Perpendicular to quadrilateral diagonal

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Perpendicular to quadrilateral diagonal

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cjquines0

510 posts

cjquines0

#1 h Jul 19, 2017, 4:36 PM • 8 Y

Y by matol.kz, tapir1729, NO_SQUARES, Adventure10, Mango247, Rounak_iitr, ohiorizzler1434, Retemoeg

Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$ . The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$ . Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$ . Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$ . Prove that $PQ \perp AC$ .

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v_Enhance

6882 posts

v_Enhance

#2 h Jul 19, 2017, 4:37 PM • 23 Y

Y by matol.kz, CeuAzul, anantmudgal09, samuel, yayups, karitoshi, HolyMath, Modesti, v4913, Hoju, Pluto04, mijail, math31415926535, michaelwenquan, HamstPan38825, nguyenducmanh2705, Adventure10, Mango247, VIATON, Rounak_iitr, bhan2025, ohiorizzler1434, Sedro

My proof proceeds in six steps. The first two steps set the stage by incorporating all the angle conditions. By the fourth step we show that $Q$ lies on $\omega$ even if $M$ is arbitrary on $\overline{AC}$ . Step 5 uses the midpoint condition for the first time by projecting $(AC;M\infty)$ , and the sixth steps solves what is essentially a completely projective problem.

$[asy] size(12cm); draw(unitcircle, red); pair P = dir(0); pair R = dir(180); pair K = dir(75); pair L = conj(K); pair B = dir(65); pair D = dir(-58); pair A = extension(K, D, B, L); pair C = extension(K, B, D, L); pair T = 2/(K+L); draw(A--B--C--D--cycle, orange); draw(P--B--K--A--L--D--P, orange); draw(K--T--L, red); draw(A--T, paleblue); pair E = extension(B, P, A, C); pair F = extension(D, P, A, C); draw(F--P, orange); pair M = midpoint(A--C); pair O = midpoint(P--R); pair X = -B+2*foot(O, B, M); pair Y = -D+2*foot(O, D, M); draw(B--M--D, yellow); draw(B--Y, yellow+dashed); draw(D--X, yellow+dashed); pair Q = extension(X, E, Y, F); draw(X--Q, brown+dotted); draw(Y--F, brown+dotted); pair Z = R*Q/B; draw(Z--K--X--L--cycle, paleblue); draw(Z--T, paleblue); draw(Z--B, dashed+paleblue); draw(T--R--Q, paleblue); dot("$P$", P, dir(P)); dot("$R$", R, dir(R)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$T$", T, dir(T)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Q$", Q, dir(Q)); dot("$Z$", Z, dir(Z)); /* TSQ Source: !size(12cm); unitcircle 0.1 yellow / red P = dir 0 R = dir 180 K = dir 75 L = conj K B = dir 65 D = dir -58 A = extension K D B L C = extension K B D L T = 2/(K+L) A--B--C--D--cycle 0.1 lightred / orange P--B--K--A--L--D--P orange K--T--L red A--T paleblue E = extension B P A C F = extension D P A C F--P orange M = midpoint A--C O := midpoint P--R X = -B+2*foot O B M Y = -D+2*foot O D M B--M--D yellow B--Y yellow dashed D--X yellow dashed Q = extension X E Y F X--Q brown dotted Y--F brown dotted Z = R*Q/B Z--K--X--L--cycle paleblue Z--T paleblue Z--B dashed paleblue T--R--Q paleblue */ [/asy]$

Step 1: Observe that $K = \overline{BC} \cap \overline{AD}$ lies on $\omega$ by all the given angle conditions. Similarly, $L = \overline{BA} \cap \overline{CD}$ lies on $\omega$ too, and actually $P$ is the arc midpoint of $\widehat{KL}$ .

Step 2: Let $T = \overline{KK} \cap \overline{LL}$ , and $S = \overline{XX} \cap \overline{YY}$ (not pictured). Then $\overline{ACST}$ are collinear by Brokard Theorem on $LKBD$ and $BDYX$ .

Step 3: Again by Brokard Theorem on $DBXY$ , we find that $\overline{XD} \cap \overline{BY}$ , $S$ and $M$ are collinear; in other words, $\overline{XD} \cap \overline{BY}$ lies on $\overline{AC}$ .

Step 4: By the converse of Pascal theorem on $BPDXQY$ , we find that $Q$ lies on $\omega$ (since we already know $\overline{XD} \cap \overline{BY}$ , $E$ , $F$ collinear).

Step 5: Let $Z$ be the point on $\omega$ such that $\overline{ZB} \parallel \overline{AC}$ . Then $-1 = (AC;M\infty) = (LK;XZ)_\omega$ . Thus $T \in \overline{XZ}$ .

Step 6: Introduce $R$ the antipode of $P$ (so $\overline{PR} \cap \overline{XZ} = T$ ). This is motivated because now the problem is ``completely projective'': we just need to check $\overline{BZ}$ , $\overline{QR}$ , $\overline{ACT}$ are parallel. Indeed, this follows by Pascal on $ZBPRQX$ .

Remark: Motivational remarks: I found the steps in the order 4 3 1 2 5 6. You basically have to realize $Q$ lies on the circle; then the obvious Pascal reverse-engineers to tell you about $DBXY$ , which leads naturally to discovery of $K$ and $L$ . In fact, I found $K$ and $L$ because I was about to try complex bashing.

Once $Q$ is on the circle, you can focus on just the upper half of the diagram (ignoring $D$ , $Y$ , $F$ and concentrating only $B$ , $X$ , $E$ ). Adding $Z$ eliminates the midpoint as usual with harmonic arguments.

Then the addition of point $R$ does two things: it changes perpendicularity to $\overline{BZ} \parallel \overline{AC} \parallel \overline{RQ}$ , and it lets one just define $T = \overline{ZX} \cap \overline{PR}$ , thus we can ignore $KLT$ . At that point the problem is completely projective.

This post has been edited 1 time. Last edited by v_Enhance, Dec 2, 2020, 2:13 AM

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EulerMacaroni

851 posts

EulerMacaroni

#3 h Jul 19, 2017, 5:08 PM • 1 Y

Y by Adventure10

Let $Y'\equiv DM\cap \odot(ABC)$ such that $Y$ lies on the arc opposite $B$ ; I claim that $Y'\equiv Y$ . Compute $\angle BPD=\angle BY'D\iff 2\pi-\angle B-\angle BCD=2\pi-\angle CBY'-\angle BCD-\angle CDM$ $\iff \angle B=\angle CBY'+\angle CDM \iff \angle ADY'=\angle CAY'$ which is obvious by reflection about $M$ . Analogously, $X$ lies on $\odot(ADC)$ ; we therefore notice that $X$ and $Y$ are the $B$ and $D$ HM points in $\triangle ABC$ and $\triangle ADC$ , so $B$ and $D$ are the $X$ and $Y$ HM points in $\triangle CXA$ and $\triangle CYA$ and hence by analogy $Q \in \odot(BPD)$ . Let $B'$ be the reflection of $B$ over $AC$ ; notice that quadrilateral $BXEB'$ is cyclic on the $B$ -Apollonius circle in $AC$ , so $\angle BPQ=\pi-\angle BXE=\angle B'BE=\frac{\pi-\angle BEB'}{2}=\frac{\pi}{2}-\angle BEC$ as desired.

This post has been edited 2 times. Last edited by EulerMacaroni, Jul 20, 2017, 9:14 PM

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anantmudgal09

1980 posts

anantmudgal09

#4 h Jul 19, 2017, 7:43 PM • 8 Y

Y by biomathematics, Yamcha, Pluto04, GuvercinciHoca, michaelwenquan, Adventure10, Funcshun840, Kingsbane2139

For anyone who has seen HM points, this is a walk in the park.

cjquines0 wrote:

Let $T_1=EX \cap PR$ and $T_2=FY \cap PR$ where $R$ is a point on $AC$ such that $PR \perp AC$ .

Claim 1: $X$ is the HM point corresponding to $B$ in triangle $ABC$ .

(Proof) Let $ABCB_1$ be a parallelogram; $L$ be midpoint of minor arc $AC$ in $(ADC)$ . Note that $\measuredangle DXM=\measuredangle BPL=\measuredangle DLB_1,$ so $D, X, L, B_1$ are concyclic, just as we desired. $\blacksquare$

Claim 2: In any triangle $ABC$ , let $X$ be the HM point opposite $B$ ; $M$ the midpoint of $AC$ and $E$ the foot of $B$ -internal bisector. Point $E'$ lies on ray $MB$ such that $EE' \perp AC$ . Then $(XM, E'B)$ depends only on $\measuredangle ABC$ .

(Proof) Let $N$ be midpoint of arc $ABC$ , $L$ the midpoint of minor arc $AC$ in $(ADC)$ . Notice that $(XM, E'B) \overset{E}{=} (NM, \infty L)$ where projection is preceded by reflection of the pencil in line $EE'$ . $\blacksquare$

To conclude, consider $(XM, E'B) \overset{E}{=} (T_1R, \infty P)$ and $(YM, F'D)\overset{F}{=}(T_2R, \infty P)$ ; hence, $\boxed{T_1=T_2},$ as desired. $\blacksquare$

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Lsway

71 posts

Lsway

#5 h Jul 19, 2017, 10:38 PM • 3 Y

Y by naw.ngs, AlastorMoody, Adventure10

Step 1
Step 2
Step 3

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Anzoteh

126 posts

Anzoteh

#6 h Jul 20, 2017, 9:28 PM • 2 Y

Y by Adventure10, Mango247

Let $\omega_1$ be the circumcircle of $ABC$ and $\omega_2$ the circumcircle of $ADC$ , then these two circles are symmetric w.r.t. $AC$ .
Also notice that $BP$ passes through $M_1$ , the midpoint of arc $AC$ of $\omega_1$ not containing $B$ , and $DP$ passes through $M_2$ , the midpoint of arc $AC$ of $\omega_2$ not containing $D$ .

We first start with a preliminary observation: $X$ lies on $\omega_2$ and $Y$ lies on $\omega_1$ . W.L.O.G. for this section we assume that $AB\le AC$ . Indeed, let $X'$ be on $BM$ satisfying $MX'\cdot MB=MA^2=MC^2$ . Then $\angle X'AC=\angle MBA$ and $\angle X'CA=\angle MBC$ . Thus $\angle ADC=\angle ABC=\angle MBA+\angle MBC=\angle X'AC+\angle X'CA=\pi - \angle AX'C$ , so $X'$ lie on $\omega_2$ . In addition, let $BM$ intersect $\omega_1$ again at $X''$ , then $X'$ and $X''$ are symmetrical w.r.t. $AC$ . Combining with the fact that $M_1$ and $M_2$ are also symmetrical w.r.t. $AC$ (being the midpoint of arc) we have $X'M_2=X''M_1$ . Knowing that the two circles have the same radius further allows us to assert $\angle X'BP=\angle X''BM_1=\angle X'DM_2=\angle X'DP$ , showing that $D, B, P, X'$ cyclic hence $X'=X$ . Similarly, $Y$ lies on $\omega_1$ .

Next, let $N_1$ be diametrically opposite $M_1$ w.r.t. $\omega_1$ and define similarly for $N_2$ . We claim that $XE$ passes through $N_2$ by claiming that $XE$ is the internal angle bisector of $\angle AXC$ . Indeed, by angle bisector theorem we have $\frac{AE}{EC}=\frac{AB}{BC}$ . Invoking our $X''$ from the previous section (i.e. the other intersection of $BM$ and $\omega_1$ ) gives $AXCX''$ parallelogram. Now invoking a little bit more trigonometric bashing we have $1=\frac{AM}{CM}=\frac{AB}{BC}\cdot\frac{\sin\angle ABM}{\sin\angle CBM}=\frac{AB}{BC}\cdot\frac{AX''}{CX''}=\frac{AB}{BC}\cdot\frac{CX}{AX},$ so $\frac{AX}{CX}=\frac{AB}{BC}=\frac{AE}{EC}$ , and the conclusion follows by the angle bisector theorem. Analogously, $YF$ passes through $N_1$ .

Finally, considering triangle $PEF$ , and letting the perpendicular from $P$ to reach $AC$ at $P_1$ we have (considering signed length) $\frac{EP_1}{FP_1}=\frac{\cot\angle FEP}{\cot\angle EFP}$ . Similarly if letting perpendcular from $Q$ to reach AC at $Q_1$ we have $\frac{EQ_1}{FQ_1}=\frac{\cot\angle FEQ}{\cot\angle EFQ}$ . Now $\cot\angle FEP=\cot\angle MEM_1=\frac{MM_1}{EM}$ , $\cot\angle EFP=\cot\angle MFM_2=\frac{MM_2}{FM}$ . Considering $MM_2=MM_1$ we have $\frac{\cot\angle FEP}{\cot\angle EFP}=\frac{FM}{EM}$ . Analogously, $\cot\angle FEQ=\cot\angle FEQ=\cot\angle MEN_2=\frac{MN_2}{EM}$ , and $\cot\angle EFQ=\cot\angle N_1FM=\frac{MN_1}{FM}$ . Therefore we have $\frac{\cot\angle FEQ}{\cot\angle EFQ}=\frac{FM}{EM}$ since again it is not hard to verify that $MN_2=MN_1$ . (For signed convention we can say that $ME<0$ if it's nearer to $A$ than $B$ , and $>0$ otherwise). Therefore, $\frac{EP_1}{FP_1}=\frac{EQ_1}{FQ_1}$ , so $P_1\equiv Q_1$ and the two perpendicular lines coincide. (Sorry for the trigo-heavy proof).

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AngleChasingXD

109 posts

AngleChasingXD

#7 h May 6, 2018, 6:40 AM • 3 Y

Y by GGPiku, Adventure10, Mango247

I would like to present my proof.

Let $AB\cap CD= \{J\}$ , $BC\cap AD=\{K\}$ . The condition of the problem implies that $BDKJ$ is cyclic and denote its circumcircle by $\gamma$ . The exterior angle bisectors of $\angle JBK$ and $\angle LDK$ both pass through the midpoint of the arc $\overarc{JBDK}$ . As their intersection is $P$ , we conclude that $P$ is the midpoint of the arc $\overarc{JBDK}~~~(1)$ . Keep this in mind.

Denote by $AA'$ and $AA"$ the tangents from $A$ to $\gamma$ . Let $M'$ the midpoint of $AA'$ and $M"$ the midpoint of $AA"$ . It is well-known that $M'M"$ is the radical axis of $\gamma$ and the degenerated circle at $A$ . Also, $A'A"$ is the polar of $A$ WRT $\gamma$ , and as $C$ belongs to this polar (well-known), we get that $C\in A'A"$ . As $M$ is the midpoint of $AC$ , we find out that $M$ lies on $M'M"$ , hence $M$ is on the radical axis of the degenerated circle at $A$ and $\gamma$ . Hence $MA^2=MC^2=MX\cdot MB=MY\cdot MD$ . Using this, we arrive at
$\Delta MXC \sim \Delta MCB$
$\Delta MYC \sim \Delta MCD$
$\Delta MXA \sim \Delta MAB$
$\Delta MYA \sim \Delta MAD$ .
Puting al these together, we get that $\frac{AX}{XC}=\frac{AB}{BC}$ , and also that $\frac{AY}{YC}=\frac{AD}{DC}$ . Also we get that $m(\angle AXC)=m(\angle AYC)=180^o-m(\angle ABC)~~(2)$ . Therefore $XE$ and $YF$ are the angle bisectors of $\angle AXC$ and $\angle AYC$ .

Here comes my favourite part of the problem:
Note that the fact that $\angle ABC=\angle ADC$ is enough, that is, it is not used at all that they are less that $90^o$ . By replacing $ABCD$ with $AXCY$ , $E$ and $F$ remain at their places, $P$ becomes $Q$ , and $M$ remains at its place. $X$ becomes $B$ and $B$ becomes $X$ . $Y$ becomes $D$ and $D$ becomes $Y$ .

Denote $AX\cap CY=\{V\}$ and $AY\cap CX=\{U\}$ . With all the observations above, doing everything we did to $ABCD$ the same for $AXCY$ , we get that $U$ and $V$ lie on $\gamma$ . Similar to (1), $Q$ is the midpoint of the arc $\overarc{VBDU}$ .
Acording to (2), we have that $m(\angle JBK)+m(\angle UXV)=180^o$ , therefore $JK=UV$ , or equivalently written $JV\parallel KU$ .
Now we have a trapezoid $JVUK$ inscribed in $\gamma$ , $P$ being the midpoint of $\overarc{JK}$ and $Q$ the midpoint of $\overarc{VJU}$ . Denote by $O$ the center of $\gamma$ . Note that the common bisector line of the segmente $JV$ and $VK$ is, actually, the external angle bisector of $\angle POQ$ . As $OP=PQ$ , we have that $PQ\perp KU$ . (3)

The final step: Let $\{W\}=BC\cap AY$ . $ABCY$ is cyclic, therefore $WY\cdot WA=WC\cdot WB$ .
$BYKU$ is cyclic, hence $WU\cdot WY=WB\cdot WK$ . From these two relations we get $\frac{WC}{WK}=\frac{WA}{WU}$ , thus $AC\parallel KU$ . From (3), we get that $PQ\perp AC$ , exactly what we wanted.

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yayups

1614 posts

yayups

#8 h Sep 14, 2018, 11:45 PM • 1 Y

Y by Adventure10

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Let $I=BC\cap \omega$ and $J=CD\cap\omega$ . The first claim is that $I,A,D$ and $J,A,B$ are collinear. Note that
$\angle PDI = \pi-\angle PBI = \angle PBC = \angle PDA,$ so $I,A,D$ collinear. Incidentally, $\angle PDI=\angle PBJ$ , so $P$ is the midpoint of arc $\widehat{IJ}$ . Let $G=IJ\cap BD$ . By Brokard on $IJDB$ , we have that $AC$ is the polar of $G$ , so $GZ$ and $GH$ are tangent to $\omega$ , where $Z$ and $H$ are the intersections of $AC$ with $\omega$ . By Brokard on $BDYX$ , $G'=BD\cap XY$ is on the polar of $M$ . But by La'Hire, $G$ is on the polar of $M$ , and $G\in BD$ , so $G=G'$ , so $G,X,Y$ are collinear. Also, we see that (by Brokard on the same quadrilateral), that $BY\cap DX$ is on the polar of $G$ , so $BY\cap DX\in AC$ . Thus, by the converse Pascal on $BPDXQY$ , we have that $Q\in\omega$ .

We claim that $X$ is the $B$ -HM point of $ABC$ . Let $X'$ be the HM point. We know that $(X'AB)$ is tangent to $AC$ , and that $\angle AX'C=\pi-\angle ABC$ (its on $ARC$ where $R$ is the orthocenter of $ABC$ ), so $X'ADC$ is cyclic. Thus,
$\begin{align*} \angle BX'D&=\angle BX'A+\angle AX'D\\ &= \pi-\angle BAC + \angle ACD\\ &= \pi-\angle BAC + (\pi-\angle CAD-\angle ADC)\\&=2\pi-\angle BAD-\angle ADC\\&=\angle BPD, \end{align*}$ so $X'$ lies on $\omega$ , so $X'=\omega\cap AM=X$ , as desired.

$[asy] unitsize(0.45inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.550513472304992, xmax = 11.10568598292777, ymin = -4.963818860356124, ymax = 10.858184671390404; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-3.1720410998173945,1.333574927567185)--(-1.9357055767518994,9.033559325606662), linewidth(2) + wrwrwr); draw((-1.9357055767518994,9.033559325606662)--(-11.9131501488594,0.8997729896494681), linewidth(2) + wrwrwr); draw((-11.9131501488594,0.8997729896494681)--(-3.1720410998173945,1.333574927567185), linewidth(2) + wrwrwr); draw((-6.469766309168775,1.1699161082445362)--(-1.9357055767518994,9.033559325606662), linewidth(2) + wrwrwr); draw(circle((-1.8111190256771401,4.210792562893538), 4.82437571673674), linewidth(2) + wrwrwr); draw((-6.469766309168775,1.1699161082445362)--(-6.255141867596467,6.0883606806294575), linewidth(2) + wrwrwr); draw((-6.047403865818704,1.9024399448072322)--(-6.255141867596467,6.0883606806294575), linewidth(2) + wrwrwr); draw(circle((-3.369607813653483,5.3145442113644545), 3.9858686688039966), linewidth(2) + wrwrwr); /* dots and labels */ dot((-3.1720410998173945,1.333574927567185),dotstyle); label("$A$", (-2.9958666793473454,1.01167567736174), NE * labelscalefactor); dot((-1.9357055767518994,9.033559325606662),dotstyle); label("$B$", (-1.874501380495403,9.183817307351209), NE * labelscalefactor); dot((-11.9131501488594,0.8997729896494681),dotstyle); label("$C$", (-12.120400754937124,0.42795127521963494), NE * labelscalefactor); dot((-6.4017999351230825,2.7274686455080905),linewidth(4pt) + dotstyle); label("$X$", (-6.851519967180737,2.7014042098783593), NE * labelscalefactor); dot((-6.469766309168775,1.1699161082445362),linewidth(4pt) + dotstyle); label("$E$", (-6.406046081335445,1.2881767099553685), NE * labelscalefactor); dot((-5.591445684467464,1.2135052211329385),dotstyle); label("$H$", (-5.5304594781222844,1.3649825523424877), NE * labelscalefactor); dot((-6.047403865818704,1.9024399448072322),linewidth(4pt) + dotstyle); label("$P$", (-5.991294532445,2.0255127968717117), E * labelscalefactor); dot((-6.255141867596467,6.0883606806294575),linewidth(4pt) + dotstyle); label("$Q$", (-6.190989722651511,6.203750622730989), NW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Note that $X$ lies on the $B$ -appolonius circle of ABC, so since $E$ is on it too and $EA$ passes through its center, we see $\angle BXE=\pi/2+\angle BEA$ . But $\angle BXE=\angle QPE$ , so $\angle QPE = \pi/2+\angle PEA$ , so $PQ\perp AC$ .

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math_pi_rate

1218 posts

math_pi_rate

#9 h Feb 22, 2019, 12:38 PM • 2 Y

Y by Adventure10, Mango247

Here's my solution (in the order I found it): By an easy angle chase, we get that $U=AB \cap CD$ and $V=AD \cap BC$ lie on $\omega$ . We forget about the condition that $M$ is the midpoint of $AC$ for some time, and define points $X,Y,Q$ similarly for any arbitrary point $M$ on $AC$ . We claim that in fact $Q$ lies on $\omega$ . Animate $M$ on $AC$ . Then $M \rightarrow X \rightarrow XE \cap \omega$ and $M \rightarrow Y \rightarrow YF \cap \omega$ are projective maps. Thus, it suffices to show our claim for three positions of $M$ . When $M=A$ , we get that $X=U,Y=V$ , and so we wish to show that $UE$ and $VF$ meet on $\omega$ . Let $UE \cap \omega=Q'$ . Then the result follows by Pascal on $UQ'VBPD$ . Similarly, we can show the claim to be true for $M=C$ . And, when $M=AC \cap \omega$ , we have $X=Y=M$ , which directly gives $Q \in \omega$ . Thus, our claim is true.

Return to the given problem (i.e. when $M$ is the midpoint of $AC$ ). Let $P'$ and $Q'$ be the antipodes of $P$ and $Q$ in $\omega$ . Then we wish to show that $AC,P'Q,PQ'$ are concurrent. Applying Pascal on $XQ'PBP'Q$ , we see that this is equivalent to proving that $T=XQ' \cap BP'$ lies on $AC$ . Let $BP' \cap AC=T'$ . Then, as $\measuredangle EBT'=90^{\circ}$ , so $\odot (EBT')$ is the $B$ -Apollonius circle of $\triangle ABC$ . Also, $\measuredangle EXT=90^{\circ}=\measuredangle EBT$ , which gives that $EBXT$ is cyclic. Thus, for proving $T=T'$ , it suffices to show that $\odot (BEX)$ is the $B$ -Apollonius circle of $\triangle ABC$ . As $X$ lies on the $B$ -median of this triangle, so our problem is reduced to showing that $X$ is the $B$ -Humpty point of $\triangle ABC$ . However, it is well known that $MA^2=MC^2=MB \cdot MX$ , i.e. $AC$ is tangent to $\odot (ABX)$ and $\odot (BCX)$ , which is how we define Humpty points in the first place

. Hence, done. $\blacksquare$

EDIT: $400^{\text{th}}$ post on HSO :showoff:

This post has been edited 2 times. Last edited by math_pi_rate, Feb 22, 2019, 12:40 PM

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ayan.nmath

643 posts

ayan.nmath

#10 h May 20, 2019, 12:49 PM • 5 Y

Y by Kayak, AlastorMoody, Pluto04, Adventure10, Funcshun840

2016 IMO Shortlist G6 wrote:

Solution with Kayak :

We assume that $ABCD$ is not a parallelogram. Let $S=AD\cap BC,~R=AB\cap CD,~L=AX\cap CY,~ K=AY\cap XC$ and $\{U,V\}=AC\cap\omega$ as shown in the diagram below.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -17.370032730607846, xmax = 6.139541560852985, ymin = -8.950203944896286, ymax = 7.400745648818856; /* image dimensions */ pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.8,-0.08)--(2,0.5), linewidth(0.4)); draw((-6.16,3.88)--(-5.8,-0.08), linewidth(0.4)); draw((-5.8,-0.08)--(-4.450656162658686,-5.33024171139331), linewidth(0.4)); draw((-4.450656162658686,-5.33024171139331)--(2,0.5), linewidth(0.4)); draw((2,0.5)--(-6.16,3.88), linewidth(0.4)); draw(circle((-8.764650673685598,-1.3671421988830121), 5.858046294391133), linewidth(0.8) + qqffff); draw((-6.16,3.88)--(-2.980235593158234,-0.44126501296832527), linewidth(0.4)); draw((-4.450656162658686,-5.33024171139331)--(-2.8045901413496854,0.14273560487399775), linewidth(0.4)); draw((-6.16,3.88)--(-1.9,0.21), linewidth(0.4)); draw((-1.9,0.21)--(-4.450656162658686,-5.33024171139331), linewidth(0.4)); draw((-2.9069134899946403,-1.4273208830994124)--(-3.9607235462495836,1.9853181724732336), linewidth(0.4)); draw((-2.9069134899946403,-1.4273208830994124)--(-2.8045901413496854,0.14273560487399775), linewidth(0.4)); draw((-2.9486521304059456,-2.0677614476898487)--(-2.9069134899946403,-1.4273208830994124), linewidth(0.4)); draw((-5.8,-0.08)--(-14.58832583745327,-0.7334908956054996), linewidth(0.4)); draw((-3.9607235462495836,1.9853181724732336)--(-4.450656162658686,-5.33024171139331), linewidth(0.4) + wrwrwr); draw((-2.9486521304059456,-2.0677614476898487)--(-6.16,3.88), linewidth(0.4) + wrwrwr); draw((2,0.5)--(-11.433793812647506,3.8474892637385096), linewidth(0.4)); draw((-2.9486521304059456,-2.0677614476898487)--(-11.433793812647506,3.8474892637385096), linewidth(0.4)); draw((-11.540362797522098,-6.525836664813006)--(-3.9607235462495836,1.9853181724732336), linewidth(0.4)); draw((-11.540362797522098,-6.525836664813006)--(2,0.5), linewidth(0.4)); draw((-5.8,-0.08)--(-6.896106866781368,4.184907010995224), linewidth(0.4)); draw((-6.16,3.88)--(-6.896106866781368,4.184907010995224), linewidth(0.4)); draw((-5.8,-0.08)--(-5.256493221081515,-6.058574568103326), linewidth(0.4)); draw((-5.256493221081515,-6.058574568103326)--(-4.450656162658686,-5.33024171139331), linewidth(0.4)); /* dots and labels */ dot((-5.8,-0.08),linewidth(2pt) + dotstyle); label("$A$", (-5.716930594788596,-0.0019230727138700656), NE * labelscalefactor); dot((-6.16,3.88),linewidth(2pt) + dotstyle); label("$B$", (-6.082996630468782,3.9637923138215188), NE * labelscalefactor); dot((2,0.5),linewidth(2pt) + dotstyle); label("$C$", (2.0721411644064722,0.5878499847708801), NE * labelscalefactor); dot((-4.450656162658686,-5.33024171139331),linewidth(2pt) + dotstyle); label("$D$", (-4.374688463961247,-5.248869584129923), NE * labelscalefactor); dot((-3.37856160483175,0.10005567553815194),linewidth(2pt) + dotstyle); label("$E$", (-3.7968273589029216,0.1811099451262248), NE * labelscalefactor); dot((-2.8045901413496854,0.14273560487399775),linewidth(2pt) + dotstyle); label("$F$", (-2.7273913034004105,0.22178394909069035), NE * labelscalefactor); dot((-2.980235593158234,-0.44126501296832527),linewidth(2pt) + dotstyle); label("$P$", (-2.890087319258271,-0.36798910839405985), NE * labelscalefactor); dot((-1.9,0.21),linewidth(2pt) + dotstyle); label("$M$", (-1.8122262141999455,0.2827949550373886), NE * labelscalefactor); dot((-3.9607235462495836,1.9853181724732336),linewidth(2pt) + dotstyle); label("$X$", (-3.8866004163876657,2.0724511294738717), NE * labelscalefactor); dot((-2.9486521304059456,-2.0677614476898487),linewidth(2pt) + dotstyle); label("$Y$", (-2.869750317276038,-1.994949266972681), NE * labelscalefactor); dot((-2.9069134899946403,-1.4273208830994124),linewidth(2pt) + dotstyle); label("$Q$", (-2.829076313311573,-1.3441652035412326), NE * labelscalefactor); dot((-14.58832583745327,-0.7334908956054996),linewidth(2pt) + dotstyle); label("$U$", (-14.502515451113059,-0.6527071361453185), NE * labelscalefactor); dot((-11.433793812647506,3.8474892637385096),linewidth(2pt) + dotstyle); label("$K$", (-11.350280143867012,3.9231183098570535), NE * labelscalefactor); dot((-11.540362797522098,-6.525836664813006),linewidth(2pt) + dotstyle); label("$L$", (-11.451965153778175,-6.4487527010816565), NE * labelscalefactor); dot((-5.256493221081515,-6.058574568103326),linewidth(2pt) + dotstyle); label("$R$", (-5.167831541268317,-5.981001655490303), NE * labelscalefactor); dot((-3.09873970725935,0.12086294484481758),linewidth(2pt) + dotstyle); label("$V$", (-3.312109331151666,0.20144694710845756), NE * labelscalefactor); dot((-6.896106866781368,4.184907010995224),linewidth(2pt) + dotstyle); label("$S$", (-6.8151287018291535,4.26884734355501), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Consider the following lemmas:

Lemma 1. Let $ABCD$ be a cyclic quadrilateral whose diagonals intersect at $Y$ and $AD\cap BC=X.$ Let $P$ be a point on $XY$ and the second intersection of $\odot (ABCD)$ with $PA$ and $PB$ be $K$ and $L$ respectively. Then it follows that $XY, BK$ and $AL$ are concurrent.
Proof. Let $XY$ intersect $\odot(ABCD)$ at $I$ and $J.$ By Brocard's Theorem, it is easy to see that $(I,J;A,B)=-1.$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.172652771186477, xmax = 10.198530855245242, ymin = -4.859595378688876, ymax = 5.831089427237665; /* image dimensions */ /* draw figures */ draw(circle((2.235675488430095,-0.6187792711653043), 3.3775654652468465), linewidth(0.4) + blue); draw((-0.08,1.84)--(5.6,-0.32), linewidth(0.4)); draw((3.32,2.58)--(-1.138688740973718,-0.7657980432597458), linewidth(0.4)); draw((1.276600275234065,5.179061053430356)--(1.5220063065799334,1.23078633411749), linewidth(0.4)); draw((1.3536390081741951,3.9396046309159387)--(-0.8973864599809147,0.6429145504281734), linewidth(0.4)); draw((1.276600275234065,5.179061053430356)--(-1.138688740973718,-0.7657980432597458), linewidth(0.4)); draw((1.276600275234065,5.179061053430356)--(5.6,-0.32), linewidth(0.4)); draw((4.845512684584302,1.5252119325475504)--(1.3536390081741951,3.9396046309159387), linewidth(0.4)); draw((1.5220063065799334,1.23078633411749)--(1.8454944779205704,-3.9737319162954723), linewidth(0.4)); /* dots and labels */ dot((-0.08,1.84),linewidth(2pt) + dotstyle); label("$A$", (-0.026763442960632506,1.8952154190856052), NE * labelscalefactor); dot((3.32,2.58),linewidth(2pt) + dotstyle); label("$B$", (3.3772356992249386,2.6265433597895353), NE * labelscalefactor); dot((5.6,-0.32),linewidth(2pt) + dotstyle); label("$C$", (5.651000751231707,-0.27217465972786026), NE * labelscalefactor); dot((-1.138688740973718,-0.7657980432597458),linewidth(2pt) + dotstyle); label("$D$", (-1.0905131748936234,-0.7109714241502183), NE * labelscalefactor); dot((1.276600275234065,5.179061053430356),linewidth(2pt) + dotstyle); label("$X$", (1.329517465253931,5.232730203025358), NE * labelscalefactor); dot((1.5220063065799334,1.23078633411749),linewidth(2pt) + dotstyle); label("$Y$", (1.568861154938854,1.2835593232241365), NE * labelscalefactor); dot((1.3536390081741951,3.9396046309159387),linewidth(2pt) + dotstyle); label("$P$", (1.4092986951489053,3.9961211396532588), NE * labelscalefactor); dot((-0.8973864599809147,0.6429145504281734),linewidth(2pt) + dotstyle); label("$K$", (-0.8378726135595381,0.6984969706609924), NE * labelscalefactor); dot((4.845512684584302,1.5252119325475504),linewidth(2pt) + dotstyle); label("$L$", (4.893079067229451,1.5760904995057083), NE * labelscalefactor); dot((1.4330467410096437,2.6620339764674266),linewidth(2pt) + dotstyle); label("$I$", (1.4890799250438795,2.7196214613336718), NE * labelscalefactor); dot((1.8454944779205704,-3.9737319162954723),linewidth(2pt) + dotstyle); label("$J$", (1.9012829461679137,-3.915517491598348), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Now inverting around $P$ with radius $\sqrt{PI\cdot PJ},$ we obtain that $(I,J;K,L)=(J,I;A,B)=-1,$ which implies that $AB, CD, KL$ are concurrent, thus again by Brocard, $XY, BK$ and $AL$ are concurrent. $\square$

Lemma 2. Let $ABCD$ be a quadrilateral inscribed in the circle $\Gamma,$ Let $AC\cap BD= Y$ and $AD\cap BC=X.$ Let $P$ be a point on line $XY,$ let $K$ be the second intersection of $PA$ with $\Gamma$ and $L$ be the second intersection of $XK$ with $\Gamma$ , finally let $DL\cap XY= Z$ then it follows that $(X,Y;P,Z)=-1.$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.680500494225463, xmax = 17.0747406351982, ymin = -14.444614950563711, ymax = 13.205224035644418; /* image dimensions */ /* draw figures */ draw(circle((-3.64,-1.16), 5.381152292957336), linewidth(0.4) + blue); draw((-3.722093369329189,11.091310229368936)--(-8.790835128644087,-2.7174650806763028), linewidth(0.4)); draw((-8.790835128644087,-2.7174650806763028)--(-0.4303602250270546,3.1591448823710016), linewidth(0.4)); draw((-6.56,3.36)--(1.6846835554896908,-1.9375249410132471), linewidth(0.4)); draw((1.6846835554896908,-1.9375249410132471)--(-3.722093369329189,11.091310229368936), linewidth(0.4)); draw((-3.722093369329189,11.091310229368936)--(-3.2083861761728762,1.2064597573855096), linewidth(0.4)); draw((-3.722093369329189,11.091310229368936)--(-5.1074720182720705,4.017192856711877), linewidth(0.4)); draw((-6.951948556963188,-5.401202277188562)--(-5.1074720182720705,4.017192856711877), linewidth(0.4)); draw((-6.56,3.36)--(-3.3947298101688306,4.79211902592437), linewidth(0.4)); draw((-8.790835128644087,-2.7174650806763028)--(-2.5295738376256147,-11.855374989077417), linewidth(0.4)); draw((-2.5295738376256147,-11.855374989077417)--(-3.2083861761728762,1.2064597573855096), linewidth(0.4)); /* dots and labels */ dot((-6.56,3.36),linewidth(2pt) + dotstyle); label("$A$", (-6.413866364279621,3.50714618227291), NE * labelscalefactor); dot((-0.4303602250270546,3.1591448823710016),linewidth(2pt) + dotstyle); label("$B$", (-0.2923845986763235,3.3008041002862822), NE * labelscalefactor); dot((1.6846835554896908,-1.9375249410132471),linewidth(2pt) + dotstyle); label("$C$", (1.8054265681877277,-1.788967255383871), NE * labelscalefactor); dot((-8.790835128644087,-2.7174650806763028),linewidth(2pt) + dotstyle); label("$D$", (-8.649238919134758,-2.5799452363326107), NE * labelscalefactor); dot((-3.722093369329189,11.091310229368936),linewidth(2pt) + dotstyle); label("$X$", (-3.5938579104623716,11.244974256771453), NE * labelscalefactor); dot((-3.3947298101688306,4.79211902592437),linewidth(2pt) + dotstyle); label("$P$", (-3.2499544404846583,4.9171504091815335), NE * labelscalefactor); dot((-5.1074720182720705,4.017192856711877),linewidth(2pt) + dotstyle); label("$K$", (-4.969471790373225,4.360562775230565), NE * labelscalefactor); dot((-3.2083861761728762,1.2064597573855096),linewidth(2pt) + dotstyle); label("$Y$", (-3.0780027054958015,1.340554321413318), NE * labelscalefactor); dot((-6.951948556963188,-5.401202277188562),linewidth(2pt) + dotstyle); label("$L$", (-6.826550528252877,-5.262392302158773), NE * labelscalefactor); dot((-2.5295738376256147,-11.855374989077417),linewidth(2pt) + dotstyle); label("$Z$", (-2.390195765540375,-11.727777537739778), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Proof. Note that the lemma is purely projective, so let us take a projective transformation fixing $\Gamma$ and taking $Y$ to the center of the circle. Now, the problem is reduced to the following :

"Let $ABCD$ be a rectangle, $P$ be a point on the perpendicular bisector of $AB,$ let $K$ be the second intersection of $PA$ with $\Gamma.$ Let $L$ be the point on $\Gamma$ such that $AKLD$ is a trapezium and $Z$ be the point at which the perpendicular bisector of $AB$ intersects $LD.$ Prove that $PY=ZY.$ "

But now this is straightforward symmetry. Hence the lemma. $\square$

Now consider the following claims:

Claim 1. $S$ and $R$ both lie on $\omega.$
Proof. Angle chasing. $\square$

Claim 2. $Q$ lies on $\omega.$
Proof. Using Lemma 1 on cyclic quadrilateral $SBDR,$ we get $BY$ and $DX$ intersect on $AC.$ Now, letting $XE\cap \omega=Q',$ and applying Pascal's Theorem on cyclic hexagon $PBYQ'XD$ we get $Q',Y,F$ are collinear. Thus $Q=Q'.$ $\square$

Claim 3. $XYLK$ is cyclic.
Proof. Applying Brocard's theorem on quadrilateral $SBDR,$ we obtain $(B,D; U,V)=-1.$ Therefore, $(U,V;A,C)\overset{S}{=}(U,V;D,B)=-1,$ now since $M$ is the midpoint of $AC,$ therefore, $MA^2=MV\cdot MU=MX\cdot MB.$ Hence, $X$ is the $B-$ Humpty point of $\triangle ABC$ and $Y$ is the $D-$ Humpty point of $\triangle ADC,$ thus, $\angle AXC=180^{\circ}-\angle ABC=180^{\circ}-ADC=\angle AYC$ So, the claim follows. $\square$

Claim 4. $K$ and $L$ lie on $\omega.$
Proof. Suppose $K$ and $L$ both doesn't lie on $\omega.$ Let the circumcircle of $XYLK$ be $\omega'.$ Let $\omega'$ intersect $AC$ at $U'$ and $V',$ again by Brocard, we have $(U',V';A,C)=-1$ also since $AM=MC$ so $MA^2=MV'\cdot MU'.$ But we also have $MA^2=MV\cdot MU,$ therefore, $M$ lies on the radical axis of $\omega'$ and $\omega$ which means $X,Y,M$ are collinear, which is a contradiction to the assumption that $ABCD$ is not a parallelogram. Thus, the claim follows. $\square$

Claim 5. $P$ is the mid-arc point of $SR$ in $\omega$ and $Q$ is the mid-arc point of $KL$ in $\omega.$
Proof. Note that $BP$ is the exterior angle bisector of $\angle SBR$ of triangle $SBR,$ therefore the first part follows. Similarly, as the $A-$ Humpty point lies on the $A-$ apollonian circle, we similarly have that $XQ$ is the exterior angle bisector of $\angle KXL$ of triangle $KXL.$ $\square$

Coming back to the original problem,

Applying Lemma 2 on quadrilateral $KXYL,$ we obtain $(A,C;M,KB\cap AC)=-1$ and $(A,C;M, RL\cap AC)=-1$ thus, it follows that $KB\parallel AC\parallel LR.$ So, it suffices to solve the following problem:

" $KBRL$ is a cyclic trapezium. Let $P$ and $Q$ be the mid-arc points of minor arc $SR$ and major arc $KL$ respectively. Prove that $PQ$ is perpendicular to the parallel sides."

Let the center of $\omega$ be $O.$ Let the common perpendicular bisector of the parallel sides of the trapezium be $\ell$ . Let $P'$ and $Q'$ be the antipodes of $P$ and $Q$ with respect to $\omega$ respectively. Note that by symmetry, $P$ and $Q'$ are reflections of each other about $\ell$ . Let the midpoint of $PQ'$ be $N,$ it is obvious that $N\in\ell.$ also midpoint of $QQ'$ is $O,$ so $PQ\parallel NO$ but $NO$ is same as $\ell.$ Thus, we are done. $\blacksquare$

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Idio-logy

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Idio-logy

#11 h Aug 23, 2019, 1:25 PM • 2 Y

Y by Adventure10, Mango247

Solution 1

Solution 2

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AlastorMoody

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AlastorMoody

#12 h Sep 10, 2019, 10:36 AM • 5 Y

Y by AmirKhusrau, RudraRockstar, aops29, Adventure10, Mango247

stoooopeeeed

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Wizard_32

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Wizard_32

#13 h Oct 15, 2019, 9:44 AM • 2 Y

Y by HolyMath, Adventure10

cjquines0 wrote:

Let $\gamma_1,\gamma_2$ be the circles $(ABC),(ADC)$ respectively. Since $\angle ABC=\angle ADC,$ hence $\gamma_1,\gamma_2$ are symmetric about $AC.$

Redefine $X,Y$ so that the rays $MB,MD$ meet $\gamma_1,\gamma_2$ again in $Y,X$ respectively.

Claim 1: $X,B,D,Y$ are concyclic, say lie on $\omega.$ Further, $P \in \omega.$
Proof: The key observation is that if $MD \cap \gamma_1=\{Y,Y'\},$ then $Y'$ is the reflection of $D$ over $M.$ This is because the orthocenter of $YAC$ is concyclic with $A,D,C.$

Similarly define $X'.$ Then we have
$MX \cdot MB=MX \cdot MX'=MY \cdot MY'=MY \cdot MD$ Hence $X,B,D,Y$ are concyclic.

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We show $P \in (XBD).$ It is easy to see that $\measuredangle MBE=\measuredangle ELM,$ since they subtend equal arcs in their circles. Hence
$\measuredangle XBP=\measuredangle MBE=\measuredangle ELM=\measuredangle XLT=\measuredangle XDP$ Hence we are done here. $\square$

Let the perpendicular bisector of $AC$ meet $\gamma_1,\gamma_2$ in $\{S,K\},\{T,L\}$ respectively, so that $K,L$ lie in the interiors of $\gamma_2,\gamma_1$ respectively.

Claim 2: The point $Q$ lies on $\omega.$ Further, $K,L$ lie on $YQ, XQ$ respectively.
Proof: Firstly, see that $D$ is the $Y$ -HM point (The Humpty point) in $\triangle YAC.$ This is because it lies on the $Y$ median and $\measuredangle ADC=-\measuredangle AYC.$ Hence, by a well-known lemma, it lies on the $Y$ Apollonius circle and so $YA:YC=DA:DC=AF:FC,$ so $YF$ bisects $\angle AYC.$ Since $K$ is the arc midpoint in $\gamma_1,$ hence $YF$ passes through $K.$

Due to the angle bisectors, $P=DT \cap BS,$ as $T,S$ are arc midpoints.

Now notice that $EMXT, EMSY$ are both cyclic, due to right angles. Hence
$\measuredangle BXQ+\measuredangle BYQ = \measuredangle MXE+\measuredangle BYM=\measuredangle MTE+\measuredangle BSM = \measuredangle BPD$ Hence $Q \in \omega,$ which is what we wanted. $\square$

To conclude, notice that $M$ is the midpoint of both $KL, TS.$ Let $PQ \cap TS=Z.$ Our goal is to show $Z=\infty.$ Then
$(T,S;M,Z) \overset{P}{=} (D,B; PM \cap \omega, Q) \overset{M}{=} (Y,X;P,MQ \cap \omega) \overset{Q}{=} (K,L;M,Z)$ Thus $ZT:ZS=ZL:ZK.$ It is not too hard to check now, say algebraically that since $\{T,S\},\{K,L\}$ have the same midpoint, this is possible if and only if $Z \in \{M,\infty\}.$ Now, $P,M,Q$ are collinear happens only if $P, Q \in TS$ (Proved formally in the comments). Even here, $PQ, TS$ are "parallel", since they coincide. Hence we are done. $\blacksquare$

Comments: We present a formal way to see that $P,M,Q$ are collinear happens only twice, which is when $P,Q \in TS$ (note that $P$ lands on $TS$ exactly twice, when $B=K,S).$ Move $B$ on its circle $\gamma_1$ while fixing the other points (in particular $D).$ Then $B \overset{M}{\mapsto} P$ is projective and so $P$ moves oN $TD$ with degree $1.$ Also, $B \overset{M}{\mapsto} X \overset{L}{\mapsto} Q$ is projective and so $Q$ moves on $YK$ with degree $1$ (note that $Y$ is fixed as $D$ is fixed).

Hence, by Zack's lemma, $\deg MP \le 1$ and $\deg MQ \le 1.$ Now assume on the contrary that $M,P,Q$ are collinear for more than three positions of $B.$ Then by Zack's lemma $0=\deg M=\deg (MP \cap MQ) \le 1+1-3=-1,$ a contradiction.

Hence, for every position of $D,$ there are exactly two positions of $D$ for which $M,P,Q$ are collinear. And even in these two positions, $PQ \parallel TS,$ and hence we are home free!

This post has been edited 1 time. Last edited by Wizard_32, Oct 15, 2019, 9:45 AM

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lminsl

544 posts

lminsl

#14 h Oct 16, 2019, 1:53 AM • 2 Y

Y by Adventure10, Mango247

We will deal with non-trivial cases where $(AB, CD)$ and $(AD, BC)$ aren't parallel.

Denote $\Omega$ by the circumcircle of triangle $BPD$ , and $R$ by the antipodal point of $P$ on $\Omega$ .

Let $U=AB \cap CD$ , and $V=AD \cap BC$ , $E' = AC \cap BR$ , $F' = AC \cap DR$ .
Since $\angle PBV=\angle PDA$ , $V$ lies on $\Omega$ . Similarly $U$ lies on $\Omega$ too. Note that $\angle PUV=\angle PBV=\angle PVU$ , so $P$ is the midpoint of arc $UV$ .

Note that $\odot (AC)$ and $\Omega$ are orthogonal, so $MC^2=MY \cdot MD$ . Since $DF \perp DF'$ , $D(AC, FF')=-1$ , so $MC^2=MF \cdot MF'$ . Combining these results give $MY \cdot MD=MF \cdot MF'$ , so $(YDFF')$ are concyclic. Similarly $(XBEE')$ are concyclic too.

Hence, $\angle QFE=\angle YFF'=\angle RXY$ , and similarly we get $\angle QEF=\angle RYX$ . This leads to $\angle XQY=\angle XRY$ , thus $Q$ lies on $\Omega$ . Then $\angle QEF=\angle RYX=\angle RQX$ , so $RQ$ is parallel to $AC$ , but since $PQ\perp QR$ , we obtain $PQ \perp AC$ , and we're done.

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Physicsknight

640 posts

Physicsknight

#15 h Dec 17, 2019, 6:57 AM • 2 Y

Y by Adventure10, Mango247

By an angle chase, notice that $G$ , $H$ lies on $\odot (PBD)$ . $BI$ is the external angle bisector of $\angle ABC$ . By Braikenridge-Maclaurin's theorem(converse of Pascal's theorem) $ME.MI=MX.MB$ , hence $IBXE$ is concyclic quadrilateral, and $\angle IXE=90^{\circ}$ . Let $Q_1$ be the second intersection of $XE$ with $\odot (PBD)$ . Now angle chasing $PQ_1\perp CA$ .

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