(AEF) passes through a fixed point on A-median

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v_Enhance

6870 posts

v_Enhance

#1 h Jul 23, 2013, 1:31 AM • 9 Y

Y by Davi-8191, Amir Hossein, HamstPan38825, megarnie, Adventure10, Mango247, Rounak_iitr, GeoKing, Nari_Tom

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

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Luis González

4145 posts

Luis González

#2 h Jul 23, 2013, 4:28 AM • 13 Y

Y by Pirkuliyev Rovsen, aops29, AlastorMoody, amar_04, myh2910, Cindy.tw, veirab, Jyx, third_one_is_jerk, Adventure10, GeoKing, khina, Funcshun840

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 $AM.$

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vslmat

154 posts

vslmat

#3 h Jul 31, 2013, 1:14 PM • 8 Y

Y by isi2003, pablock, AlastorMoody, HolyMath, Han1728, Infinityfun, Adventure10, Mango247

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 $(AHK)$ ).

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thecmd999

2860 posts

thecmd999

#4 h Apr 23, 2014, 9:27 PM • 3 Y

Y by Tafi_ak, Adventure10, Mango247

Solution

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leader

339 posts

leader

#5 h Apr 23, 2014, 11:14 PM • 2 Y

Y by Adventure10, Mango247

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.

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leminscate

109 posts

leminscate

#6 h Jun 23, 2014, 3:30 AM • 6 Y

Y by MathPassionForever, myh2910, Tafi_ak, Adventure10, Mango247, k_rs52

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.

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Wolstenholme

543 posts

Wolstenholme

#7 h Jul 28, 2014, 4:41 PM • 3 Y

Y by Adventure10, Mango247, k_rs52

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.

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XmL

552 posts

XmL

#8 h Jul 28, 2014, 6:21 PM • 4 Y

Y by RC., gemcl, Adventure10, Mango247

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.

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andria

824 posts

andria

#9 h Apr 22, 2015, 4:59 PM • 4 Y

Y by microsoft_office_word, myh2910, Adventure10, Stuffybear

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

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Dukejukem

695 posts

Dukejukem

#10 h May 15, 2015, 11:10 PM • 4 Y

Y by FISHMJ25, zephyr7723, Adventure10, k_rs52

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. $\square$

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K6160

276 posts

K6160

#11 h Sep 14, 2016, 4:30 AM • 10 Y

Y by JasperL, tworigami, Pluto1708, myh2910, IAmTheHazard, Tafi_ak, RopuToran, Adventure10, Mango247, CyclicISLscelesTrapezoid

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.

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tenplusten

1000 posts

tenplusten

#12 h Mar 10, 2017, 1:09 PM • 2 Y

Y by Adventure10, Mango247

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!!!

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trumpeter

3332 posts

trumpeter

#13 h Apr 6, 2017, 10:18 PM • 3 Y

Y by anser, Adventure10, Mango247

Solution

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wu2481632

4233 posts

wu2481632

#16 h May 27, 2017, 3:26 AM • 2 Y

Y by Adventure10, Mango247

After a quick glance, it looks like no one else has used this inversion, so I will post my solution.

Solution

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reveryu

218 posts

reveryu

#17 h Jun 23, 2017, 5:50 PM • 1 Y

Y by Adventure10

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!

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Bigtaitus

72 posts

Bigtaitus

#49 h Jan 18, 2024, 9:59 PM • 1 Y

Y by Vahe_Arsenyan

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.

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Leo.Euler

577 posts

Leo.Euler

#50 h Jan 19, 2024, 12:42 AM

Y by

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.
:yoda:

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.

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MagicalToaster53

159 posts

MagicalToaster53

#51 h Feb 2, 2024, 1:47 AM

Y by

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 $(AEF)$ . $\blacksquare$

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Math4Life7

1703 posts

Math4Life7

#52 h Feb 19, 2024, 4:32 AM

Y by

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. $\blacksquare$

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Sleepy_Head

563 posts

Sleepy_Head

#53 h Apr 17, 2024, 3:13 PM • 1 Y

Y by GeoKing

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 $ACCDBB$ . $\blacksquare$
$[asy] import geometry; import olympiad; size(8cm,0); point A = (0.179,0.888); point B = (0,0); point C = (1,0); point D = (0.648,-0.238); circle w = circumcircle(A,B,C); line AB = line(A,B); line AC = line(A,C); line BD = line(B,D); line CD = line(C,D); point EE = intersectionpoint(AB,CD); point F = intersectionpoint(AC,BD); line Btan = tangent(w,B); line Ctan = tangent(w,C); point P = intersectionpoint(Btan,Ctan); line AP = line(A,P); point M1 = intersectionpoints(w,AP)[0]; point M = midpoint(B--C); point H = orthocenter(A,B,C); point Q = foot(H,A,M); draw(w); draw(B--C); draw(B--P--C--P--A); draw(A--EE--C); draw(A--F--B); draw(EE--F,dashed); draw(A--M,dashed); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,E); dot("$D$",D,S); dot("$E$",EE,SW); dot("$F$",F,SE); dot("$P$",P,S); dot("$M'$",M1,SW); dot("$M$",M,S); dot("$Q$",Q,NE); [/asy]$

proof desired point is A-HM point

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Aiden-1089

277 posts

Aiden-1089

#54 h May 16, 2024, 2:59 PM • 1 Y

Y by GeoKing

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. $\square$

This post has been edited 1 time. Last edited by Aiden-1089, May 16, 2024, 2:59 PM

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shendrew7

792 posts

shendrew7

#55 h Jun 18, 2024, 12:13 AM

Y by

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. $\blacksquare$

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shendrew7

792 posts

shendrew7

#56 h Jun 18, 2024, 12:34 AM

Y by

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. $\blacksquare$

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john0512

4175 posts

john0512

#57 h Aug 17, 2024, 9:42 PM

Y by

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.

This post has been edited 1 time. Last edited by john0512, Aug 17, 2024, 9:42 PM

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Eka01

204 posts

Eka01

#58 h Sep 7, 2024, 9:38 AM

Y by

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.

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cursed_tangent1434

559 posts

cursed_tangent1434

#59 h Sep 11, 2024, 1:04 AM

Y by

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.

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falcon311

100 posts

falcon311

#60 h Sep 26, 2024, 3:59 PM

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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.

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Saucepan_man02

1300 posts

Saucepan_man02

#61 h Nov 14, 2024, 3:50 PM

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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 $\triangle ABC$ .

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SimplisticFormulas

84 posts

SimplisticFormulas

#62 h Nov 14, 2024, 8:01 PM

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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. $\blacksquare$

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eg4334

617 posts

eg4334

#63 h Mar 8, 2025, 4:25 AM

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20 minute bary. Do it wrt to $\triangle ABC$ . Let $D = (0, m, 1-m)$ . The circle $ABD$ can be computed into $-a^2yz-b^2xz-c^2xy+(\lambda z)(x+y+z)=0$ for some $\lambda$ . Plugging in $D$ gives $\lambda = a^2m$ . Now letting $y=0$ gives us $x =\frac{\lambda}{b^2} = \frac{a^2m}{b^2}$ . Now this gives us $F = (a^2m : 0 : b^2-a^2m)$ and similarly $E = (a^2(1-m) : c^2-a^2(1-m): 0)$ . Then the circle $AEF$ can be written as $a^2yz+b^2xz+c^2xy = (x+y+z)(\lambda_1 y + \lambda_2 z)$ for some $\lambda_1, \lambda_2$ . Plugging in $F$ gives $\lambda_2 = a^2m$ and plugging in $E$ gives $\lambda_1 = a^2(1-m)$ . Now its not hard to check that the point $(a^2: b^2+c^2-a^2: b^2+c^2-a^2)$ , indepdent of $D$ , always lies on this circle because both sides equal $a^6 - 3a^4b^2 - 3a^4c^2 + 2a^2b^4 + 4a^2b^2c^2 + 2a^2c^4$ .

Remark: I checked $D = B, C$ and intersected those circles to get the point to plug in.

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