Circumcircle of ADM

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6877 posts

#1 h Jul 19, 2012, 8:12 PM • 16 Y

Y by Davi-8191, baladin, anantmudgal09, Tumon2001, doxuanlong15052000, tenplusten, Durjoy1729, AlastorMoody, snoofmath, HamstPan38825, Adventure10, Mango247, ehuseyinyigit, and 3 other users

Triangle $ABC$ is inscribed in circle $\Omega$ . The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$ ), respectively. Let $M$ be the midpoint of side $BC$ . The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$ ), respectively. Let $N$ be the midpoint of segment $PQ$ , and let $H$ be the foot of the perpendicular from $L$ to line $ND$ . Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$ .

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v_Enhance

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v_Enhance

#2 h Jul 19, 2012, 9:02 PM • 39 Y

Y by SMOJ, joey8189681, Mediocrity, vsathiam, Davi-8191, Pure_IQ, baladin, 62861, tenplusten, niyu, AlastorMoody, Pluto1708, myh2910, HolyMath, kiyoras_2001, llplp, Aryan-23, HamstPan38825, SSaad, rayfish, IAmTheHazard, Adventure10, Mango247, Sedro, Mathandski, Ritwin, and 13 other users

$[asy]import graph; size(6cm); real lsf=0.2; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-13.31,xmax=14.17,ymin=-6.54,ymax=9.19; pen zzttqq=rgb(0.6,0.2,0); pair A=(-5,6), B=(-6,-1), C=(5,-1), M=(-0.5,-1), D=(-1.97,-1), Q=(-5.56,2.11), P=(2.43,0.8), L=(-0.5,-4.38), H=(-2.47,-4.06); D(A--B--C--cycle,zzttqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR((-0.5,1.79),6.17)); D(CR((-1.23,3.48),4.54)); D(A--L); D(P--Q); D(M--(-1.56,1.45)); D(M--L); D(H--(-1.56,1.45)); D(H--M); D(H--L); D(A); MP("A",(-5.23,6.16),NE*lsf); D(B); MP("B",(-6.34,-1.22),N*lsf); D(C); MP("C",(5.17,-0.92),NE*lsf); D(M); MP("M",(-0.4,-0.86),NE*lsf); D(D); MP("D",(-1.87,-0.86),NE*lsf); D(Q); MP("Q",(-5.95,1.83),NE*lsf); D(P); MP("P",(2.48,0.36),N*lsf); D(L); MP("L",(-0.4,-4.24),NE*lsf); D((-1.56,1.45)); MP("N",(-1.46,1.58),NE*lsf); D(H); MP("H",(-2.75,-3.84),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

By some angle chasing, we see the problem is equivalent to showing that $AD \parallel MN$ .

Claim: $QB = PC$ .
Proof: Use Power of a Point to get $BM \cdot BD = AB \cdot QB$ and $CM \cdot CD = AC \cdot QC$ . Now use the angle bisector theorem.

Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$ , which must be parallel to the angle bisector since they have the same magnitude.

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pi37

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pi37

#3 h Jul 19, 2012, 11:10 PM • 10 Y

Y by Binomial-theorem, Pluto1708, rashah76, SSaad, Adventure10, and 5 other users

Let $X$ be the midpoint of arc $BAC$ . Then $\angle DAX=\angle XMD=90$ , so $ADMX$ is cyclic with diameter $XD$ . Then clearly $X,N,D,H$ are collinear, so $\angle LHD=\angle LHX=90$ , and $H$ lies on $\Omega$ . Also $X$ is the spiral symmetry center that brings $QP$ to $BC$ , so it brings $NP$ to $MC$ , and $\triangle XNM\sim\triangle XPC$ . Finally $\angle LHD=\angle LMD=90$ , so $DMLH$ is cyclic. Then
$\angle NMX=\angle PCX=\angle ACX=\angle ALX=\angle DLM=\angle DHM=\angle NHM$
and we are done.

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leader

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leader

#4 h Jul 21, 2012, 12:09 PM • 2 Y

Y by Adventure10 and 1 other user

$CP*CA=CM*CD$ $BM*BD=BQ*BA$ $BA/CA=BD/CD$ $BM=MC$ so $CP=BQ$ if $K,L$ are midpoints of $BP$ and $CQ$ resp. than $KNLM$ is a rhombus because $CP=BQ$ and $MN$ is the bisector of $\angle KNL$ because $KN\parallel BA$ and $NL\parallel CA$ we have $MN\parallel AL$ and since $\angle LMD=\angle LHD=90$ $LMDH$ is cyclic so
$\angle LPH=\angle LDH=\angle MNH$ so $ML$ is tangent to $\odot MHN$

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Babai

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Babai

#5 h Jul 21, 2012, 1:41 PM • 5 Y

Y by vsathiam, Adventure10, Mango247, and 2 other users

See as angle LMN is right angle, $L,H,D,M$ cyclic.So the problem reduces to show that $AD||MN$ .
The circum-circle of $ADM$ is $-a^2yz-b^2zx-c^2xy+(\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z)(x+y+z)=0$ .
So, $Q=(a^2,2(b+c)c-a^2,0)$ , $P=(a^2,0,2(b+c)b-a^2)$ .
The equation of a line parallel to $AD$ which passes through $M$ is nothing but $(b-c)x+(b+c)y-(b+c)z=0$ ......(i)
The point $N=(a^2(b+c),b(2(b+c)c-a^2),c(2(b+c)b-a^2))$ obviously lies on (i).
So done.

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AnonymousBunny

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AnonymousBunny

#6 h Jun 23, 2014, 10:28 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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By PoP, we have that $BQ \times AB = BD \times BM \implies BQ = \dfrac{\dfrac{ac}{b+c} \times \dfrac{a}{2}}{c} = \dfrac{a^2}{2(b+c)}.$ Note that this expression is symmetric in $b,c.$ Similarly we get that $CP= \dfrac{a^2}{2(b+c)},$ so $BQ=CP.$ Let $X,Y$ be the midpoints of $BP,CQ$ respectively. Since $XN= MY = \dfrac{1}{2} BQ = \dfrac{1}{2} CP = XM = NY,$ $\triangle NXM \cong \triangle NYM.$ Consequently, $MN$ bisects $\angle XNY.$ However, since $XN \parallel AB$ and $NY \parallel AC,$ $MN \parallel AD.$ Since $\angle LBC = \angle LAC = \angle LAB = \angle LCB,$ $\triangle LBM \cong \triangle LCM \implies LM \perp BC,$ which in turn implies quadrilateral $HLMD$ is cyclic. It follows that $\angle HML = \angle HDL = \angle HNM,$ so we're done. $\blacksquare$

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junioragd

314 posts

junioragd

#7 h Nov 2, 2014, 12:27 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

First,let $R$ be the midpoint of the arc $BAC$ .Now,from PoP we easy get that $BQ=CP$ . Then,from an easy angle chase we obtain that it is enough to prove that $MN$ is parallel to $AD$ and this is true by an simple angle chase using that $RQB$ and $RMN$ are similar(cause $RQN$ is similar to $RMN$ ),so we are finished.

This post has been edited 3 times. Last edited by junioragd, Nov 5, 2014, 5:05 PM

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sayantanchakraborty

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sayantanchakraborty

#8 h Nov 4, 2014, 9:50 AM • 1 Y

Y by Adventure10

Let $\angle{CAM}=\theta,\angle{APQ}=x$ .Then we have $\angle{QMB}=\angle{AMB}-\angle{AMQ}=C+\theta-x$ .So $\angle{PMC}=A-C-\theta+x$ .Also by PoP we have $BQ \cdot AB=BD \cdot BM$ and $CP \cdot AC=CM \cdot CD$ and equating these we get $BQ=CP$ .Now sine rule in $\triangle{BQM}$ and $\triangle{CPM}$ gives $\frac{BM}{\sin{\angle{BQM}}}=\frac{BQ}{\sin(C+\theta-x)}$ and $\frac{CM}{\sin{\angle{CPM}}}=\frac{CP}{\sin(A-C-\theta+x)}$ .Dividing these two relations and using the fact that $\sin{\angle{BQM}}=\sin{\angle{CPM}}$ we get $\sin(C+\theta-x)=\sin(A-C-\theta+x)$ .Note that these angles cannot be supplementary so $C+\theta-x=A-C-\theta+x \implies x=C+\theta-\frac{A}{2}$ .Now $\angle{HDL}=\angle{ADN}=\angle{QDN}-\angle{QDA}=\angle{QDN}-\angle{QPA}=90-\frac{A}{2}-x=90-C-\theta$ .
$\angle{LMD}=\angle{LHD}=90^{\circ} \implies LMDH$ is cyclic $\implies \angle{HML}=\angle{HDL}=90-C-\theta$ .Let $AM$ meet $\Omega$ at $X$ .Then $\triangle{XBC} \sim \triangle{MQP}$ so $\triangle{XMC} \sim \triangle{MNP}$ .Thus $\angle{MNP}=\angle{XMC}=C+\theta$ .Finally $\angle{HNM}=90-\angle{MNP}=90-C-\theta=\angle{HML}$ and we are through.

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Sardor

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Sardor

#9 h Dec 28, 2014, 3:47 PM • 2 Y

Y by Adventure10, Mango247

We must prove that $MN || AD$ , but it's true :
Lemma: (from SHARYGIN'S book )
Let $ABCD$ be a convex quadrilateral, and $M \in AD , N \in BC$ , such that $frac{AB}{CD}=frac{AM}{MD}=frac{BN}{NC}$ and $BA \cap CD=P$ .Then angle bisector of $\angle BPC$ and $MN$ are parallel.
Proof : (hint) Let $ABNX$ and $DCNY$ are parallelogramm ...

But in the our problem we have $frac{BQ}{CP}=frac{QN}{NP}=frac{BM}{CM}=1$ , so we are done.

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john10

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john10

#10 h Apr 10, 2016, 6:48 PM • 2 Y

Y by Adventure10, Mango247

Sorry to revive this topic . can you explain more how did you got this ?
Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$ , which must be parallel to the angle bisector since they have the same magnitude.[/quote]

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tenplusten

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tenplusten

#12 h Feb 17, 2017, 12:00 PM • 2 Y

Y by Ferid.---., Adventure10

Here is a solution by Ferid and Murad (me)
A little angle-chasing gives that we need to prove $MN\parallel AD$ .
Let $MN\cap AC=K$ .
We will prove that $\frac{CM}{CD}=\frac{CK}{AC}$
Let's use Barycentric coordinates
It's easy to get $K=(\frac{b+c}{2b}:0:\frac{b-c}{2b})$
The rest is clear by Distance formula.

This post has been edited 1 time. Last edited by tenplusten, Feb 17, 2017, 12:01 PM

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rezareza14

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rezareza14

#13 h Feb 17, 2017, 1:33 PM • 1 Y

Y by Adventure10

This question was nice

. Thank you for it

HINT: (attachment figure)
Prove $L,M,F$ are aligned.
Prove $D,N,F$ are aligned.
Prove $H$ lies on $\Omega$ .
Prove $\angle DHM=\angle DLM$
Prove $NM \parallel DL$ by Thales theorem.

GOOD LUCK

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Delray

348 posts

Delray

#14 h May 6, 2017, 3:45 PM • 2 Y

Y by Adventure10, Mango247

v_Enhance wrote:

Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$ , which must be parallel to the angle bisector since they have the same magnitude.

Why does this work?

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Delray

348 posts

Delray

#15 h May 6, 2017, 4:14 PM • 1 Y

Y by Adventure10

It suffices to show that $\measuredangle{NHM}=\measuredangle{NML}$ . However we have that $\measuredangle{NML}=\measuredangle{NMD}+\measuredangle{DML}=\measuredangle{NMH}+\measuredangle{HML}$ . We also have that $\measuredangle{NHM}=\measuredangle{NMH}+\measuredangle{HNM}$ . We must now show that $\measuredangle{HNM}=\measuredangle{HML}$ , but since $LM$ bisects $BC$ , we have that $\angle{BML}=90$ , so $DMLH$ is cyclic, so we have that $\measuredangle{HML}=\measuredangle{HDL}$ , implying it suffices to show that $AD\parallel MN$ .

We show this by using barycentrics on $\triangle{ABC}$ . We have that $A=(1,0,0)$ , $D=(0:b:c)$ and $M=(0,\frac{1}{2},\frac{1}{2})$ . Plugging in these point into the general circle formula $-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$ we get that the equation of the circle passing through these points is $-a^2yz-b^2xz-c^2xy+\frac{a^2}{2(b+c)}(cy+bz)(x+y+z)=0$ We let $P=(n:0:m)$ , where $n+m=b$ . Plugging in and simplifying yields $P=\bigg(\frac{a^2}{2(b+c)}:0:b-\frac{a^2}{2(b+c)}\bigg)$ Exploiting symmetry, we find that $Q=\bigg(\frac{a^2}{2(b+c)}:c-\frac{a^2}{2(b+c)}:0 \bigg)$ Now we have that $N$ is the midpoint of $PQ$ , but homogenizing both and taking the average seems like a pain, so instead, we find that $N=\frac{1}{2}(cP+bQ)$ , which evaluates to $N=\bigg(\frac{a^2}{4}:\frac{b}{2}(c-\frac{a^2}{2(b+c)}):\frac{c}{2}(b-\frac{a^2}{2(b+c)})\bigg)=\bigg(\frac{a^2}{4bc},\frac{1}{2}-\frac{a^2}{4c(b+c)},\frac{1}{2}-\frac{a^2}{4b(b+c)}\bigg)$ We now easily compute
$\vec{NM}=\bigg(-\frac{a^2}{4bc},\frac{a^2}{4c(b+c)},\frac{a^2}{4b(b+c)}\bigg)=\bigg(-1:\frac{b}{b+c}:\frac{c}{b+c}\bigg)=\vec{AD}$ implying that $AD\parallel MN$ as desired. $\square$

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v_Enhance

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v_Enhance

#16 h May 7, 2017, 2:26 PM • 4 Y

Y by v4913, HamstPan38825, Adventure10, Mango247

Delray wrote:

v_Enhance wrote:

Now notice that the vector $\vec{MN}$ is half of $\vec{BQ} + \vec{CP}$ , which must be parallel to the angle bisector since they have the same magnitude.

Why does this work?

If $\vec v$ and $\vec w$ are two unit vectors then $\vec v + \vec w$ is parallel to the angle bisector of the two vectors (at the origin).

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IAmTheHazard

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IAmTheHazard

#56 h Aug 16, 2023, 5:02 PM • 1 Y

Y by centslordm

Let $K$ be the antipode of $L$ , hence the midpoint of arc $BAC$ . Since $\angle KAD=\angle KMD=90^\circ$ , $K$ lies on $(ADM)$ . Therefore, it is the center of the spiral similarity sending $\overline{QP}$ to $\overline{BC}$ , which also sends $N$ to $M$ , and sends $D$ to $L$ as well since $D$ is the midpoint of arc $PQ$ (and $L$ is the midpoint of arc $BC$ ). Thus it follows that $\frac{KN}{KD}=\frac{KM}{KL}$ , so $\triangle KMN \sim \triangle KLD \implies \measuredangle KMN=\measuredangle KLD=\measuredangle MLD$ . Furthermore, since $DMLH$ is clearly cyclic, it follows that $\measuredangle MLD=\measuredangle MHD=\measuredangle MHN$ , implying the desired tangency. $\blacksquare$

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eibc

600 posts

eibc

#57 h Aug 22, 2023, 3:41 AM

Y by

Let $K$ be the antipode of $L$ wrt $(ABC)$ , so that $K$ , $M$ , and $L$ are collinear. Then since $\overline{KM} \perp \overline{KD}$ and $\overline{KA} \perp \overline{AD}$ , $K$ also lies on $(ADM)$ , and thus must be the center of the spiral sim which sends $\overline{BC}$ to $\overline{QP}$ . Evidently this spiral similarity also takes $M$ to $N$ and $L$ to $D$ (the latter because $\overline{KL}$ and $\overline{KD}$ are diameter in their respective circles). So, $K$ is also the center of the spiral similarity taking $\overline{MN}$ to $\overline{LD}$ , and thus $\triangle KMN \sim \triangle KLD$ . To finish, note that $HDML$ is cyclic since $\overline{LH} \perp \overline{HD}$ and $\overline{LM} \perp \overline{MD}$ , so
$\measuredangle KMN = \measuredangle KLD = \measuredangle MLD = \measuredangle MHD = \measuredangle MHN,$ and $\overline{KML}$ is indeed tangent to $(HMN)$ .

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john0512

4187 posts

john0512

#58 h Oct 7, 2023, 11:39 PM

Y by

First, note that since $\angle QAD=\angle PAD$ , $D$ is the arc midpoint of $PQ$ . Hence, $ND\perp PQ$ , and as a result $PQ\parallel HL$ .

Next, consider the antipode of $L$ on $(ABC)$ , which we call $T$ . Since $\angle TAD=\angle TMD=90,$ $T$ also lies on $(ADM)$ . This also means that it is the antipode of $D$ on $(ADM)$ since $\angle TAD=90$ , and hence $T$ also lies on $ND$ .

Note that this further means that $H$ is on $(ABC)$ , since line $DN$ goes through the antipode of $L$ .

Thus, $DMLH$ is cyclic since $\angle DML=\angle DHL=90$ , so $\angle DHM=\angle DLM.$ As such, it suffices to show that $NM\parallel AL$ since then we would have $\angle TMN=\angle TLA=\angle THM$ by cyclic $DMLH$ which would imply the circumcircle tangency since $T,M,L$ are collinear.

Unfortunately, $NM$ is a terrible line to angle chase with, so we turn to barycentric coordinates to prove that $AL\parallel NM$ .

We know that $D=(0:b:c)$ and $M=(0:1:1)$ . Obviously, the equation of $(ADM)$ has $u=0$ . Plugging in $D$ and $M$ into this equation reveals that $2(v+w)=a^2$ and $vb+wc=\frac{a^2bc}{b+c}.$ Solving this system of equations yields $v=\frac{a^2c}{2(b+c)}, w=\frac{a^2b}{2(b+c)}.$ Thus, intersecting this circle with $AB$ and $AC$ yields $P=(\frac{a^2}{2b(b+c)},0,1-\frac{a^2}{2b(b+c)})$ and symmetrically $Q=(\frac{a^2}{2c(b+c)},1-\frac{a^2}{2b(b+c)},0).$ Let $m_b=2b(b+c)$ and $m_c=2c(b+c)$ ( $m$ for "messy"). Then, we rewrite this as $P=(\frac{a^2}{m_b},0,1-\frac{a^2}{m_b})$ and $Q=(\frac{a^2}{m_c},1-\frac{a^2}{m_c},0).$ Thus, we have $N=(\frac{a^2}{2}(\frac{1}{m_b}+\frac{1}{m_c}), \frac{1}{2}-\frac{a^2}{2m_c},\frac{1}{2}-\frac{a^2}{2m_b}).$ Thus, $N-M=(\frac{a^2}{2}(\frac{1}{m_b}+\frac{1}{m_c}), -\frac{a^2}{2m_c},-\frac{a^2}{2m_b}).$ If we multiply this by $m_bm_c\frac{2}{a^2},$ this becomes $(m_b+m_c,-m_b,-m_c).$ However, expanding out the definition and dividing by $2(b+c)$ , this is $(b+c,-b,-c).$ Hence, $N-M$ is a real multiple of $(b+c,-b,-c)$ . However, $A-D=(\frac{b+c}{a+b+c},\frac{-b}{a+b+c},\frac{-c}{a+b+c})$ is also a real multiple of $(b+c,-b,-c)$ , hence $NM\parallel AD$ and we are done.

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EpicBird08

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EpicBird08

#59 h Oct 27, 2023, 12:55 AM

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Solved with r00tsOfUnity.

We first note that $\angle NHL = 90^\circ$ , so $DHLM$ is cyclic. Thus
$\measuredangle LMH = \measuredangle LDH.$ We wish to prove that this is equal to $\measuredangle MNH,$ so it suffices to show that $AD \parallel MN.$

To do so, we will first prove that $BQ = CP.$ By Power of a Point on $B,$ we get
$BQ \cdot AB = BD \cdot BM,$ so
$BQ = \frac{BD \cdot BM}{AB}.$ Similarly,
$CP = \frac{CD \cdot CM}{CA}.$ However, by the Angle Bisector Theorem, $\frac{BD}{AB} = \frac{CD}{CA},$ and since $BM = CM,$ this implies the claim.

Thus if we set the circumcenter of $\triangle ABC$ (or any point on that matter) to be the origin, then the vector $b - q - (p - c) = b + c - p - q$ is parallel to the angle bisector of $\angle BAC.$ But this implies that the vector $\frac{b+c-p-q}{2} = m - n$ is parallel to $AD,$ so we are done.

This post has been edited 3 times. Last edited by EpicBird08, Oct 28, 2023, 3:57 AM

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starchan

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starchan

#60 h Oct 27, 2023, 8:20 AM • 1 Y

Y by mxlcv

solution

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shendrew7

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shendrew7

#61 h Dec 21, 2023, 6:26 PM

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The key step is to notice that $(ADM)$ passes through $K$ , where $K$ is the top point of $(ABC)$ wrt $BC$ , as $\angle KAD = \angle KMD = 90$ .

Claim 1: Point $H$ lies on $(ABC)$ .

The previous result tells us $KD$ , the diameter of $(ADM)$ , passes through $N$ and $KL$ , the diameter of $(ABC)$ , passes through $M$ . We finish by noting $\angle KHL = 90$ . ${\color{blue} \Box}$

Claim 2: $\angle KMN = \angle MHN$ , which gives the desired result.

Note $K$ is the center of spiral similarity sending $QB \rightarrow PC$ , and by the Gliding Principle, it is also the center of spiral similarity mapping $NM \rightarrow PC$ . Thus $\triangle KMN \sim \triangle KCP$ .

This congruency along with the cyclicity of $DMLH$ then gives the angle equality
$\angle KMN = \angle KPC = \angle KLA = \angle MHN. \quad \blacksquare$

This post has been edited 1 time. Last edited by shendrew7, Dec 21, 2023, 6:47 PM

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OronSH

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OronSH

#62 h Mar 13, 2024, 5:55 PM

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Let $R$ be the arc midpoint of $(BAC).$ We have $\measuredangle ADM=\measuredangle DCA+\measuredangle CAD=\measuredangle BLA+\measuredangle LAB=\measuredangle ABL=\measuredangle ARM$ so $ARMD$ is cyclic. Furthermore it is the miquel point of $BCPQ,$ and the spiral similarity taking $BC$ to $QP$ takes $M$ to $N.$ Also notice $LHDM$ is cyclic with diameter $LD.$

Thus $\measuredangle HNM=\measuredangle RNM=\measuredangle RPC=\measuredangle RPA=\measuredangle RDA=\measuredangle HDL=\measuredangle HML.$

This post has been edited 2 times. Last edited by OronSH, Mar 13, 2024, 5:56 PM

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dolphinday

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dolphinday

#63 h Mar 14, 2024, 7:59 PM • 1 Y

Y by peppapig_

First note that $LMDH$ is cyclic as $\angle LHD = \angle LMD = 90^{\circ}$ .
Let $L'$ be the antipode of $L$ . Then notice that $L' \in (ADM)$ since $\angle L'AL = 90^{\circ}$ and $L' - O - M - L$ with $\angle L'MD = 90^{\circ}$ . So it follows that $L'$ is the Miquel point of quadrilateral $PQBC$ . So it follows that the spiral similarity at $L'$ sends $MN \to BQ$ by the gliding lemma(mean geometry theorem?), from which we get $\angle L'MN = \angle L'BQ = \angle L'LD = \angle MHD$ which gives our tangency as desired.

This post has been edited 1 time. Last edited by dolphinday, Mar 14, 2024, 8:00 PM

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MagicalToaster53

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MagicalToaster53

#64 h May 11, 2024, 10:16 PM

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Observe that as $DHLM$ is cyclic, what we want to show is $\angle HML = \angle HNM = \angle MHD$ , or rather $\triangle NMH$ is isosceles. This is however also equivalent to showing $NM \parallel DL$ . Therefore it suffices to show $NM \parallel DL$ .

Also take note that $DN \perp QP$ , as $AD$ bisects $\angle QAP$ , so $D$ is the midpoint of minor arc $\hat{QP}$ . From this, we make the following claim:

Claim: $X$ is the midpoint of major arc $\hat{BC}$ .
Proof: If $X' = LM \cap (ABC)$ , then $\angle X'AD = 90^{\circ} = \angle X'MD \implies AX'MD \text{ is cyclic, so that } X = X'. \phantom{c} \square$
Now observe the spiral similarity at $X$ sending $QD \stackrel{X}{\mapsto} BL, DP \stackrel{X}{\mapsto} LC$ , and $QP \stackrel{X}{\mapsto} BC$ . From these spiral similarites, we find that $D \stackrel{X}{\mapsto} L$ and $N \stackrel{X}{\mapsto} M$ under $X$ . Hence $X$ is also the midpoint of major arc $\hat{QP}$ . Therefore, the spiral similarity taking $N \mapsto M$ takes $D \mapsto L$ . Thus $\triangle XNM \sim \triangle XDL \implies NM \parallel DL$ , as desired. $\blacksquare$

Remark

This post has been edited 2 times. Last edited by MagicalToaster53, May 11, 2024, 10:24 PM

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ihatemath123

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ihatemath123

#65 h Jun 20, 2024, 3:15 AM

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Let $L'$ be the midpoint of major arc $BC$ . Since $\angle L'AD = \angle DML' = 90^{\circ}$ , it follows that $L'$ lies on $(ADM)$ . Since lines $AL'$ and $AD$ are the external and internal angle bisectors of $\angle QAP$ respectively, it follows that $L'$ lies on line $DN$ .

Since $\angle LHN = \angle LHL' = 90^{\circ}$ , it follows that $H$ is the second intersection of $(DML)$ and $(ABC)$ .

Since $\angle QL'P = \angle BAC$ , it follows that $\tfrac{L'N}{L'D} = \tfrac{L'M}{L'L}$ . Since $DHLM$ is a cyclic quadrilateral, $L'D \cdot L'H = L'M \cdot L'L$ . Multiplying these two equations together gives us
$L'N \cdot L'H = L'M^2,$ so $(HMN)$ is tangent to line $ML$ as desired.

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AN1729

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AN1729

#66 h Oct 22, 2024, 12:48 PM

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Let $T$ be the midpoint of arc $BAC$
Thus, $TD$ is diameter of circle $(APDMQ)$
Let $TD \cap PQ = N' \implies TN' \perp PQ$
Spiral Similarity centered at $T$ takes
$P \rightarrow B$
$Q \rightarrow C$
$D \rightarrow L$
$\implies N' \rightarrow M$
So, $N \equiv N' \implies MN \parallel DL$
Also, $H \in (ABC)$ as $\angle LHT = 90$
Now, $\angle LMH = \angle LDH = \angle MNH$
$\implies LM$ is tangent to $(MNH)$ $\blacksquare$

This post has been edited 5 times. Last edited by AN1729, Oct 22, 2024, 1:01 PM

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cj13609517288

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cj13609517288

#67 h Nov 13, 2024, 2:13 PM

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Note that $\angle DML=90^{\circ}$ so $DHLM$ is cyclic. Then we want to prove
$\angle HML=\angle HNM$ $\angle HDL=\angle DNM$ $\angle ADN=\angle DNM,$ so we want to prove $AD$ and $NM$ are parallel.

Note that by power of a point,
$(BQ)(BA)=(BD)(BM)\Longrightarrow BQ=(BM)\cdot\frac{BD}{BA}=(CM)\cdot\frac{CD}{CA}=CP.$ Now we claim that the concyclicity is actually irrelevant: for any $Q$ , $P$ in this orientation such that $BQ=CP$ , we have $AD\parallel NM$ .

In fact, if we vary $Q$ linearly (not in the projective sense, just literally $Q=B+t\overrightarrow{BA}$ ), $P$ will also vary linearly, so $N$ will also vary linearly. Thus $N$ will always lie on a line, and we claim that this is exactly the line through $M$ parallel to $AD$ .

To do this, it suffices to check two points $Q$ . If $Q=B$ , then $P=C$ and $N=M$ . If $Q=A$ , then $P$ is the point on $AC$ such that $BA=PC$ , and their midpoint $Z$ has barycentric coordinates $(b+c:0:b-c)$ . Thus the displacement vector
$\overrightarrow{ZM}=\left(-\frac{b+c}{2b},\frac12,\frac{c}{2b}\right)$ will be parallel to
$\frac{1}{b+c}(-b-c,b,c)$ which is exactly the displacement vector $\overrightarrow{AD}$ . $\blacksquare$

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Mathandski

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Mathandski

#68 h Dec 13, 2024, 5:43 AM

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

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mananaban

36 posts

mananaban

#69 h Dec 29, 2024, 11:41 PM

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Again spiral very nice. Also stealing/using Dr. Chen's diagram from #2.

Claim. $\odot ABC \cap \odot AQP = X$ is the antipode of $D$ wrt $\odot AQP$ and the antipode of $L$ wrt $\odot ABC$ .
Proof. Let $X'$ be the antipode of $L$ wrt $\odot ABC$ . Then since $\angle LMD = \angle LAX'$ , $MDAX'$ is cyclic and $X'$ lies on $\odot AQP$ . Thus $X=X'$ .
Since $\angle DMX =90^{\circ}$ , $DX$ is a diameter of $\odot AQP$ . But $\overleftrightarrow{DN}$ is also a diameter of $\odot AQP$ as well, so $X$ is the antipode of $D$ wrt $\odot AQP$ as well. $\Box$

We now show that it is sufficient to prove that $NM \parallel DL$ through a simple angle chase (noting that $HDML$ is cyclic). $\angle HML = \angle HDL$ , but we seek $\angle HNM = \angle HML$ , so it is sufficient to prove that $\angle HDL = \angle HNM$ , which is $NM \parallel DL$ .

Now consider the spiral similarity centered at $X$ sending $QP$ to $BC$ . Since $N$ and $M$ are midpoints of $QP$ and $BC$ , this spiral similarity also sends $N$ to $M$ . Similarly, since $D$ and $L$ are both arc midpoints of arcs $QP$ and $BC$ , $D$ is sent to $L$ . Now the spiral itself implies that
$\frac{DX}{DN} = \frac{LX}{LM} \implies \frac{XN}{XD} = \frac{XM}{XL}.$ This then implies that $\triangle NXM \sim \triangle DXL$ (by SAS) since $XND$ and $XML$ are collinear, so $NM \parallel DL$ , as desired. $\blacksquare$

This post has been edited 1 time. Last edited by mananaban, Dec 29, 2024, 11:42 PM
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Saucepan_man02

1339 posts

Saucepan_man02

#70 h Jan 23, 2025, 3:41 PM

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We bary it (for practice):

Take $\triangle ABC$ as reference triangle.
Note that: $D = (0:b:c)$ with $(ADM)$ given by: $(ADM): a^2 yz+ b^2 zx+c^2xy - \left( \tfrac{a^2c}{2(b+c)} y + \tfrac{a^2b}{2(b+c)} z \right )(x+y+z) = 0.$
Let $P = (p, 0, 1-p)$ . Plugging it into the circle equation, we get: $p=\tfrac{a^2}{2b(b+c)}$ , Similarly, $Q=(q, 1-q, 0)$ with $q=\tfrac{a^2}{2c(b+c)}$ .
Therefore, $N = (a^2(b+c):b(2c(b+c)-a^2): c(2b(b+c)-a^2))$ .
Note that, the point at infinity of $AD$ is $P_\infty = (-(b+c): b:c)$ . Therefore, line $M P_\infty$ is given by: $M P_\infty: (b-x)x+(b+c)y-(b+c)z=0$ . Plugging $N$ into the line, gives that $N$ lies on $M P_\infty$ or $MN \parallel AD$ .

Therefore: $\angle DNM = \angle ADN = \angle HDL = \angle HML$ and we are done.

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