intersecting median $AM$ of $ABC$ and $(AEF)$ we get $(AEF)\cap AM=(a^2,b^2+c^2-a^2,b^2+c^2-a^2)$ which is fixed. So Done!!!
07.04.2017 01:18
27.05.2017 06:26
After a quick glance, it looks like no one else has used this inversion, so I will post my solution.
23.06.2017 20:50
Dukejukem wrote: we find that $E' \equiv A'B' \cap C'D'$ and $F' \equiv A'C' \cap B'D'.$ Isn't A was sent to a point at infinity? so,what did these intersection mean? thankyou!
23.06.2017 23:27
Let $M$ be midpoint of $BC$. Denote $BD=t$. We can find that $BF=\frac{at}{c}$, so $AF=\frac{c^2-at}{c}$. Similarly, $AE=\frac{b^2-a(a-t)}{b}$. Let $(AEF)$ intersect A-median at $G$. Let diameter of $(AEF)$ be $1$. By Ptolemy's, $AEsinBAM+AFsinCAM=AGsinBAC$. We know that $\frac{sinBAM}{b}=\frac{sinCAM}{c}=n$. Rearranging, we get $AG=\frac{n(b^2+c^2-a^2)}{sinBAC}$, and hence $G$ doesn't depend on $D$.
09.03.2018 18:48
Invert around $A$ with radius $\sqrt{bc}$, and reflect the figure around the angle bisector of $\angle A$. Let $X^*$ denote image any point $X$ under this transformation, and let $BB \cap CC=K$ It is required to show there exists a unique point on $A-$symmedian that is always collinear with $E^*$ and $F^*$. By Pascal's theorem on $ABBD^*CC$, $E^*,F^*$ and $K$ are collinear. Since $AK$ is $A-$symmedian, we are done. $\square$
08.07.2018 19:58
Diagram
22.07.2018 11:29
Am I missing something because G1 looks harder to me? Anyways... Perform a $\sqrt{bc}$ inversion followed by reflection along the angle bisector of $\angle BAC$. I now claim the desired point is the unique point $X$ on the $A$-symmedian such that $XB, XC$ are tangent to $\omega_{ABC}$ (it's well known that $X$ satisfying this condition lies on $A$-symmedian). Now the inverted problem reads Quote: In $\Delta ABC$, let $D$ be a varible point on the arc $BC$ not containing $A$. $E_D :\overset{\text{def}}{=} BD \cap AC$ and let $F_D :\overset{\text{def}}{=} AB \cap CD$. Let $X$ be the point (on the $A$-symmedian) where the tangents from $B$ and $C$ from $\omega_{ABC}$ meet. Then $X, E_D, F_D$ are colinear Let $G :\overset{\text{def}}{=} AC \cap BD$. By Brocard, $\Delta EFG$ is self-polar w.r.t $\omega_{ABC}$, and hence the pole of $EF$ is $G$. On the other hand, clearly the polar of $X$ w.r.t $\omega_{ABC}$ is $BC$. As clearly $G \in BC$ so we finally get $$G \in BC \Leftrightarrow \text{Pole of EF} \in \text{Polar of X} \overset{\text{La-Hire}}{\Leftrightarrow} X \in EF \Leftrightarrow E, F, X \text{colinear} \blacksquare $$
06.11.2018 05:05
[asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -14.746637581395362, xmax = 31.396346604651207, ymin = -14.510337674418569, ymax = 16.20171348837207; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-2.84,5.09)--(-4.86,-5.25), linewidth(2) + wrwrwr); draw((-4.86,-5.25)--(7.16,-5.21), linewidth(2) + wrwrwr); draw((7.16,-5.21)--(-2.84,5.09), linewidth(2) + wrwrwr); draw(circle((-2.5978372866007646,-0.3246197950741255), 5.420032315860253), linewidth(2) + wrwrwr); draw(circle((3.4073187570611343,1.1509890845253732), 7.385445053925736), linewidth(2) + wrwrwr); draw(circle((-0.3266219383926623,1.8804436497881094), 4.076557523910895), linewidth(2) + wrwrwr); draw(circle((-2.8261034088530366,0.9140743603368577), 4.175948762046959), linewidth(2) + wrwrwr); draw((-5.940877778631267,-10.782810015369954)--(-4.86,-5.25), linewidth(2) + wrwrwr); draw((-2.84,5.09)--(1.15,-5.23), linewidth(2) + wrwrwr); draw((-2.84,5.09)--(-0.30294380615529565,-5.234835087541282), linewidth(2) + wrwrwr); draw((1.3475879470134404,0.7767844145761567)--(-5.940877778631267,-10.782810015369954),linewidth(2) + wrwrwr); draw((-4.410597499008975,-2.9495931384914815)--(4.575253560782421,-2.5477111676058937),linewidth(2) + wrwrwr); /* dots and labels */ dot((-2.84,5.09),dotstyle); label("$A$", (-2.7314839069767416,5.3850788837209285), NE * labelscalefactor); dot((-4.86,-5.25),dotstyle); label("$B$", (-4.739003348837209,-4.952148093023238), NE * labelscalefactor); dot((7.16,-5.21),dotstyle); label("$C$", (7.27615032558141,-4.922185116279052), NE * labelscalefactor); dot((-0.30294380615529565,-5.234835087541282),dotstyle); label("$D$", (-0.2745198139534828,-4.952148093023238), NE * labelscalefactor); dot((-3.863131891834357,-0.14721968394418253),linewidth(4pt) + dotstyle); label("$E$", (-3.7502251162790685,0.08163200000000838), NE * labelscalefactor); dot((2.807261975181314,-0.7266798344367534),linewidth(4pt) + dotstyle); label("$F$", (2.931518697674428,-0.4876645581395254), NE * labelscalefactor); dot((-0.027241070881577307,-2.1851058016296045),linewidth(4pt) + dotstyle); label("$X$", (0.08503590697675023,-1.9558504186046388), NE * labelscalefactor); dot((1.15,-5.23),linewidth(4pt) + dotstyle); label("$M$", (1.2835549767441936,-4.982111069767424), NE * labelscalefactor); dot((1.3475879470134404,0.7767844145761567),linewidth(4pt) + dotstyle); label("$P$", (1.46333283720931,1.0104842790697741), NE * labelscalefactor); dot((-4.410597499008975,-2.9495931384914815),linewidth(4pt) + dotstyle); label("$Q$", (-4.289558697674418,-2.7049248372092887), NE * labelscalefactor); dot((-0.9030289700871873,-2.792721124473026),linewidth(4pt) + dotstyle); label("$D'$", (-0.5741495813953436,-3.5438881860464964), NE * labelscalefactor); dot((4.575253560782421,-2.5477111676058937),linewidth(4pt) + dotstyle); label("$F'$", (4.699334325581407,-2.3154061395348706), NE * labelscalefactor); dot((-5.940877778631267,-10.782810015369954),linewidth(4pt) + dotstyle); label("$E'$", (-6.7165598139534906,-10.375446883720901), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P$ and $Q$ be the feet of the altitudes from $B$ and $C$ to their respective opposite sides in $\triangle ABC$. Let $X=AM\cap (APQ)$ where $M$ is the midpoint of $BC$. It is well known that $(QXMD)$ cyclic, so the inversion $\psi$ at $A$ with power $AQ\cdot AB=AP\cdot AC$ swaps $(X,M)$. Note $\psi(E)=D'P\cap AB\equiv E'$ and similarly $F'\equiv QD'\cap AC=\psi(F)$. Now, we will show that $(AEFX)$ cyclic. Applying $\psi$, we just want to show that $E'MF'$ collinear, but this follows from the fact that $M$ is the pole of $PQ$ wrt $(APQ)$ and Brokard on $APD'Q$.
06.10.2023 19:37
We will use barycentric coordinates. First, we compute the coordinates of $F$. To do this, we need the equation of $(ABD)$. Let $D=(0,p,q)$ with $p+q=1$. Clearly, since it passes through $A$ and $B$, $u=v=0.$ Plugging in $D$ gives $$wq=a^2pq$$$$w=a^2p.$$Thus, the equation of $(ABD)$ is $$(x+y+z)(a^2pz)=a^2yz+b^2xz+c^2xy.$$Letting $y=0$ (line $AC$), we have $$(x+z)(a^2p)=b^2x.$$Thus, the ratio of $x$ to $x+z$ is $a^2p$ to $b^2$, so we have $$F=(a^2p:0:b^2-a^2p).$$ By symmetry, we have $$E=(a^2q:c^2-a^2q:0).$$ Now, we find the equation of $(AEF)$. Clearly, $u=0$ since it passes through $A$. Plugging in the point $F$, we have $$b^2(w(b^2-a^2p))=b^2a^2p(b^2-a^2p)$$$$w=a^2p.$$Similarly $$v=a^2q.$$Thus, the equation of $(AEF)$ is $$(x+y+z)(a^2qy+a^2pz)=a^2yz+b^2xz+c^2xy.$$We will intersect this with the A-median by setting $y=z=1$. This becomes $$(x+2)(a^2q+a^2p)=a^2+b^2x+c^2x$$$$a^2(x+2)(p+q)=a^2+b^2x+c^2x.$$However, since $p+q=1$, this is just $$a^2x+2a^2=a^2+b^2x+c^2x$$so $$x(a^2-b^2-c^2)=-a^2$$$$x=\frac{a^2}{b^2+c^2-a^2}.$$This means that the intersection of $(AEF)$ with the A-median is $$(\frac{a^2}{b^2+c^2-a^2}:1:1).$$Since this does not depend on $p$ and $q$, this is the desired fixed point, hence done.
19.01.2024 00:59
We will show the desired point is the $A-Humpty$. Take $\sqrt{bc}$ inversion. Now it is easy to check that the problem turns into the following. Problem. Let $AE'F'$ be a triangle and $B,C$ be points in $AF'$, $AE'$ such that $\angle (BE', CF')=180-\angle BAC$. Show that the intersection of the tangents to $ABC$ at $B$ and $C$ intersects at $E'F'$. This is just straightforward by Brokards Theorem on $ABC$, which ends the problem.
19.01.2024 03:42
postfarming ig. Lemma: For any $\bullet$, we have \[ \text{Pow}(\bullet, (AEF))-\text{Pow}(\bullet, (AEC))=\text{Pow}(\bullet, (AFB))-\text{Pow}(\bullet, (ABC)). \]Proof. Let $f(\bullet)$ denote LHS and $g(\bullet)$ denote the RHS. Note that $f$ and $g$ are linear functions. It suffices to show that $f(\bullet)=g(\bullet)$ holds for three distinct non-collinear positions of $\bullet$, since every point in the plane can be described a linear combination of those points. This task is easy: choose $\bullet = A, E, F$. It is easy to verify that each of these points all work, by the means of radical axis. It suffices to show that $\text{Pow}(M, (AEF))$ is constant as point $D$ varies, where $M$ is the midpoint of side $BC$. By the lemma, we can reduce this to proving that $\text{Pow}(M, (AEC))+\text{Pow}(M, (AFB))$ is constant. Let $f(\bullet) := \text{Pow}(M, (AEC))-\text{Pow}(M, (ABC))$ and $g(\bullet) := \text{Pow}(M, (AFB))-\text{Pow}(M, (ABC))$. It is well-known that $f$ and $g$ are linear, so \begin{align*} \text{Pow}(M, (AEC))+\text{Pow}(M, (AFB)) &= \frac{f(B)+g(B)+f(C)+g(C)}{2} \\ &= \frac{\text{Pow}(B, (AEC))+\text{Pow}(C, (AFB))}{2} \\ &= \frac{\text{Pow}(B, (ADC))+\text{Pow}(C, (ADB))}{2} \\ &= \frac{BD \cdot BC + DC \cdot BC}{2} \\ &= \frac{BC^2}{2}, \end{align*}which is indeed constant.
02.02.2024 04:47
We are prompted to invert with respect to $A$ as there are many circles that contain $A$ as a common point. Inverting about $A$, we let $X'$ denote the image of a point $X$ under this inversion. Let $R = AD' \cap B'C'$. Then the polar of $R$ is $E'F'$, and as $B'C'$ passes through $R$, we have the pole of $B'C'$ lies on the polar of $R$ which is $E'F'$. Thus $E', P', F'$ are collinear, where $P'$ is the pole of $B'C'$. Now observe that the pole of $B'C'$ is just $B'B' \cap C'C' = P'$, which is really the symmedian $AP'$ of triangle $AB'C'$. Then under inversion $P' \mapsto P$, as the symmedian is mapped to the median of $A$ under inversion with $A$. Hence $P$ is invariant with regards to $(AEF)$. $\blacksquare$
19.02.2024 07:32
Very simple with bary + trivial synthetic observations We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. We let $D = (0, m, n)$ with $m+n = 1$. Notice that since $\triangle ABC \sim \triangle DEB$ we can see that $E = (ma^2:0:b^2-ma^2)$. Similarly $F = (na^2:c^2-na^2:0)$. Now we consider $(AEF)$. We get that $u = 0$, $v = na^2$, and $w = mb^2$. Now we can see that the point $\left(\frac{a^2}{2S_a}:1:1\right)$ always lies on $(AEF)$ and obviously this lies on the median. $\blacksquare$
17.04.2024 18:13
Perform a force-overlaid inversion about $A$ in $ABC$. Then $ABC$ maps to itself with $B$ and $C$ swapping, and $\overline{BC}$ and $(ABC)$ are swapped. Furthermore, we have that the $M$ of $BC$ maps to $\overline{AM} \cap (ABC)$ under the inversion, so after reflection $M$ maps to the intersection of the $A$-symmedian and $(ABC)$, denoted $M'$ below. Thus, it is equivalent to solve this following: "$ABC$ is a triangle and $D$ is a variable point on $(ABC)$. Let $E=\overline{AB} \cap \overline{CD}$, $F=\overline{AC} \cap \overline{BD}$, $P$ be the intersection of the tangents at $B$ and $C$ to $(ABC)$, and $M'=\overline{AP} \cap (ABC) \neq A$. Prove that $P=\overline{EF} \cap \overline{AM}$ for any choice of $D$." This is sufficient because $P$ reflecting over the $\angle BAC$ bisector maps $P$ to a point on the median, and $P$ is fixed so its image after reflection and inversion must be as well. However, this result follows immediately by Pascal's on $ACCDBB$. $\blacksquare$
16.05.2024 17:59
Let $M$ be the midpoint of $BC$, $X=AM \cap (AEF)$ such that $X \neq A$. Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, such that $f(P)=Pow_{(AEF)}(P)-Pow_{(ABC)}(P)$, where $Pow_\omega (P)$ denotes the power of point $P$ to circle $\omega$. By linearity of pop, we have $2f(M)=f(B)+f(C)$. Computing, we have $f(B)+f(C) = BA \cdot BE + CA \cdot CF = (BD+CD) \cdot BC = BC^2$ and $f(M)=MA \cdot MX + \frac{BC^2}{4}$. It follows that $MX = \frac{BC^2}{4MA}$, so $X$ is a fixed point. $\square$
18.06.2024 03:13
Define the function $f(X) = \operatorname{pow}(X,A) - \operatorname{pow}(X,(AEF))$ for all points $X$, where $A$ represents the point circle $A$. It suffices to show \begin{align*} f(M) &= \frac 12 f(B) + \frac 12 f(C) \\ &= \frac 12 \left(AB^2+AC^2-BE \cdot BA - CF \cdot CA\right) \\ &= \frac 12 \left(AB^2+AC^2-BD \cdot BC - BE \cdot BC\right) \\ &= \frac 12 \left(AB^2+AC^2-BC^2\right) \end{align*} is fixed, where $M$ is the midpoint of $BC$, as this implies $\operatorname{pow}(M,(AEF))$ is fixed. $\blacksquare$
18.06.2024 03:34
We claim the desired point is $H_A$, the $A$-Humpty point. Force overlay at $A$ with radius $\sqrt{AB \cdot AC}$. Then $D^*$ is a point on $(ABC)$, $E^* = BD^* \cap AC$, $F^* = CD^* \cap AB$, and $H_A^* = BB \cap CC$. Hence Pascal on $ABBD^*CC$ says $H_A^* \in E^*F^* = (AEF)^*$, as desired. $\blacksquare$
18.08.2024 00:42
me, a year ago, post 48 wrote: We will use barycentric coordinates. noooooooooo We claim that the answer is the $A$-humpty point, which inverts to the intersection of the tangents to $B$ and $C$. Perform a $\sqrt{bc}$ inversion. The problem now becomes: In triangle $\triangle ABC$, $D$ is a point on the circumcircle. Let $AC$ and $BD$ meet at $E$, and let $AB$ and $CD$ meet at $F$. Let $T$ be the intersection of the tangents at $B$ and $C$. Show that $E,F,T$ are collinear. Let $P$ be the intersection of $AD$ and $BC$. By Brocard, $EF$ is the polar of $P$. Since $P$ clearly lies on the polar of $T$ which is $BC$, $T$ lies on the polar of $P$, which is $EF$, done.
07.09.2024 12:38
Taking $D$ to be the foot of $A$ altitude, we get that the fixed point, if it exists, must be $A$ humpty point. We now $\sqrt{bc}$ invert. Inverted problem wrote: In cyclic quadrilateral $ABDC$, prove that $CD \cap AB=E, BD \cap AC=F, BB\cap CC=T $ are collinear. Now since the problem is purely projective, the result can follow by numerous ways. I list three: Pascal's on $ABBDCC$. Taking a homography sending $AD \cap BC=X$ to the center of the circle. By brokard's theorem, $EF$ is the polar of $X$ and obviously $BC$ is the polar of $T$ so the required follows by La Hire's Theorem.
11.09.2024 04:04
We present two approaches. In my opinion it makes sense to just barybash here since it's so clean. We set $A=(1,0,0)$ and etc. Then, let $M$ be the midpoint of $BC$. We know, $M=(0:1:1)$. Let $D=(0:r:s)$ for $r,s\in \mathbb{R}$. We consider the equation of circle $(ABD)$, \[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]Since this circle passes through $A$ and $B$, we have that $u=0$ and $v=0$. Further, this it passes through $D$, \begin{align*} -a^2yz - b^2xz -c^2xy + wz(x+y+z) &= 0\\ -a^2rs &= ws(r+s)\\ w &= \frac{a^2r}{r+s} \end{align*}Now, let $F=(t_1:0:t_2)$. Since $F$ is given to be on $(ABD)$, we also have that \begin{align*} -a^2yz - b^2xz - c^2xy + \frac{a^2r}{r+s}z (x+y+z) &= 0\\ -b^2t_1t_2 + \frac{a^2r}{r+s}t_2(t_1+t_2) &= 0\\ t_2 &= \frac{b^2r-a^2r+b^2s}{a^2r} \end{align*}Thus, $F=(a^2r:0:b^2r-a^2r+b^2s)$. Via an exactly similar calculation, we also have that $E=(a^2s:c^2r+c^2s-a^2s:0)$. Now, we similarly consider the equation of the circle $(AEF)$, \[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]Since this circle passes through $A$, $u=0$. Since $F$ must lie on this circle, we have that \begin{align*} -a^2yz-b^2xz-c^2xy + (vy+wz)(x+y+z) &=0\\ -b^2xz + wz(x+z) &= 0\\ b^2x &= w(x+z)\\ a^2b^2r &= w(b^2r+b^2s\\ w &= \frac{a^2r}{r+s} \end{align*}with a similar calculation we can also see that $v= \frac{a^2s}{r+s}$. Now, let $P$ be the intersection of $(AEF)$ and $AM$. Clearly $P=(t:1:1)$. Now, plugging this into our circle equation we have \begin{align*} -a^2-b^2t-c^2t + (v+w)(t+2) &=0\\ -a^2 - b^2t -c^2t + a^2(t+2) &=0\\ t &= \frac{a^2}{b^2+c^2-a^2} \end{align*}Thus, $P=(a^2 : S_A : S_A)$. But clearly, this is a fixed point which does not depend on the position of $D$ along the $BC$ line segment. Thus, the desired result must be true. For those of you allergic to bash, we present the following equally straightforward synthetic solution. Let $P$ denote the intersection of lines $\overline{BF}$ and $\overline{CE}$, and let $H_A$ denote the $A-$Humpty Point in $\triangle ABC$. We can start off by making the following simple observation. Claim : Point $P$ lies on circles $(AEF)$ and $(BH_AC)$. Proof : First of all note that, \[\measuredangle CPB = \measuredangle PCB + \measuredangle CBP = \measuredangle ECD + \measuredangle DBF = \measuredangle EAD + \measuredangle DAF = \measuredangle EAF \]Thus, $\measuredangle EPF = \measuredangle CPB = \measuredangle EAF$, so $P$ lies on $(EAF)$. Further, $\measuredangle CPB = \measuredangle EAF = \measuredangle BAC$ which implies that $P$ lies on $(BHC)$ ($H$ is the orthocenter of $\triangle ABC$). We also know that $H_A$ lies on $(BHC)$ which implies that $P$ lies on $(BH_AC)$ as desired. Now, let $H_A' = (BPC) \cap (AEF)$. Note that, \[\measuredangle BCH_A' = \measuredangle FPH_A' = \measuredangle FAH_A' = \measuredangle CAH_A'\]and similarly, $\measuredangle H_A'BC = \measuredangle H_A'AB$. But, this means $H_A'$ lies on the intersection of the two circles passing through $A$ and tangent to side $BC$ at $B$ and $C$ respectively. We know that this point is simply the $A-$Humpty Point $H_A$ so $H_A' \equiv H_A$. Thus, the point $H_A$ lies on $(AEF)$, implying that the circumcircle of $AEF$ always passes through a fixed point other than $A$, and since it is well known that $H_A$ lies on the $A-$median, we are done.
26.09.2024 18:59
People would have posted this already multiple times, but I'm gonna do this. Do $\sqrt{bc}$ inversion. Now A, B, C, D lie on the circle Apply pascals on ABBDCC to get that E, F, and the tangents from B and C intersect at a fixed point. The tangents from B and C intersect on the symmedian. When reflected across the angle bisector, this is the median.
14.11.2024 18:50
Normal inversion suffice: Invert around point $A$ with variable radius. Then: it suffice to show that $E'F'$ passes though a fixed point. As $D'$ varies on $(AB'C'$), let $R = AD'\cap B'C'$. Then $R$ lies on $B'C'$ implies $E'F'$ passes through point $X' = B'B' \cap C'C'$ (due to Brokard's theorem). Thus, $AX'$ is the symmedian of $\triangle AB'C'$ which implies line $AX$ is median of $\triangle ABC$.
14.11.2024 23:01
Let the circle passing through $A$ and $B$ and tangent to $BC$ meet $(AEF)$ in $X \neq A$. Let $BX$ meet $(AEF)$ in $H \neq X$. Let $EF$ meet $BC$ in $G$. Let $DF $ meet $(AFE)$ in $H'$ Let $EX$ meet $BC$ in $Y$. Let $AX$ meet $BC$ in $M$. Observe that $\angle XBG=\angle XAB=\angle EAX$ $\implies \angle BHA=\angle XHF=\angle XAF= \angle A-x=\angle H'AG-\angle H'BA=\angle BH'A$ $\implies H' \equiv H$ Note that by Power of Point, $BD \cdot BC=BE \cdot BA=BX \cdot BH=$, so $DXHC$ is cyclic. Hence, observe that $\angle A =\angle FAC= \angle HAC=\angle HXC$ Also, $\angle YXC=180-\angle HXC-\angle YXB=180-\angle A-\angle EXH=180-\angle A-(180-\angle XHF-\angle FEH-\angle EHX)=\angle EGC $ (i havent written the full angle chase but its easy and i have outlined it) Hence $EXCG$ is concyclic $\implies BCG=\angle XEF=\angle XAF$ $\implies (AXC)$ is tangent to $BC$ $MB^2=MX \cdot MA=MC^2$ $\implies M$ is the midpoint of $BC$, and the proof follows upon noting that $X$ is fixed for a fixed triangle. $\blacksquare$