4 concyclic points

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4 concyclic points

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Source: Japan Mathematical Olympiad Finals 2015 Q4

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178 posts

#1 h Apr 1, 2016, 11:08 AM • 5 Y

Y by Davi-8191, superagh, Adventure10, Mango247, Rounak_iitr

Scalene triangle $ABC$ has circumcircle $\Gamma$ and incenter $I$ . The incircle of triangle $ABC$ touches side $AB,AC$ at $D,E$ respectively. Circumcircle of triangle $BEI$ intersects $\Gamma$ again at $P$ distinct from $B$ , circumcircle of triangle $CDI$ intersects $\Gamma$ again at $Q$ distinct from $C$ . Prove that the $4$ points $D,E,P,Q$ are concyclic.

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#2 h Apr 1, 2016, 1:29 PM • 4 Y

Y by mathematiculperson, Bongcloud, Adventure10, Mango247

Hello.

My solution.

$[asy]import graph; size(13cm); 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 = -7.206468043275083, xmax = 18.668867444856577, ymin = -5.178703775799412, ymax = 9.461288671433072; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); draw((-0.28841398607392854,3.5125679783443826)--(0.,0.)--(6.34,0.)--cycle, aqaqaq); /* draw figures */ draw((-0.28841398607392854,3.5125679783443826)--(0.,0.), sqsqsq); draw((0.,0.)--(6.34,0.), sqsqsq); draw((6.34,0.)--(-0.28841398607392854,3.5125679783443826), sqsqsq); draw(circle((3.17,2.028410722499191), 3.7634226522076535)); draw(circle((1.1813942940691942,1.2823733195711395), 1.2823733195711395)); draw(circle((-3.7020066357528836,4.595866375681703), 5.901427020160467)); draw(circle((1.3777771808530928,-8.944590232727533), 10.228848896052138)); draw((-0.09667793668340248,1.1774318895876272)--(1.7818559729119043,2.4154783910403728)); draw((-0.4806487506386131,1.1140181598596959)--(2.106592378854092,5.6384681029903225)); draw((2.106592378854092,5.6384681029903225)--(1.7818559729119043,2.4154783910403728)); draw((-0.4806487506386131,1.1140181598596959)--(-0.09667793668340248,1.1774318895876272)); draw((-0.4806487506386131,1.1140181598596959)--(6.34,0.)); draw((-0.28841398607392854,3.5125679783443826)--(2.106592378854092,5.6384681029903225)); draw((2.106592378854092,5.6384681029903225)--(0.,0.)); draw((-0.09667793668340248,1.1774318895876272)--(1.1813942940691942,1.2823733195711395)); draw((1.1813942940691942,1.2823733195711395)--(6.34,0.)); draw((1.1813942940691942,1.2823733195711395)--(1.7818559729119043,2.4154783910403728)); draw((1.1813942940691942,1.2823733195711395)--(0.,0.)); /* dots and labels */ dot((-0.28841398607392854,3.5125679783443826),linewidth(3.pt) + dotstyle); label("$A$", (-0.6160395495874147,3.697703604997692), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.1296610629683985,-0.6067460015806296), NE * labelscalefactor); dot((6.34,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.6553188253668765,0.0012271066931445816), NE * labelscalefactor); dot((1.1813942940691942,1.2823733195711395),linewidth(3.pt) + dotstyle); label("$I$", (1.4753879428743548,1.3874057935573498), NE * labelscalefactor); dot((-0.09667793668340248,1.1774318895876272),linewidth(3.pt) + dotstyle); label("$D$", (0.016252483017306345,1.4603625665502027), NE * labelscalefactor); dot((1.7818559729119043,2.4154783910403728),linewidth(3.pt) + dotstyle); label("$E$", (2.0590421268171744,2.311524918133487), NE * labelscalefactor); label("$h$", (-6.671451707994166,9.120823730799758), NE * labelscalefactor); dot((2.106592378854092,5.6384681029903225),linewidth(3.pt) + dotstyle); label("$P$", (2.1319988998100268,5.910725719114231), NE * labelscalefactor); dot((-0.4806487506386131,1.1140181598596959),linewidth(3.pt) + dotstyle); label("$Q$", (-0.9808234145516768,1.2171733232406932), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Let $\angle{A},\angle{B},\angle{C}$ be the angles of $\triangle{ABC}$ .We will prove that $\angle{PQD}+\angle{PED}=180^{\circ}$ .

$\angle{PQD}=\angle{PQC}-\angle{CQD}=\angle{PAC}-\left(180^{\circ}-\angle{DIC}\right)$ .

We easily see that $\angle{DIC}=180^{\circ}+\frac{\angle{A}-\angle{B}}{2}$ .Hence $\angle{PQD}=\angle{PAC}+\frac{\angle{A}-\angle{B}}{2} \ (1)$ .

$\angle{PED}=\angle{PEI}-\angle{DEI}=180^{\circ}-\angle{PBI}-\frac{\angle{A}}{2}$ .We have $\angle{PBI}=\angle{PBC}-\frac{\angle{B}}{2}$ .

Thus, $\angle{PED}=180^{\circ}-\angle{PBC}+\frac{\angle{B}-\angle{A}}{2} \ (2)$ .

Adding $(1),(2)$ and taking in account that $\angle{PAC}=\angle{PBC}$ ,we have the desired result,which ends the proof.

This post has been edited 1 time. Last edited by gavrilos, Apr 1, 2016, 1:29 PM

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ABCDE

#3 h Apr 1, 2016, 1:55 PM • 5 Y

Y by mhq, Davi-8191, qweDota, Adventure10, Mango247

Invert about the incircle and the problem becomes:

Let $DEF$ be a triangle with midpoints $A'$ , $B'$ , and $C'$ . Lines $EB'$ and $FC'$ intersect the circumcircle of $D'E'F'$ at $P'$ and $Q'$ . Prove that $DEP'Q'$ is cyclic.

But $P'Q'$ is parallel to $DE$ , so the result follows from anti-parallels or the converse of Reim's theorem.

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#4 h Apr 1, 2016, 2:28 PM • 2 Y

Y by mathematiculperson, Adventure10

Consider the inversion $\Psi$ with center $I$ that fixed $\Gamma.$ I will use $\Psi(A)$ to denote the image of $A$ after inversion.

It's clear that $\Psi(D)$ is antipode of $I$ WRT $\odot(I\Psi(A)\Psi(B)),$ this is because $ID$ is perpendicular to $\overline{AB}.$ Similarly we can find $\Psi(E).$

At last $\Psi(P)\Psi(Q)$ is anti-parallel to $\Psi(B)\Psi(C)$ since they're concyclic. Notice that $I$ is the orthocenter of $\Psi(\triangle ABC)$ and $I\Psi(D),I\Psi(E)$ is diameter, it follows that $\Psi(D)\Psi(E) \parallel \Psi(B)\Psi(C).$ Hence we conclude that $\Psi(P),\Psi(Q),\Psi(D),\Psi(E)$ are concyclic, as desired. $\square$

The diagram after inversion:
$[asy] size(7cm); pointpen=black; pointfontpen=fontsize(9pt); pathpen=black; void b(){ pair A=D("A",dir(-75),dir(-90)*2); pair B=D("B",dir(20),dir(20)); pair C=D("C",dir(160),dir(160)); pair I=D("I",IP(A--foot(A,B,C),B--foot(B,C,A)),dir(180)); pair D=D("D",2*circumcenter(I,A,B)-I,dir(-45)); pair E=D("E",2*circumcenter(I,A,C)-I,dir(-135)); pair P=D("P",IP(L(B,E,-0.01,0),unitcircle),dir(-100)*1.5); pair Q=D("Q",IP(L(C,D,-0.01,0),unitcircle),dir(-80)); D(A--B--C--cycle,magenta); D(circumcircle(I,D,B),red+linetype("4 4")); D(circumcircle(I,E,C),red+linetype("4 4")); D(unitcircle); D(B--E); D(C--D); D(A--I); D(P--Q); D(D--E,dashed); } b(); pathflag=false; b(); [/asy]$

This post has been edited 1 time. Last edited by Complex2Liu, Apr 1, 2016, 2:46 PM
Reason: add a diagram!

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241 posts

#5 h Apr 2, 2016, 6:00 AM • 2 Y

Y by mathematiculperson, Adventure10

My solution:
Let $M$ and $N$ the midpoints of arcs $\overarc {ACB}$ and $\overarc {ABC}$ and let $M'=CI\cap \Gamma$ and let $P'=MD\cap \Gamma$ $\Longrightarrow$ $MM'\parallel ID$ $\Longrightarrow$ $\angle M'CP=\angle M'MD =\angle MDI$ $\Longrightarrow$ $IDP'C$ are cyclic $\Longrightarrow$ $P'=P$ $\Longrightarrow$ $P,D,M$ are collinear similarly $Q,E,N$ are collinear, but $MN\parallel DE$ $\Longrightarrow$ $\angle DEQ=\angle MNQ=\angle MPQ$ $\Longrightarrow$ $\angle DEQ=\angle MPQ$ hence we get $DEPQ$ are cyclic.

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#6 h May 1, 2018, 7:07 AM • 2 Y

Y by Adventure10, Mango247

After an inversion through $I$ (call F the intersection of the incircle and $BC$ ), the circumcircle of $ABC$ maps to the nine-point circle of $DEF$ , $P', Q'$ are the intersections of the medians to $E$ and $D$ to the nine-point circle. Let $G$ be the centroid of $DEF$ , $B', C'$ be the midpoints of $DF, EF$ respectively. Then we can see that by power of a point, $GB'*GP' = GC' * GQ'$ but since $GB' = \frac{GE}{2}$ $GC' = \frac{GD}{2}$ we get that $GP'*GE = GQ'*GD$ and so $DEP'Q'$ are concyclic by power of a point and the conclusion follows by inverting through $I$ again.

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jayme

#7 h May 1, 2018, 12:10 PM • 3 Y

Y by mira74, Adventure10, Mango247

Dear Mathlinkers,

all this problem can be managed only with the Reim's theorem and its converse...

Sincerely
Jean-Louis

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#9 h May 7, 2020, 9:18 PM • 1 Y

Y by H.M-Deadline

So me and my good friend Deadline have found a beautiful , yet overkill solution to this problem.
We know that
$\angle{BIE}+\angle{BPE}=180^{\circ}$ & $\angle{BIE}=\angle{DIE}+\angle{DIB}=(180^{\circ}-\angle{A})+(90^{\circ}-\frac{\angle{B}}{2})$ $\implies \angle{BPE}=\angle{A}+\frac{\angle{B}}{2}-90^{\circ}$ $\implies \angle{APE}=\angle{BPE}+\angle{APB}=\angle{A}+\frac{\angle{B}}{2}-90^{\circ}+\angle{C}=\frac{\angle{A}+\angle{C}}{2}=90^{\circ}-\frac{\angle{B}}{2}$ Also we have that:
$\angle{APC}=180^{\circ}-\angle{B}$ $\implies \angle{EPC}=(\angle{APC})-(\angle{APE})=(180^{\circ}-\frac{\angle{B}}{2})-(90^{\circ}-\frac{\angle{B}}{2}) \implies \angle{APE}=\angle{EPC}=90^{\circ}-\frac{\angle{B}}{2}$
So $PE$ bisects $\angle{APC}$ .
Lemma.(1)
There exists circle $\omega$ and $\omega'$ such that $\omega$ is tangent to $\Gamma$ and $AC$ at $P$ and $E$ respectively and similarly for $\omega'$ we have tangency to $\Gamma$ and $AB$ at $Q$ and $D$ respectively.
Proof.
Lemma(2)
It is very well known that two points $E$ and $F$ are Antihomologous $\iff$ $IE$ = $IF$ where $IE$ and $IF$ are tangents to the two circles
Proof.
(Statement 1) (still in the figure below) We know that for two Antihomologous points $E$ and $F$ we have that $ABFE$ is concyclic so we have that $SA \cdot SB =SE \cdot SF$

https://i.imgur.com/W6Qztt5.png

And now in the original question assume that the Exsimilicenter of the circles $\omega$ and $\omega'$ is $S$
(Recall $\omega$ and $\omega'$ from the first lemma)
Now we have by Monge's Theorem that $P$ (Exsimilicenter of $\omega$ and $\Gamma$ ) and $Q$ (Exsimilicenter of $\omega'$ and $\Gamma$ ) and $S$ ( Exsimilicenter of $\omega$ and $\omega'$ ) are collinear.
We can say that the tangents to $P$ and $Q$ at $\Gamma$ are also tangent to the circles $\omega$ and $\omega'$ so $P$ and $Q$ are Antihomologous .
We can also see that the tangents to $D$ and $E$ to the circles $\omega$ and $\omega'$ are also tangents to the incircle so $D$ and $E$ are also Antihomologous thus $S$ , $D$ , $E$ are also collinear .
By the statement above we have that $SE \cdot SD=SP \cdot SQ= \sqrt{P_{\omega}^{S}}\cdot \sqrt{P_{\omega'}^{S}}$ $\implies DEPQ$ is concyclic .
Q.E.D

This post has been edited 4 times. Last edited by Alireza_Amiri, Jun 7, 2020, 8:12 AM

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#11 h Jan 12, 2021, 11:38 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -36.16032070482368, xmax = 12.410545019947582, ymin = -23.148661982794046, ymax = 7.200192679219777; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw(circle((-8.084985197033781,-9.389784130129987), 6.867564972007016), linewidth(0.8) + red); draw(circle((-9.829137155351777,-7.660493935096696), 2.9945794601179148), linewidth(0.8) + red); draw((-11.580471648679685,-3.4783525682240515)--(-14.444490168347334,-11.982108408567374), linewidth(0.8) + blue); draw((-14.444490168347334,-11.982108408567374)--(-1.2764735712637427,-8.491108389669614), linewidth(0.8) + blue); draw((-1.2764735712637427,-8.491108389669614)--(-11.580471648679685,-3.4783525682240515), linewidth(0.8) + blue); draw(circle((-20.647002002422514,-0.7326784730841313), 12.846043240048244), linewidth(0.8) + linetype("4 4") + ffxfqq); draw(circle((-3.123158263901362,16.941767644510236), 25.499831915602535), linewidth(0.8) + linetype("4 4") + ffxfqq); draw(circle((-11.835344446663854,-2.869722934192979), 3.924124723290979), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-12.667084104221008,-6.704688903532239)--(-8.519111299179933,-4.967663008521221), linewidth(0.8) + blue); draw((-12.667084104221008,-6.704688903532239)--(-9.06174770854159,-10.55507846139481), linewidth(0.8) + blue); draw((-9.06174770854159,-10.55507846139481)--(-8.519111299179933,-4.967663008521221), linewidth(0.8) + blue); /* dots and labels */ dot((-11.580471648679685,-3.4783525682240515),dotstyle); label("$A$", (-11.462194212096858,-3.1488937431824264), NE * labelscalefactor); dot((-14.444490168347334,-11.982108408567374),dotstyle); label("$B$", (-14.319303960612814,-11.656731660985466), NE * labelscalefactor); dot((-1.2764735712637427,-8.491108389669614),dotstyle); label("$C$", (-1.1448534535670134,-8.164708635021531), NE * labelscalefactor); dot((-12.667084104221008,-6.704688903532239),linewidth(4pt) + dotstyle); label("$D$", (-12.541546783758442,-6.450442785911964), NE * labelscalefactor); dot((-8.519111299179933,-4.967663008521221),linewidth(4pt) + dotstyle); label("$E$", (-8.382864816474104,-4.704431272929996), NE * labelscalefactor); dot((-9.829137155351777,-7.660493935096696),linewidth(4pt) + dotstyle); label("$I$", (-9.716182699114885,-7.402812702083946), NE * labelscalefactor); dot((-7.92647451186035,-2.524048699207559),linewidth(4pt) + dotstyle); label("$P$", (-7.811442866770913,-2.260015154755243), NE * labelscalefactor); dot((-14.099486260710892,-6.0747840396677555),linewidth(4pt) + dotstyle); label("$Q$", (-13.97010165801642,-5.8155295084639755), NE * labelscalefactor); dot((-9.06174770854159,-10.55507846139481),linewidth(4pt) + dotstyle); label("$F$", (-8.922541102304896,-10.29166811447229), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Another favorite geometry problem

$\color{black}\rule{25cm}{1pt}$
Let $F$ be the touch point of $BC$ with the incircle of $ABC$ .
To solve this problem we shall use inversion around the incircle of $ABC$ .
First let's write down the inverted problem.

Inverted problem statement wrote:

Let $DEF$ be a scalene triangle. Let $A',B'$ and $C'$ be the midpoints of $DE,DF$ and $FE$ , respectively. Let $P'$ and $Q'$ be the intersections of the nine point circle of $DEF$ with $EB'$ and $DC'$ , respectively. Prove that $DQ'P'E$ is a cyclic quadrilateral.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.131199820393046, xmax = -0.12019751476062866, ymin = -12.10481146430267, ymax = 0.3987952050989597; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw(circle((-9.829137155351777,-7.660493935096696), 2.9945794601179148), linewidth(0.8) + red); draw(circle((-10.209403061170875,-7.283468241147083), 1.4972897300589572), linewidth(0.8) + red); draw((-12.667084104221008,-6.704688903532239)--(-8.519111299179933,-4.967663008521221), linewidth(0.8) + blue); draw((-12.667084104221008,-6.704688903532239)--(-9.06174770854159,-10.55507846139481), linewidth(0.8) + blue); draw((-9.06174770854159,-10.55507846139481)--(-8.519111299179933,-4.967663008521221), linewidth(0.8) + blue); draw((-10.593097701700469,-5.836175956026731)--(-9.06174770854159,-10.55507846139481), linewidth(0.8) + blue); draw((-9.260459750849632,-6.1252872839834)--(-11.674617750283346,-6.975211123550312), linewidth(0.8) + blue); draw((-9.260459750849632,-6.1252872839834)--(-9.670146561772933,-8.680278523038332), linewidth(0.8) + blue); draw((-9.670146561772933,-8.680278523038332)--(-11.674617750283346,-6.975211123550312), linewidth(0.8) + blue); draw((-8.519111299179933,-4.967663008521221)--(-10.864415989256797,-8.629883704434821), linewidth(0.8) + linetype("4 4") + ffxfqq); draw((-12.667084104221008,-6.704688903532239)--(-8.79042958673626,-7.761370756929311), linewidth(0.8) + linetype("4 4") + ffxfqq); draw(circle((-11.38305824230203,-3.949770664728511), 3.039456681147177), linewidth(0.8) + linetype("4 4") + qqwuqq); /* dots and labels */ dot((-12.667084104221008,-6.704688903532239),linewidth(4pt) + dotstyle); label("$D$", (-12.610725097688054,-6.598516058635217), NE * labelscalefactor); dot((-8.519111299179933,-4.967663008521221),linewidth(4pt) + dotstyle); label("$E$", (-8.464654685344604,-4.858997557557375), NE * labelscalefactor); dot((-9.829137155351777,-7.660493935096696),linewidth(4pt) + dotstyle); label("$I$", (-9.77256333277156,-7.55328937125689), NE * labelscalefactor); dot((-9.06174770854159,-10.55507846139481),linewidth(4pt) + dotstyle); label("$F$", (-9.013976317263925,-10.456846568544716), NE * labelscalefactor); dot((-10.593097701700469,-5.836175956026731),dotstyle); label("$A'$", (-10.544229434753463,-5.709138178384891), NE * labelscalefactor); dot((-8.79042958673626,-7.761370756929311),dotstyle); label("$C'$", (-8.739315501304265,-7.631763890102506), NE * labelscalefactor); dot((-10.864415989256797,-8.629883704434821),dotstyle); label("$B'$", (-10.805811164238854,-8.494983597404293), NE * labelscalefactor); dot((-11.67461775028334,-6.975211123550318),linewidth(4pt) + dotstyle); label("$H$", (-11.616714525643568,-6.873176874594876), NE * labelscalefactor); dot((-9.260459750849632,-6.1252872839834),linewidth(4pt) + dotstyle); label("$P'$", (-9.21016261437797,-6.023036253767359), NE * labelscalefactor); dot((-11.674617750283346,-6.975211123550312),linewidth(4pt) + dotstyle); label("$Q'$", (-11.616714525643568,-6.873176874594876), NE * labelscalefactor); dot((-9.670146561772933,-8.680278523038332),linewidth(4pt) + dotstyle); label("$J$", (-9.615614295080325,-8.57345811624991), NE * labelscalefactor); dot((-10.082647759231175,-7.40914347246362),linewidth(4pt) + dotstyle); label("$G$", (-10.03414506225695,-7.304786728245769), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $G$ be the centroid of $DEF$ .
Since we have that $GP'.GB'=GC'.GQ'$ , this implies that:
$\frac{GP'}{GQ'}=\frac{GB'}{GC'}=\frac{\frac{1}{2} GD}{\frac{1}{2}GE}=\frac{GD}{GE}$ thus we have that $GP'.GE=GD.GQ'$ , thus we have that $DQ'P'E$ is a cyclic quadrilateral.

Thus this implies that $DQPE$ is a cyclic quadrilateral.

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559 posts

#12 h Apr 16, 2022, 8:33 PM

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Let incircle meet $BC$ at $F.$ Invert wrt incircle, images denote with hatches. We see that $P',Q'$ are second intersections of $DEF$ nine-point circle with $B'E,C'F$ respectively, and $B'C'\parallel DE$ so by Reim $D,E,P',Q'$ are concyclic, done.

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8874 posts

#13 h Aug 13, 2023, 8:26 PM

Y by

Invert about the incircle. The inverted problem reads the following:

inverted problem wrote:

Let triangle $DEF$ have medians $\overline{DB}$ and $\overline{EC}$ , and let them intersect the nine-point circle at $P$ and $Q$ . Show that $DEPQ$ is cyclic.

This is just power of a point at the centroid $G$ : $GP \cdot GD = 2GP \cdot GB = 2 GQ \cdot GC = GQ \cdot GE.$

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#16 h Dec 24, 2023, 10:02 PM

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Invert with respect to the incircle. We can now think in terms of $\triangle DEF$ . Points $A$ , $B$ , and $C$ are mapped to the respective midpoints of the sides. Then points $P$ and $Q$ are mapped to the intersection of the nine-point circle with the $E$ and $D$ median, respectively.

Note that $P^*Q^*$ is then antiparallel to $B^*C^*$ , which in turn makes it antiparallel to $DE$ , giving the desired. $\blacksquare$

This post has been edited 1 time. Last edited by shendrew7, Dec 24, 2023, 10:03 PM

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#17 h Dec 30, 2023, 6:18 PM

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Loved this problem.

Perform an inversion w.r.t. the incircle. The problem statement changes to as follows:

Inverted problem statement wrote:

Let $ABC$ be a triangle. $M$ and $N$ be the midpoints of $AB$ and $AC$ respectively. Let $\odot(N_9)$ be the nine point circle of $\triangle ABC$ . Define $P = BN\cap \odot(N_9)\; (\neq N)$ , $M=CM\cap \odot(N_9)\; (\neq M)$ . Prove $BCPQ$ is cyclic.

Let $G$ be the centroid.

We use directed lengths throughout. Now note that,
$\begin{align*} GP\cdot GB&=-3\cdot GP\cdot \left(\dfrac{-GB}{3}\right)\\ &= -3 \cdot GP \cdot GN\\ &= -3 \cdot \operatorname{Pow}_{\odot(N_9)}(G)\\ &= -3 \cdot GQ \cdot GM\\ &= -3 \cdot GQ \cdot \left(\dfrac{-GC}{3}\right)\\ &= GQ \cdot GC .\end{align*}$
This gives us that $BCPQ$ is cyclic. So inverting back, we are done. :yoda:

:yoda:

This post has been edited 1 time. Last edited by kamatadu, Dec 30, 2023, 6:19 PM

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#18 h May 2, 2024, 5:21 PM

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Consider inversion $\Psi$ around the incircle. Let the incircle touch $AB$ at $F$ and let $\Omega$ denote the nine-point circle of $\triangle DEF$ .

Clearly, $\Psi(E)=E$ and $\Psi(D)=D.$ We also have that $\Psi(\Gamma)=\Omega.$

Let $M,N$ be the midpoints of $DF,EF$ . Now $\Psi((BIE))=ME$ and $\Psi((CID))=DN.$

Hence, $P^*=\Psi(P)=\Omega\cap ME$ and $Q^*=\Psi(Q)=\Omega\cap DN$ . Now $DEPQ$ is cyclic iff $DEP^*Q^*$ is cyclic.

If $G$ is the centroid of $\triangle DEF$ , from PoP in $\Omega$ it follows that $GM\cdot GP^*=GN\cdot GQ^* \iff \frac{GP^*}{GQ^*}=\frac{GN}{GM}=\frac{GD}{GE}\iff GP^*\cdot GE = GQ^*\cdot GD.$ Hence, $DEP^*Q^*$ is cyclic and we are done.

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#19 h May 3, 2024, 3:53 AM

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Let $X$ be the $A$ -intouch point.
Inverting about the incircle, $\Delta A'B'C'$ is the medial triangle of $\Delta XED$ , and $(ABC)'$ is the nine-point circle of $\Delta XED$ .
$P'$ and $Q'$ are the second intersections of $(ABC)'$ with $EB'$ and $DC'$ .
Let $G$ be the centroid of $\Delta XED$ . $GB' \cdot GP' = GC' \cdot GQ' \implies 2GB' \cdot GP' = 2GC' \cdot GQ' \implies GE \cdot GB' = GD \cdot GP' \implies D,E,P',Q'$ concyclic. Inverting back gives the desired result. $\square$

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#20 h May 3, 2024, 4:29 AM

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Invert about the incircle, so that our restated problem is

Quote:

Suppose that in some $\triangle DEF$ , the medians from $D$ and $E$ intersect the nine-point circle again at points $P$ and $Q$ . Prove that $DPQE$ is cyclic.

This restatement is now trivial by Reim's theorem.

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#21 h Jul 26, 2024, 9:51 PM

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Switch noatation such that $\triangle DEF$ is the intouch triangle of $\triangle ABC$ with the usual nomenclature. Just invert about incircle, note that $A^*$ is midpoint of $\overline{EF}$ (and analogous for others) and then do power of a point from centroid of the intouch triangle.

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#22 h Feb 19, 2025, 11:02 PM

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Invert about incircle. Then, circumcircle becomes 9 point circle, and $P^*, Q^*$ are the intersections from $B, C$ -medians with the 9 point circle. From POP on centroid, this is done.

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#23 h Apr 30, 2025, 4:24 PM

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Invert about the incircle, $P'$ , and $Q'$ go where the $D$ , and $E$ medians intersect the Nine Point Circle of $\triangle DEF$ . Cyclicity follows from PoP.

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