Constant Angle Sum

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Source: AMASCIWLOFRIAA1PD (mock oly geo contest) P3

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#1 h May 11, 2021, 1:06 PM • 4 Y

Y by amar_04, sotpidot, centslordm, MS_asdfgzxcvb

Let $ABC$ be a triangle with circumcircle $\Omega$ , $A$ -angle bisector $l_A$ , and $A$ -median $m_A$ . Suppose that $l_A$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$ . A line $l$ parallel to $\overline{BC}$ meets $l_A$ , $m_A$ at $G$ , $N$ respectively, so that $G$ is between $A$ and $D$ . The circle with diameter $\overline{AG}$ meets $\Omega$ again at $J$ .

As $l$ varies, show that $\angle AMN + \angle DJG$ is constant.

MP8148

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#2 h May 11, 2021, 1:26 PM • 4 Y

Y by Mathematicsislovely, Bumblebee60, centslordm, MS_asdfgzxcvb

$[asy] defaultpen(fontsize(9pt)); size(5cm); pair A,B,C,I,X,D,M,O,A1,M1,V,J,P1,Q1,N; A=dir(135); B=dir(210); C=dir(330); I=incenter(A,B,C); X=(B+C)/2; O=circumcenter(A,B,C); D=extension(A,I,B,C); M=-A+2*foot(O,A,I); A1=-A+2O; M1=-M+2O; V=-M1+2*foot(O,M1,D); J=-A1+2*foot(O,A1,I); P1=(D+X)/2; Q1=extension(A,P1,I,X); N=extension(D,Q1,A,X); draw(A--B--C--A); draw(circumcircle(A,B,C)); fill(A--V--D--cycle,0.8*black+0.5*white); fill(A--X--M--cycle,0.8*black+0.5*white); draw(A--M); draw(A--A1--J); draw(M--M1--V); draw(A--X); draw(I--N); draw(arc(circumcenter(J,D,V),circumradius(J,D,V),-40,100)); draw(circumcircle(A,D,X)); draw(A--V--D--A); draw(A--X--M--A); draw(V--I); draw(M--N); dot("$A$" , A , dir(A)); dot("$B$" , B , dir(B)); dot("$C$" , C , dir(C)); dot("$G$" , I , dir(250)); dot("$X$" , X , dir(310)); dot("$M$" , M , dir(M)); dot("$D$" , D , dir(220)); dot("$A^*$" , A1 , dir(A1)); dot("$M_A$" , M1 , dir(M1)); dot("$V$" , V , dir(270)); dot("$J$" , J , dir(J)); dot("$O$" , O , dir(0)); dot("$N$" , N , dir(80)); [/asy]$
Claim: As $G$ varies over $\ell_A$ , $\measuredangle NMA+\measuredangle DJG=\angle\left(\frac{B-C}{2}\right)$ which is constant as $\Delta ABC$ is fixed.

Proof: Let $A^*$ be the $A-$ antipode of $\odot(ABC)$ and $X$ be the midpoint of $BC$ . Clearly $A^*,G,J$ are collinear. Let $M_A$ be the midpoint of the major arc $\widehat{BAC}$ and let $\overline{M_AD}\cap\odot(ABC)=V$ . Now notice that $OM_A=OM$ and $AO=A^*O$ , this $AM_A^*M$ is a rectangle $\implies \overline{AD}\parallel\overline{A^*M_A}$ , thus by the converse of Reims we have $J,G,D,V$ are concyclic and $-1=(AB,AC;AD,AM_A)\overset{A}{=}(BC;MM_A)\overset{D}{=}(AV;BC)\implies\overline{AV}$ is the $A-$ Symmedian of $\Delta ABC$ . Now, notice that $\measuredangle MAX=\measuredangle VAD$ and $\measuredangle DVA=\measuredangle XMA$ and as $GN\parallel\overline{BC}$ , $\frac{GA}{GD}=\frac{NA}{NX}\implies\Delta AVD\cup G\stackrel{+}{\sim}\Delta AMX\cup N$ . Thus, $\measuredangle NMA+\measuredangle DJG=\measuredangle NMA+\measuredangle DVG=\measuredangle NMA+\measuredangle XMN=\measuredangle M_AMA=\measuredangle M_ABA=\angle\left(\frac{B-C}{2}\right)$ which is constant as $\Delta ABC$ is fixed. $\blacksquare$

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MP8148

#3 h May 12, 2021, 7:56 AM • 2 Y

Y by centslordm, Mango247

Here is an extension for this problem. I wasn't able to prove it, but geogebra tells me that it's true.

Let $l$ meet $\overline{AC}$ , $\overline{AB}$ at $E$ , $F$ , and let $\overline{JD}$ meet $\Omega$ again at $K$ . Show that $\overline{AK}$ passes through the $M$ -Dumpty point in $\triangle MEF$ .

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#4 h May 12, 2021, 9:00 PM • 1 Y

Y by centslordm

Epic problem.

Let $m_A$ meet $\Omega$ again at $Y$ , and let $X$ be the reflection of $Y$ over $OM$ . Let $m_A$ meet the midpoint of $BC$ at $E$ and let $AB < AC$ WLOG. Here, $\overarc{AB}$ will denote the angle subtending any minor arc $AB$ , and $\overbrace{AB}$ will denote the angle subtending any major arc $AB$ .

Lemma 1: $DEMX$ is cyclic.

Proof: Notice $\angle EYM = \angle AYM = \overarc{AM}$ , and that $\angle EDM = \overarc{AB} + \overarc{CM}$ . Since $M$ is the midpoint of $\overarc{BC}$ , $\overarc{BM} = \overarc{CM}$ and thus $\angle EDM = \overarc{AM} = \angle EYM$ . Then, since $X$ is the reflection of $Y$ over $OM$ , we have $\angle EXM = \angle EDM$ and thus $DEMX$ is cyclic. $\square$

Lemma 2: $XDGJ$ is cyclic.

Proof: Since $DEMX$ is cyclic and $\angle DEM = 90^\circ$ , $\angle DXM = 90^\circ$ . Then, we have $\angle AMX + \angle MDX + 90^\circ = 180^\circ$ . We see that $\angle AMX = \overarc{AX}$ , and thus $\angle MDX + 90^\circ = \overbrace{AX} = \angle AJX$ . Since $AG$ is the diameter of the circumcircle of $\triangle GAJ$ , $\angle AJG = 90^\circ$ , and thus $\angle GJX = \angle MDX$ . It then follows that $\angle GDX + \angle GJX = 180^\circ$ . $\square$

Lemma 3: Let $MN$ meet $\Omega$ again at $R$ . Then, $R$ , $G$ , $X$ are collinear.

Proof: We first claim that $RNGA$ is cyclic. Let the two points at which $GN$ meets $\Omega$ be $P$ and $Q$ . Then $M$ is the midpoint of $\overarc{PMQ}$ , since $PQ \parallel BC$ . Then there exists an inversion about $M$ w.r.t some arbitrary circle with radius $r$ that transforms $\Omega$ to the line $PQ$ . This implies that $MN \cdot MR = MG \cdot MA = r^2$ , and thus $RNGA$ is cyclic. Then let $RG$ meet $\Omega$ again at a point $X'$ . Since $\angle NRG = \angle NAG$ , $\angle MRX' = \angle MAY$ , and thus $X' = X$ . $\square$

Now, we consider $\triangle XRM$ , where we see trivially that $X$ and $M$ are fixed as $\ell$ varies. Since $XDGJ$ is cyclic, we have $\angle DJG = \angle DXG$ , and we also see that $\angle AMN = \angle GMN$ . Then, we see that: $\angle AMN + \angle DJG = \angle GMN + \angle DXG = 180^\circ - \angle XRM - \angle DXM - \angle DMX.$ Since $D$ , $X$ , $M$ are fixed, and $\angle XRM$ is constant since $R$ lies on $\Omega$ , we can conclude that $\angle AMN + \angle DJG$ is constant as $\ell$ varies. $\blacksquare$

This post has been edited 1 time. Last edited by sotpidot, May 12, 2021, 9:00 PM

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#5 h Jun 3, 2021, 7:52 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $E$ be the midpoint of $BC$ . Let $L$ and $O$ be points on $AB$ and $AC$ respectively such that $AOGLJ$ is cyclic. We will prove $\angle GJD= \angle NME$ and $\angle AMN=\angle MJD$ . This finishes the solution because $\angle AMN + \angle DJG=\angle AME$ is constant.

Let the foot of perpendicular from $A$ to $HF$ be $I$ . $I$ is on $AOGLJ$ . Let $A'$ be the antipode of $A$ . Observe that $J-G-A'$ is collinear. We have $\angle IJG=\angle IAG=\angle A'AM=\angle A'JM=\angle MJG \implies \textbf{J-I-M}$ are collinear.

$\angle EMA=\angle MAI=\angle GJM$ . Proving $\frac{sin(\angle MJD)}{sin(\angle DJG)}=\frac{sin(\angle AMN)}{sin(\angle NME)}$ finishes the solution because $\frac{sin(x)}{sin(k-x)}$ is injective where k is constant.( $0<x<k<90$ ).

Observe that $J$ is a spiral similarity center which takes triangle $LGO$ to triangle $BMC$ . We have $MD\cdot MA=MB^2$ , $sin(A)=OL/AG$ , $ME\cdot BC=sin(A)\cdot MB^2$ . The last two comes from the area of $BMC$ and $ALGO$ .

$\frac{sin(\angle MJD)}{sin(\angle DJG)} =\frac{DM\cdot JG }{DG\cdot JM} = \frac{DM\cdot OL }{DG\cdot BC} =\frac{\frac{MB^2}{MA}\cdot AG \cdot ME }{DG\cdot MB^2}=\frac{AG\cdot ME }{DG\cdot MA}=\frac{AN\cdot ME }{DG\cdot AM}=\frac{sin(\angle AMN)}{sin(\angle NME)}$ $\square$

This post has been edited 5 times. Last edited by SerdarBozdag, Jun 3, 2021, 9:11 AM

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#6 h Jun 16, 2023, 10:28 PM • 2 Y

Y by GeoKing, MS_asdfgzxcvb

Let $A'$ be the $A$ -antipode on $\Omega$ . Let $JD$ hit $\Omega$ again at $T_1$ , let $T_1'$ be the reflection of $T_1$ over $AA'$ , and let $T_1''$ be the $T_1'$ -antipode on $\Omega$ . Let $MN$ hit $\Omega$ again at $T_2$ . Note that $\measuredangle GJA = \pi /2 = \measuredangle A'JA$ , so $A'-G-J$ . Furthermore,
$\measuredangle DJG = \measuredangle T_1JA' = -\measuredangle T_1'JA' = -\measuredangle T_1''JA = \measuredangle AMT_1'',$ while
$\measuredangle NMA = \measuredangle T_2MA,$ so it suffices to show that $\measuredangle T_2MT_1''$ is constant.

Animate $G$ on $l_A$ . We have that $G \stackrel{\infty_{BC}}\mapsto N \mapsto MN = MT_2$ is projective. In addition, $G \stackrel{A'}\mapsto J \stackrel{D}\mapsto T_1 \stackrel{\infty_{AA}}\mapsto T_1' \stackrel{O}\mapsto T_1'' \mapsto MT_1''$ is projective. Thus, $\deg MT_2 = \deg MT_1'' = 1$ . The following lemma finishes the problem:

Lemma: Let $k{}$ be a constant, and let $A{}$ , $B{}$ , $C{}$ , and $D{}$ be moving points on a line $\ell$ . Then, the statement `` $(A, B; C, D) = k$ '' has degree at most $\deg A + \deg B + \deg C + \deg D$ .
Proof: This is just writing it out. Take a projective transformation onto $\mathbb P^1$ . Then, the statement `` $\deg A = d$ '' is equivalent to $A = n_A(t) / d_A(t)$ for polynomials $n_A, d_A$ with $\max \{\deg n_A, \deg d_A\} = d$ . Similarly, we can write $B = n_B(t) / d_B(t)$ , $C = n_C(t) / d_C(t)$ , and $D = n_D(t) / d_D(t)$ . Then,
$(A, B; C, D) = \frac{A-C}{B-C} \div \frac{A-D}{B-D} = \frac{n_A(t) d_C(t) - n_C(t) d_A(t)}{n_B(t) d_C(t) - n_C(t) d_B(t)} \div \frac{n_A(t) d_D(t) - n_D(t) d_A(t)}{n_B(t) d_D(t) - n_D(t) d_B(t)},$ so `` $(A, B; C, D) = k$ '' is just,
$(n_A(t) d_C(t) - n_C(t) d_A(t))(n_B(t) d_D(t) - n_D(t) d_B(t)) - k(n_B(t) d_C(t) - n_C(t) d_B(t))(n_A(t) d_D(t) - n_D(t) d_A(t)) = 0,$ which has at most the desired degree. $\square$

Observe that this also implies that if $(A, B; C, D)$ is the same for $\deg A + \deg B + \deg C + \deg D + 1$ values of $t{}$ , then it is always constant (alternatively, you can just pick that one value, let's say $k{}$ , and then the above lemma will tell you that $(A, B; C, D) = k$ is always true since it is true for more than $\deg A + \deg B + \deg C + \deg D$ values of $t{}$ ). Finally, observe that the dual will also be true.

Now for the problem, let $I$ and $J$ be the circular points at infinity. By Laguerre's formula,
$\measuredangle T_2 MT_1'' = \frac1{2i} \log (MT_2, MT_1''; MI, MJ),$ so it suffices to show that $\measuredangle T_2 MT_1''$ is constant for $3{}$ values of $t{}$ .

if $G = A$ , then $N = T_2 = A$ , and $J = A$ as well. Thus, $T_1 = M$ , so $T_1'$ is the reflection of $M$ over $AA'$ , and $T_1''$ is the point such that $AA'T_1T_1''$ is a cyclic isosceles trapezoid. Thus, $\measuredangle T_2MT_1'' = \measuredangle MAA'$ .
if $G = D$ , then $N$ is the midpoint of $BC$ , so $MT_2$ is the perpendicular bisector of $BC$ . In addition, $J-D-A'$ , so $T_1=A'$ , and $T_1' = A'$ , so $T_1'' = A$ . Thus, $\measuredangle T_2MT_1'' = \measuredangle OMA = \measuredangle MAO = \measuredangle MAA'$ .
if $G = M$ , then $N = MM \cap m_A$ , so $MT_2$ is the tangent to $\Omega$ at $M$ . In addition, $J = G$ , so $T_1 = A$ , and $T_1'=A$ , so $T_1''=A'$ . Thus, $\measuredangle T_2MT_1'' = \measuredangle MAA'$ (by tangency).

We are done. $\blacksquare$

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#7 h Apr 16, 2025, 6:35 AM

Y by

Let $S$ be the intersection of $A-$ symmedian and $(ABC)$ , $K$ be the midpoint of $BC$ , $L$ be the midpoint of arc $BAC$ , $A'$ be the antipode of $A$ , $AK\cap (ABC)=W,SG\cap (ABC)=P,JD\cap (ABC)=Q$ .
Pascal at $SPMMAW$ gives $G,PM\cap AW,BC_{\infty}$ are collinear thus, $P,M,N$ are collinear. Pascal at $PQJA'LS$ implies $PQ\cap A'L,D,G$ are collinear and since $A'L\parallel DG$ we get $PQ\parallel A'L$ . Hence $\measuredangle GJD+\measuredangle AMN=\measuredangle A'JQ+\measuredangle AMP=\measuredangle PML+\measuredangle AMP=\frac{\measuredangle B-\measuredangle C}{2}$ as desired. $\blacksquare$

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