The reflection of AD intersect (ABC) lies on (AEF)

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alifenix-

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alifenix-

#1 h Jan 27, 2020, 5:00 PM • 10 Y

Y by Aryan-23, bobjoe123, v4913, samrocksnature, centslordm, megarnie, pog, Adventure10, Rounak_iitr, ItsBesi

Let $ABC$ be a triangle. Distinct points $D$ , $E$ , $F$ lie on sides $BC$ , $AC$ , and $AB$ , respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$ . Let $Q$ denote the reflection of $P$ across $BC$ . Show that $Q$ lies on the circumcircle of $\triangle AEF$ .

Proposed by Ankan Bhattacharya

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62861

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62861

#2 h Jan 27, 2020, 5:01 PM • 13 Y

Y by Aryan-23, Toinfinity, anonman, samrocksnature, HamstPan38825, 554183, aopsuser305, megarnie, mathleticguyyy, CyclicISLscelesTrapezoid, john0512, Adventure10, starchan

I'm told that this problem was proposed by Brandon Wang and Eric Gan. I can't confirm whether that's true or not, though.

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v_Enhance

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v_Enhance

#3 h Jan 27, 2020, 5:06 PM • 24 Y

Y by 62861, Aryan-23, amar_04, AlastorMoody, Toinfinity, Imayormaynotknowcalculus, v4913, awin, Jc426, samrocksnature, centslordm, HamstPan38825, ASweatyAsianBoie, 554183, BVKRB-, ike.chen, math31415926535, megarnie, mathleticguyyy, CyclicISLscelesTrapezoid, john0512, aidan0626, ihatemath123, Adventure10

This problem was proposed by Ankan Bhattacharya.

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62861

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62861

#4 h Jan 27, 2020, 5:22 PM • 22 Y

Y by Aryan-23, amar_04, richrow12, Richangles, AlastorMoody, Frestho, Imayormaynotknowcalculus, mathlogician, mlgjeffdoge21, tree_3, awin, anonman, centslordm, HamstPan38825, ASweatyAsianBoie, 554183, BVKRB-, math31415926535, aopsuser305, megarnie, CyclicISLscelesTrapezoid, Adventure10

v_Enhance wrote:

This problem was proposed by Ankan Bhattacharya.

Oh, thanks for the clarification.

Solution with Brandon Wang, Eric Gan, and 0 others:

Let $Q^* = \overline{BE} \cap \overline{CF}$ . Then
$\angle BCQ^* = \angle DCF = \angle DAF = \angle PAB = \angle PCB = \angle BCQ$ and analogously $\angle CBQ^* = \angle CBQ$ , where we direct angles mod $180^\circ$ .

It follows that $Q = Q^*$ . Finally, to show that $Q$ lies on the circumcircle of $\triangle AEF$ , note that
$\angle AFQ = \angle AFC = \angle ADC = \angle ADB = \angle AEB = \angle AEQ$ and we are done.

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math_pi_rate

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math_pi_rate

#6 h Jan 27, 2020, 5:36 PM • 5 Y

Y by Aryan-23, amar_04, AmirKhusrau, Imayormaynotknowcalculus, Adventure10

Moving points solution: Animate $D$ linearly on $BC$ . Then $D \mapsto P \mapsto Q$ is a projective map (last map from reflection in $BC$ ). Also note that $Q$ moves on $\odot (BHC)$ , where $H$ is orthocenter of $\triangle ABC$ . Now consider $\sqrt{bc}$ inversion (denoting inverse of object $Z$ with $Z'$ ). Then $D \mapsto D' \mapsto E'$ is projective (last map by perspectivity at $B$ from $\odot (ABC)$ to $AC$ ). This gives $\text{deg}(E')=1$ . Similarly, we get $\text{deg}(F')=1$ . Also, $\text{deg}(Q')=2$ since $Q'$ moves projectively on a circle. Thus, it suffices to show that $E',F',Q'$ are collinear (i.e. prove the original problem) for $(1+1+2)+1=5$ positions of $D$ . $D=B,C$ are obvious. Take $D$ as the foot of the $A$ -altitude for the third position (and use the fact that reflection of orthocenter in $BC$ lies on circumcircle). Then take $D$ as the foot of the $A$ -tangent. Then $\sqrt{bc}$ inversion gives that $E'$ and $F'$ lie on the perpendicular bisector of $BC$ , and $Q'$ is the circumcenter of $\triangle ABC$ . These are collinear, as desired. Finally consider $D$ as the foot of the $A$ -symmedian. Then $Q$ is well known to be the $A$ -Humpty point. In this case, the result follows from here. Hence, done. $\blacksquare$

This post has been edited 4 times. Last edited by math_pi_rate, Jan 27, 2020, 6:24 PM
Reason: \sqrt{bc} not \sqrt{BC} :P

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Gems98

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Gems98

#7 h Jan 27, 2020, 5:40 PM • 2 Y

Y by Aryan-23, Adventure10

By Angle Chasing, we get that $\angle BQD = \angle BPD = \angle BCA = \angle BED$ So, $BFDQ$ is conncyclic. Similarly, we get that $CEQD$ is concyclic.
Therefore
$\begin{align*} \angle EQF &= 360^\circ - \angle DQF - \angle DQE \\ &= 360^\circ - (180^\circ - \angle CBA) - (180^\circ - \angle QCB) \\ &= 180^\circ - \angle BAC \end{align*}$ which implies that $Q$ lies on the circumcircle of $\triangle AEF$

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Aryan-23

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Aryan-23

#8 h Jan 27, 2020, 5:55 PM • 4 Y

Y by amar_04, Richangles, Adventure10, Mango247

Note that we have ...
$\angle FCB = \angle FAD = \angle FAP = \angle BAP=\angle BCP = \angle BCQ$ So , $Q$ lies on $CF$
$\angle CFA = \angle ADC = \angle BDP= \angle BDQ \implies BDQF \text{ is cyclic}$ Similarly , $CEQD$ is cyclic...
We are done by Miquel on ${{(D,E,F)}}$

PS

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Pluto1708

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Pluto1708

#9 h Jan 27, 2020, 6:13 PM • 2 Y

Y by Aryan-23, Adventure10

alifenix- wrote:

Let $ABC$ be a triangle. Distinct points $D$ , $E$ , $F$ lie on sides $BC$ , $AC$ , and $AB$ , respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$ . Let $Q$ denote the reflection of $P$ across $BC$ . Show that $Q$ lies on the circumcircle of $\triangle AEF$ .

Let $Q'=BE\cap CF$ .We will first show that $Q'\in \odot{AEF}$ .Indeed since $\measuredangle{FQ'E}=-\measuredangle{BQ'C}=\measuredangle{Q'CB}+\measuredangle{CBQ'}=\measuredangle{FCB}+\measuredangle{CBE}=\measuredangle{FCD}+\measuredangle{DBE}=\measuredangle{FAD}+\measuredangle{DAE}=\measuredangle{FAE}$ which proves this Claim $\square$ .

Back to the main problem we have $\measuredangle{CPD}=\measuredangle{CBA}=-\measuredangle{CED}=-\measuredangle{CQD}$ and similarly $\measuredangle{BPD}=-\measuredangle{BQD}$ . hence $Q'$ is reflection of $P$ about $BC$ which implies $Q'\equiv Q$ and we are done $\square$

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tworigami

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tworigami

#10 h Jan 27, 2020, 6:26 PM • 5 Y

Y by blacksheep2003, mira74, amar_04, Adventure10, ehuseyinyigit

It's quite interesting that both EGMO TST geometry problems were basically extensions of ELMO 2013 G3.

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wu2481632

4233 posts

wu2481632

#11 h Jan 27, 2020, 7:07 PM • 1 Y

Y by Adventure10

Solution:

We use $\sqrt{bc}$ inversion. $D \mapsto D'$ on $(ABC)$ such that $\angle{BAP} = \angle{D'AC}$ ; $F \mapsto F' = BD' \cap AC$ , $E \mapsto E' = CD' \cap AB$ . We identify $Q'$ as follows: let $A_1$ be the reflection of $A$ in $BC$ ; then $Q, D, A_1$ are collinear and $QBA_1C$ is cyclic. Let $O$ be the circumcenter of $ABC$ ; we know that $A_1$ and $O$ swap under this inversion. Then $O, D', A,$ and $Q'$ must be concyclic; so must $B, O, C,$ and $Q'$ .

Thus $Q'$ , which we want to show lies on $E'F'$ , is the intersection of $(BOC)$ and $(AOD')$ . But then $Q'$ is just the Miquel point of $ABD'C$ so we are done.

@above haha, solved both the same way

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blacksheep2003

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blacksheep2003

#12 h Jan 27, 2020, 8:20 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Morally Questionable Solution

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NJOY

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NJOY

#14 h Jan 28, 2020, 12:13 PM • 2 Y

Y by Adventure10, ehuseyinyigit

alifenix- wrote:

Let $ABC$ be a triangle. Distinct points $D$ , $E$ , $F$ lie on sides $BC$ , $AC$ , and $AB$ , respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$ . Let $Q$ denote the reflection of $P$ across $BC$ . Show that $Q$ lies on the circumcircle of $\triangle AEF$ .

The problem is quite trivial for any TST

. Anyways, here goes the solution:

Let $BE$ meet $CF$ at point $T$ . We first prove the following.
Claim.The points $A,E,F,T$ are concyclic.
Proof. This simply follows because $\begin{align*}\measuredangle{ETF} & =-\measuredangle{CTB}\\& =\measuredangle{BCT}+\measuredangle{TBC}\\& =\measuredangle{BCF}+\measuredangle{CBE}\\& =\measuredangle{FCD}+\measuredangle{DBE}\\& =\measuredangle{FAD}+\measuredangle{DAE}\\& =\measuredangle{EAF}. ~~~~~~~~~~~~~\square\end{align*}$
Therefore, we have $\measuredangle{DPC}=\measuredangle{ABC}=-\measuredangle{DEC}=-\measuredangle{DQC}$ and similarly $\measuredangle{DPB}=-\measuredangle{DQB}$ . Hence, we obtain that $T$ is reflection of $P$ about $BC$ , which implies $Q\equiv T$ and hence, the conclusion follows. $\blacksquare$

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IndoMathXdZ

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IndoMathXdZ

#15 h Jan 28, 2020, 12:35 PM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Nice

, but quite trivial for a #4.
Claim. $BDQF$ is cyclic.
Proof. Notice that
$\measuredangle BFD = \measuredangle BCA = \measuredangle BPA \equiv \measuredangle BPD = \measuredangle BQD$
Similarly, $DQEC$ is cyclic.
Hence, by Miquel Theorem, $AQFE$ is cyclic as well.

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Mathematicsislovely

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Mathematicsislovely

#17 h Jan 28, 2020, 5:33 PM • 1 Y

Y by Adventure10

wu2481632 wrote:

Solution:

We use $\sqrt{bc}$ inversion. $D \mapsto D'$ on $(ABC)$ such that $\angle{BAP} = \angle{D'AC}$ ; $F \mapsto F' = BD' \cap AC$ , $E \mapsto E' = CD' \cap AB$ . We identify $Q'$ as follows: let $A_1$ be the reflection of $A$ in $BC$ ; then $Q, D, A_1$ are collinear and $QBA_1C$ is cyclic. Let $O$ be the circumcenter of $ABC$ ; we know that $A_1$ and $O$ swap under this inversion. Then $O, D', A,$ and $Q'$ must be concyclic; so must $B, O, C,$ and $Q'$ .

After this we can also proceed in the following way:
Let $X$ be the intersection of $AD'$ and $BC$ .Then by brocards theorem $E'F'$ is the polar of $X$ .
Again $Q'$ is the intersection of the circumcircle of $(BOC)$ and $OD'$ .So inversion in $(ABC)$ sends $X$ to $Q'$ .So $Q'$ also lies in polar of $X$ .So $E',Q',F'$ are collinear.So inverting back $A,E,Q,F$ lies on a circle.

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AlastorMoody

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AlastorMoody

#18 h Jan 30, 2020, 7:53 AM • 3 Y

Y by Arefe, Adventure10, Mango247

Solution

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ihatemath123

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ihatemath123

#51 h Sep 30, 2023, 4:06 AM

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Claim: $\triangle FDQ \sim \triangle ADE$ . Note that
$\angle FDQ = 180^{\circ} - \angle FDB - \angle QDC = 180^{\circ} - \angle A - \angle CDP = 180^{\circ} - \angle EDC - \angle ADB = \angle ADE.$ Furthermore,
$\begin{align*} & \frac{FD}{DQ} \\ = ~ & \frac{FD}{DP} \\ = ~ & \frac{FD \cdot AD}{BD \cdot DC} \\ = ~ & \frac{AD}{BD \cdot \frac{DC}{FD}} \\ = ~ & \frac{AD}{DE}, \end{align*}$ so by SAS similarity our claim is proven.

So, $\angle FQD = \angle AED = 180^{\circ} - \angle B$ ; similarly, $\angle EQD = 180^{\circ} - \angle C$ , so
$\angle FQE = 360^{\circ} - \angle FQD - \angle EQD = 180^{\circ} - \angle A$ as desired.

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asdf334

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asdf334

#52 h Oct 22, 2023, 3:12 PM • 1 Y

Y by ehuseyinyigit

good example of phantom points

Let $Q'$ be the Miquel point of $\triangle DEF$ w.r.t $\triangle ABC$ . Notice that
$\measuredangle BQ'D=\measuredangle BFD=\measuredangle ACB=\measuredangle DPB$ $\measuredangle CQ'D=\measuredangle CED=\measuredangle ABC=\measuredangle DPC$ which implies $Q'$ is the reflection of $P$ over $BC$ , thus $Q'\equiv Q$ , done.

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cj13609517288

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cj13609517288

#53 h Dec 13, 2023, 8:29 PM

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Note that
$\angle CBP=\angle CAP=\angle EAD=\angle EBC,$ so $Q$ lies on $BE$ . Similarly, $Q$ lies on $CF$ , so $Q=BE\cap CF$ . Then
$\angle AFQ+\angle AEQ=\angle AFC+\angle AEB=\angle ADC+\angle ADB=180^{\circ}\;\blacksquare$

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dolphinday

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dolphinday

#54 h Jan 4, 2024, 12:59 PM

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We will prove that $Q = \overline{CF} \cap \overline{BE}$ .
Note that if and only if $Q$ lies on $CF$ , then $\angle{BCP} =\angle{FCB}$ .
$\angle BCP = \angle BAP = \angle FCD$ and then by symmetry $Q$ lies on $BE$ .
Now we can prove that $AEFQ$ is cyclic.
$\angle AEQ + \angle AFQ = \angle ADB + \angle ADC = 180^{\circ}$ so we are done.

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lelouchvigeo

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lelouchvigeo

#55 h Jan 4, 2024, 2:20 PM

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Nice

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qwerty123456asdfgzxcvb

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qwerty123456asdfgzxcvb

#56 h Feb 21, 2024, 12:37 AM

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deleted $\quad$

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ItsBesi

139 posts

ItsBesi

#57 h Apr 9, 2024, 4:26 PM • 1 Y

Y by Rounak_iitr

Lovely problem

Here is my solution without phantom points

$\textbf{Claim:}$ Points $B-E-Q$ -are collinear

$\textbf{Proof:}$

$\angle PBC=\angle PAC \equiv \angle PAE=\angle DBE \equiv \angle CBE \equiv \angle EBC \implies \angle PBC=\angle EBC$ $...(1)$

Since $Q$ -is the reflection of $P$ over $BC$ we get:

$\angle PBC=\angle QBC$ $...(2)$

Combining $(1)$ with $(2)$ we get:

$\angle EBC= \angle QBC$
So Points $B-E-Q$ -are collinear

$\textbf{Claim:}$ Points $C-Q-F$ -are collinear

$\textbf{Proof:}$

$\angle QCB=\angle BCP=\angle BAP=180-\angle PAF=\angle QCB=\angle FCB \implies \angle QCB \equiv \angle FCB$

So Points $C-Q-F$ -are collinear

$\textbf{Claim:}$ $Q \in \odot$ $( \triangle AEF )$

$\textbf{Proof:}$

$\angle FQE=\angle FQB=180-\angle BQC \equiv 180-\angle BPC=180-(180- \angle A)=\angle A \implies \angle FQE=\angle A$ $...(*)$

$\angle A + \angle FAC=180 \implies \angle FAC= 180-\angle A , \angle FAE \equiv \angle FAC=180-\angle A \implies \angle FAE=180-\angle A$ $...(**)$

$(*) + (**) \implies \angle FQE+\angle FAE=\angle A + 180-\angle A=180 \implies \angle FQE+\angle FAE=180$

So Points $A,F,Q,E$ -are concylic $\iff$ $Q$ lies on the circumcircle of $\triangle AEF$ $\blacksquare$

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Rounak_iitr

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#60 h May 18, 2024, 8:02 AM • 1 Y

Y by ehuseyinyigit

I Love Colorful Geo diagrams. :love:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -19.68117321949262, xmax = 22.89738128743351, ymin = -15.250512502480996, ymax = 9.540656309645739; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen ffqqtt = rgb(1,0,0.2); pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqccqq = rgb(0,0.8,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen wwqqcc = rgb(0.4,0,0.8); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-2.5886072783962946,6.149935911574594)--(-9.774306840329475,-7.809119271185058)--(7.045384433858,-10.037595717607374)--cycle, linewidth(1) + ffvvqq); /* draw figures */ draw((-2.5886072783962946,6.149935911574594)--(-9.774306840329475,-7.809119271185058), linewidth(1) + ffvvqq); draw((-9.774306840329475,-7.809119271185058)--(7.045384433858,-10.037595717607374), linewidth(1) + ffvvqq); draw((7.045384433858,-10.037595717607374)--(-2.5886072783962946,6.149935911574594), linewidth(1) + ffvvqq); draw(circle((-0.6681263925632243,-3.6676876138441528), 10.003698242135544), linewidth(1) + blue); draw((-2.5886072783962946,6.149935911574594)--(-0.11930894731573694,-13.656320049342908), linewidth(1) + ffqqtt); draw(circle((-4.356422343706356,-1.7690643722696051), 8.113922331441293), linewidth(1) + ffdxqq); draw(circle((4.34477874864532,-0.684262611748228), 9.735405038115452), linewidth(1) + qqccqq); draw(circle((0.6564827975021906,1.2143606298263208), 5.906819191620764), linewidth(1) + fuqqzz); draw((-9.774306840329475,-7.809119271185058)--(1.1851603191548814,-4.668751900169902), linewidth(1) + ffqqtt); draw((1.1851603191548814,-4.668751900169902)--(7.045384433858,-10.037595717607374), linewidth(1) + ffqqtt); draw((7.045384433858,-10.037595717607374)--(-0.11930894731573694,-13.656320049342908), linewidth(1) + ffqqtt); draw((-9.774306840329475,-7.809119271185058)--(-0.11930894731573694,-13.656320049342908), linewidth(1) + ffqqtt); draw((-0.11930894731573694,-13.656320049342908)--(1.1851603191548814,-4.668751900169902), linewidth(1) + wwqqcc); draw((-5.245984907664572,0.9876715212154584)--(1.1851603191548814,-4.668751900169902), linewidth(1) + ubqqys); draw((1.1851603191548814,-4.668751900169902)--(3.447153510064016,-3.991661838099557), linewidth(1) + ubqqys); /* dots and labels */ dot((-2.5886072783962946,6.149935911574594),dotstyle); label("$A$", (-2.477436117803798,6.427863813055835), NE * labelscalefactor); dot((-9.774306840329475,-7.809119271185058),dotstyle); label("$B$", (-9.675768766167941,-7.524116841302483), NE * labelscalefactor); dot((7.045384433858,-10.037595717607374),dotstyle); label("$C$", (7.166662063595267,-9.747540053152415), NE * labelscalefactor); dot((-0.11930894731573694,-13.656320049342908),dotstyle); label("$P$", (-0.0038777946207523622,-13.388395562556676), NE * labelscalefactor); dot((-0.6983774444780035,-9.011608339405127),linewidth(4pt) + dotstyle); label("$D$", (-0.5875263877313587,-8.802585188116193), NE * labelscalefactor); dot((3.447153510064016,-3.991661838099557),linewidth(4pt) + dotstyle); label("$E$", (3.5535993443391334,-3.772090171305724), NE * labelscalefactor); dot((-5.245984907664572,0.9876715212154584),linewidth(4pt) + dotstyle); label("$F$", (-5.145543972023712,1.2028192652084968), NE * labelscalefactor); dot((1.1851603191548814,-4.668751900169902),dotstyle); label("$Q$", (1.3023833423410807,-4.383531554564455), NE * labelscalefactor); label("$m$", (2.4418877384141693,-4.744837826490069), NE * labelscalefactor,ubqqys); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

$\color{red}\textbf{Claim:-}$ $AFQE$ are cyclic points.
$\color{blue}\textbf{Proof:-}$ Since $Q$ is the reflection of $P$ over $DC$ therefore we get, $BP=BQ$ and $BC=CQ.$ Therefore we get some angles equal $\angle PBD=\angle PBQ=\angle CBQ=\angle DBQ$ Now since $ABPC$ is cyclic we get, $\angle CBP=\angle DBP=\angle DAC=\angle PAC$ Also we get, $\angle BAD=\angle BAP=\angle BCP=\angle DCP$ Therefore we get, $B,Q,E$ are collinear and $C,Q,F$ are collinear. Now we get, $\angle EQF=\angle BQC=\angle CPB=180-\angle BAC=180-\angle FAE.$ Q.E.D

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Mathandski

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Mathandski

#61 h Aug 28, 2024, 11:56 PM • 1 Y

Y by ehuseyinyigit

Subjective Rating (MOHs) $\text{[math]}$

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LLL2019

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LLL2019

#62 h Oct 1, 2024, 2:55 AM

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Good problem not hard!

Note that $\angle BDF = \angle A = \angle CDE$ and $\angle QDC = \angle PDC = \angle ADC$ . Hence, $\angle ADF = \angle QDE$ . We now claim that $\triangle DFQ \sim \triangle DAE$ . indeed, $DF\cdot DE = DB \cdot \frac{AC}{AB} \cdot DC \cdot \frac{AB}{AC} = DB \cdot DC = DA \cdot DP = DA\cdot DQ$ , which combined with $\angle QDF = \angle ADE$ , proves our claim. We can thus get $\angle FQE + \angle FAE = 2\pi - \angle AED - \angle AFD + \angle A = \angle A + \angle FDE + \angle A = \pi$ as wanted.

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AshAuktober

958 posts

AshAuktober

#63 h Oct 6, 2024, 10:26 AM

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Simple enough with the power of wishful thinking.
Let $Q' = BE \cap CF$ . We claim that $Q' = Q$ . Indeed notice that
$\angle Q'BC = \angle EBD = \angle EAD = \angle CAP = \angle CBP.$ Similarly, $\angle Q'CB = \angle PCB.$ This suffices.

From here, $\angle AFQ = \angle AFC = \angle ADC = \pi - \angle ADB = \pi - \angle AEB = \pi - \angle AFQ,$ yielding the result. $\square$

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cosdealfa

27 posts

cosdealfa

#64 h Nov 10, 2024, 7:38 AM • 2 Y

Y by pb_ana, Kaus_sgr

Claim 1: $B - Q - E$ collinear.
Proof: $\angle DBE = \angle DAE =\angle PAC =\angle PBC =\angle DBQ$ $\square$
Similarly we prove $C - Q - F$ collinear.

Claim 2: $BDQF$ cyclic
Proof: $\angle QBD = \angle EBD =\angle DAC =\angle DFC =\angle DFQ$ $\square$
Similarly, we prove $DQEC$ cyclic.

We are done because this means $Q$ is Miquel’s point in triangle $\triangle ABC$ for the points $D$ , $E$ , $F$ . $\blacksquare$

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EeEeRUT

55 posts

EeEeRUT

#65 h Feb 2, 2025, 7:56 AM

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Let $BE$ intersect $CF$ at $Q’$
By angle chasing, we have $\triangle DBF \sim \triangle DEC$ which gives $BD \cdot DC = DE \cdot DF$
Moreover, $\angle FDB = \angle EDC$
So, an involution $\mathcal{S}$ ( $\sqrt{DB \cdot DC}$ inversion of $\triangle DBC$ ) sends $E \rightarrow F, B\rightarrow C$ Applying, DDIT on complete quadrilateral $AEQ’F$ gives that $\mathcal{S}$ also send $Q’ \rightarrow A$ Since $\triangle DBF \sim \triangle DEC$ , $D$ is the miquel point of $\square AEQ’F$ , which yield that $\square AEQ’F$ is cyclic.
So, it is remained to show that $DQ’ = DP$
Take power of point on point $D$ wrt $(ABC)$ , we have $DB \cdot DC = DA \cdot DP = DA \cdot DQ’$ which yields the desired results.

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SimplisticFormulas

85 posts

SimplisticFormulas

#66 h Feb 2, 2025, 12:20 PM

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took me like 5 mins
solution

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blueprimes

324 posts

blueprimes

#67 h Mar 31, 2025, 2:18 AM

Y by

Here's a fun solution:

Let $A', E', F'$ be the reflections of $A, E, F$ over $BC$ , we wish to show $P \in (A'E'F')$ . Now $\angle BF'D = \angle ACB = \angle BPD$ so $BF'PD$ is cyclic. Similarly, $CE'PD$ is cyclic, then
$\angle F'PE' = (180^\circ - \angle F'PD) + (180^\circ - \angle E'PD) = \angle A'BC + \angle A'CB = 180^\circ - \angle F'A'E'$ implying $A'F'PE'$ is cyclic as needed.

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