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An epitome of config geo

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Source: The Golden Digits Contest, December 2024, P3

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#1 h Dec 22, 2024, 8:34 AM • 3 Y

Y by trigadd123, Funcshun840, Rounak_iitr

Let $ABC$ be a scalene acute triangle with incenter $I$ and circumcircle $\Omega$ . $M$ is the midpoint of small arc $BC$ on $\Omega$ and $N$ is the projection of $I$ onto the line passing through the midpoints of $AB$ and $AC$ . A circle $\omega$ with center $Q$ is internally tangent to $\Omega$ at $A$ , and touches segment $BC$ . If the circle with diameter $IM$ meets $\Omega$ again at $J$ , prove that $JI$ bisects $\angle QJN$ .

Proposed by David Anghel

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#2 h Dec 22, 2024, 2:10 PM • 1 Y

Y by AndreiVila

Truly an epitome. Here is a completely synthetic solution that is not very neatly written. Rename $J$ to $T$ (I will change this eventually).

Let $AM$ meet $BC$ at $Q$ and let $D$ be the projection of $I$ onto $BC$ . Notice that $\omega$ is tangent to $BC$ at $Q$ . Observe that $T$ is simply the $A$ -mixtilinear touchpoint.

Claim. $QI, BC$ and $AT$ concur.
Proof. Denote the incircle of $\triangle ABC$ by $\gamma$ , its circumcircle by $\Omega$ and its $A$ -mixtilinear incircle by $\omega_{A}$ . Let $QI$ an $BC$ meet at $W$ , the exsimilicenter of $\gamma$ and $\omega$ .

By Monge on $\Omega, \omega$ and $\gamma$ we find that $A, W$ and the exsimilicenter of $\Omega$ and $\gamma$ are collinear.
By Monge on $\Omega, \omega_{A}$ and $\gamma$ we find that $A, T$ and the exsimilicenter of $\Omega$ and $\gamma$ are collinear.

Combining the above, $A, T$ and $W$ are collinear, as desired.

We will show that $TM, NQ$ and $BC$ concur at a point $R$ .

First, let's see how this finishes. Let $L$ be the antipode of $M$ in $\left(ABC\right)$ and let $TD$ meet $\left(ABC\right)$ again at $A'$ . Since $IT\perp TR$ , the desired reduces to showing that
$-1=\left(TI\cap NQ, R; N, Q\right)=\left(R, IT\cap BC; D, AT\cap BC\right)=\left(M, L; A', A\right)$ The latter is trivial because it's well-known that $AA'\parallel BC$ and thus $ALA'M$ is a kite.

We now end the solution by showing that $TM, NQ$ and $BC$ concur indeed. Let $BC$ and $TM$ meet at $R$ . Let $E$ and $F$ be the tangency points of the incircle with sides $AC$ and $AB$ respectively. Finally, let $\left(AEF\right)$ meet $\left(ABC\right)$ at $S$ . It's well-known (and follows by radical axes) that $AS$ passes through $R$ . Furthermore, it is also well-known that $RI\perp AI$ .

Let $D'$ be the reflection of $D$ in $N$ . Then $D'$ lies on $\left(AEF\right)$ and $AD'\parallel BC$ . Similarly, let $K'$ be the reflection of $K$ across $Q$ . Then $\angle KAK'=90^{\circ}$ and by, say, homothety, the desired reduces to showing that $D'K'$ passes through $R$ . Let $RD'$ meet $\left(AEF\right)$ at $P$ .

Claim. $KP\perp A'R$ .
Proof. Note that
$\angle RPI=180^{\circ}-\angle D'PI=\angle D'AI=\angle IKR,$ so $RPIK$ is cyclic. Thus $\angle RPK=\angle RIK=90^{\circ}$ .

Now let $D'R$ meet the line perpendicular to $BC$ at $K$ at $K_{0}$ . We show that $K_{0}A$ is tangent to $\left(AEF\right)$ , which implies that $K_0A\perp AI$ and thus $K_0\equiv K'$ , as desired.

Notice that since $KP\perp PK_0$ , we have
$\angle K_0KP=90^{\circ}-\angle PK_0K=90^{\circ}-\angle PD'I=90^{\circ}-\angle PAI=\angle PIA,$ so $\triangle PKK_0\simeq\triangle PIA$ . But then by spiral similarity $\triangle PAK_0\simeq\triangle PIK$ , so
$\angle PAK_0=\angle PIK=180^{\circ}-\angle PIA,$ which implies the desired tangency and ends our proof.

Remark. Here is a very nice additional property, courtesy of AndreiVila: if $O$ is the circumcenter of $\triangle ABC$ and $S$ is defined as above, then $S, N$ and $O$ are collinear. I will leave the proof below (modulo the messy angle chase at the end).

Observe that this reduces to showing that $\angle NSA=90^{\circ}-\angle SBA$ . Let $R'$ be the midpoint of $AR$ , where $R$ is defined as above. Then $R'SNI$ is cyclic, so
$\angle NSA=\angle NIR'=\angle AIR'+\angle AIN=\left(90^{\circ}-ARI\right)+\left(180^{\circ}-\angle AID\right).$
Now $\angle AID=270^{\circ}-\angle ABC-\frac{1}{2}\angle BAC$ while
$\angle ARI=\angle ARC-\angle IRC=\left(\angle ABC-\angle SAB\right)-\frac{1}{2}\left(\angle ABC-\angle ACB\right),$ where we used the (well-known) fact that $RI$ is tangent to $\left(BIC\right)$ . Combining this with the above, the desired follows.

This post has been edited 5 times. Last edited by trigadd123, Feb 1, 2025, 10:57 AM

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#3 h Dec 22, 2024, 5:24 PM • 3 Y

Y by qwerty123456asdfgzxcvb, ihatemath123, trigadd123

Here is an alternate(?) approach to show the concurrence.

The main claim is that $N$ is the center of the circle through $D'$ and $S$ (intouch and sharkydevil) tangent to the circumcircle and line $BC$ . This finishes since by monge, $NQ\cap BC$ , which is the exsimilicenter of this circle and $\omega$ , lies on $AS$ and $BC$ . From there it is well known that $AS, BC, JM$ concur, say by radical axis. To show the claim, let $A_1$ be the reflection of $D'$ over $N$ , then it suffices to show $\angle D'SA_1$ is right, but this is true since if $P=(AEF)\cap SD'$ , then by reim $PI \parallel BC$ , implying $AA_1IP$ is a rectangle.

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#4 h Dec 22, 2024, 7:12 PM • 2 Y

Y by Double07, wizixez

This one deserves a complex bash solution, so here it is.

Before we begin the bash, we note some simple facts about $J, Q$ . Let $O$ be the center of $\Omega$ , and $L$ be the intersecting point of line $MO$ and $\Omega$ .(So $L$ is the midpoint of arc $BAC$ ) We easily find that $J,I,L$ are collinear. Let $S$ be the intersecting point of lines $AI$ and $BC$ . We easily find that $QS\perp BC$ .proof

We use complex coordinates with $\Omega$ as the unit circle. To avoid confusion with the imaginary unit $i$ , I will write $r$ for the complex number corresponding to $I$ . Otherwise I will use small letters for complex numbers corresponding to points.
It is easy to see that there are complex numbers $x,y,z$ such that $|x|=|y|=|z|=1$ , $a=x^2, b=y^2, c=z^2$ and the midpoints of the small arcs $BC, CA, AB$ are $-yz, -zx, -xy$ respectively. We have $r=-(xy+yz+zx)$ , because $I$ is the orthocenter of the points corresponding to $-yz, -zx, -xy$ . Also $m=-yz$ and $l=yz$ .
First we calculate $j$ . $I$ is on $JL$ , so we have $r+jl\bar{r}=j+l$ . This gives $-(xy+yz+zx)-jyz\frac{x+y+z}{xyz}=j+yz$ . Solving for $j$ gives $j=-\frac{x(xy+xz+2yz)}{2x+y+z}$ Next, we calculate $n$ . $NI\perp BC$ , so $\frac{\bar{n}-\bar{r}}{n-r}=-\frac{\bar{y^2}-\bar{z^2}}{y^2-z^2}=\frac{1}{y^2z^2}$ . Plugging in $r=-(xy+yz+zx)$ gives $xn-xy^2z^2\bar{n}=-x^2y-x^2z+y^2z+yz^2$ . Let $K$ be the midpoint of segment $AB$ . $NK//BC$ , so $\frac{\bar{n}-\bar{k}}{n-k}=\frac{\bar{y^2}-\bar{z^2}}{y^2-z^2}=-\frac{1}{y^2z^2}$ . Plugging in $k=\frac{x^2+y^2}{2}$ gives $2x^2n+2x^2y^2z^2\bar{n}=x^4+x^2y^2+x^2z^2+y^2z^2$ . Solving for $n$ gives $n=\frac{x^4-2(y+z)x^3+(y^2+z^2)x^2+2(y^2z+yz^2)x+y^2z^2}{4x^2}$ Next, we calculate $s$ . $S$ is on $AM$ and $BC$ , so $s+y^2z^2\bar{s}=y^2+z^2$ and $s-x^2yz\bar{s}=x^2-yz$ . Solving for $s$ gives $s=\frac{x^2y^2+x^2z^2+x^2yz-y^2z^2}{x^2+yz}$ Next, we calculate $q$ . $Q$ is apparently on $AO$ , so $\bar{q}=\frac{q}{x^4}$ . Also $QS\perp BC$ , so $\frac{\bar{q}-\bar{s}}{q-s}=-\frac{\bar{y^2}-\bar{z^2}}{y^2-z^2}=\frac{1}{y^2z^2}$ . Solving for $q$ gives $q=\frac{x^4}{x^4-y^2z^2}(s-y^2z^2\bar{s})=\frac{x^4}{x^4-y^2z^2}(2s-y^2-z^2)$ . We plug in the previously calculated formula for $s$ and we get $q=\frac{x^4(y^2+2yz+z^2)}{(x^2+yz)^2}$ Now we are ready to prove that $JL$ bisects $\angle{QJN}$ , solving the problem. This is equivalent to proving that $\frac{(n-j)(q-j)}{(l-j)^2}$ is a real number. Therefore we must prove that $\frac{(n-j)(q-j)}{(l-j)^2}=\frac{(\bar{n}-\bar{j})(\bar{q}-\bar{j})}{(\bar{l}-\bar{j})^2}$ . This is equivalent to $\frac{\bar{n}-\bar{j}}{n-j}\frac{\bar{q}-\bar{j}}{q-j}=\frac{1}{l^2j^2}$ .
We have
$n-j=\frac{x^4-2(y+z)x^3+(y^2+z^2)x^2+2(y^2z+yz^2)x+y^2z^2}{4x^2}+\frac{x(xy+xz+2yz)}{2x+y+z}$ $=\frac{2x^5+x^4(y+z)+4x^3yz+x^2(y^3+5y^2z+5yz^2+z^3)+x(2y^3z+6y^2z^2+2yz^3)+y^3z^2+y^2z^3}{4x^2(2x+y+z)}$ and
$q-j=\frac{x^4(y^2+2yz+z^2)}{(x^2+yz)^2}+\frac{x(xy+xz+2yz)}{2x+y+z}$ $=\frac{x(x^5(y+z)+x^4(2y^2+6yz+2z^2)+x^3(y^3+5y^2z+5yz^2+z^3)+4x^2y^2z^2+x(y^3z^2+y^2z^3)+2y^3z^3)}{(x^2+yz)^2(2x+y+z)}$
Define $u=2x^5+x^4(y+z)+4x^3yz+x^2(y^3+5y^2z+5yz^2+z^3)+x(2y^3z+6y^2z^2+2yz^3)+y^3z^2+y^2z^3$ and $v=x^5(y+z)+x^4(2y^2+6yz+2z^2)+x^3(y^3+5y^2z+5yz^2+z^3)+4x^2y^2z^2+x(y^3z^2+y^2z^3)+2y^3z^3$ . It is easy to see that $x^5y^3z^3\bar{u}=v$ and $x^5y^3z^3\bar{v}=u$ , so $\frac{\bar{u}\bar{v}}{uv}=\frac{1}{x^{10}y^6z^6}$ .

We have $\frac{\bar{n}-\bar{j}}{n-j}=x^4\cdot\frac{xyz(2x+y+z)}{xy+xz+2yz}\cdot\frac{\bar{u}}{u}$ and $\frac{\bar{q}-\bar{j}}{q-j}=\frac{1}{x^2}\cdot (x^2yz)^2\cdot\frac{xyz(2x+y+z)}{xy+xz+2yz}\cdot\frac{\bar{v}}{v}$ . Therefore,
$\frac{\bar{n}-\bar{j}}{n-j}\frac{\bar{q}-\bar{j}}{q-j}=x^8y^4z^4(\frac{2x+y+z}{xy+xz+2yz})^2\frac{\bar{u}\bar{v}}{uv}$ $=x^8y^4z^4(\frac{2x+y+z}{xy+xz+2yz})^2\frac{1}{x^{10}y^6z^6}=\frac{1}{y^2z^2}(\frac{2x+y+z}{x(xy+xz+2yz)})^2=\frac{1}{l^2j^2}$ which is what we need to show. This completes the solution.

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#5 h Dec 22, 2024, 7:29 PM • 1 Y

Y by ehuseyinyigit

Is there a way to prove that $BN$ and $CQ$ meet on $JI$ , and $BQ$ and $CN$ meet on $JI$ , because these surpirising facts complete the problem through DDIT or isogonal line lemma

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#6 h Dec 22, 2024, 11:09 PM

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Let $S_A$ be the A-sharkydevil point, let $D$ be the intouch point of incircle with $BC$ , also let $A'$ a point such that $AA' \parallel BC$ also $JM \cap BC=T$ and $AJ \cap BC=S$ , also let $L$ midpoint of arc $BAC$ and let $AA' \cap (AS_AI)=U$ . It's clear that $J$ is A-mixtilinear intouchpoint of $\triangle ABC$ from $\sqrt{bc}$ invert and that from easy radax+angle chase we get $A,S_A,T$ colinear.
Claim 1: $Q,I,S$ are colinear.
Proof: From Monge at $\Omega, \omega$ and A-mixtilinear incircle we get that the exsimilicenter of $\omega$ and A-mixtilinear incircle lie in $AJ$ , but also from Monge at $\omega$ . the A-mixtilinear incircle and the incircle of $\triangle ABC$ we get that the exsimilicenter of the incircle and $\omega$ lies on $AJ$ (combined with the first one), but this one also lies in $BC$ due to the common tangent they share, thus the claim is proven as it means $Q$ is such exsimilicenter.
Claim 2: $N$ is center of $(S_ADU)$ and is tangent to $(ABC), BC$ .
Proof: From spiral sim ratios it's easy to see that $S_A,D,M$ are colinear and also $U,I,D$ colinear simply from $\angle AUI=90$ so we have $\angle S_AUD=\angle S_AIM=\angle S_ADB$ so $(S_ADU)$ is tangent to $BC$ so from invert at $(BIC)$ (recall I-E Lemma) we get that it's also tangent to $(ABC)$ as desired, now it's clear that from Reim's we get $S_A,U,L$ colinear so $\angle US_AD=90$ which means the center of $(S_ADU)$ is the midpoint of $UD$ but in fact since $UD=\text{dist}(A, BC)$ we have that the center is our point $N$ as desired.
Claim 3: $NQ, BC, JM$ are concurrent.
Proof: Monge at $(AS_AU), \omega, \Omega$ tells us that the exsimilicenter of $\omega, (AS_AU)$ lies in $AS_A$ , but from common tangent it also lies on $BC$ so it has to be $T$ , thus done.
Finish: From $\sqrt{bc}$ invert note $J,I,L$ colinear and also let $IL \cap BC=V$ then notice that using $\sqrt{bc}$ invert ratios and Ratio Lemma one can easly get $J,D,A'$ colinear thus:
$-1=(A, A'; L, M) \overset{J}{=} (S, D; V, T) \overset{I}{=} (Q, N; LI \cap NQ, T)$ Which implies the desired conclusion as $\angle TJI=90$ thus we are done :cool:

:cool:

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#7 h Dec 23, 2024, 12:06 PM

Y by

MathLuis wrote:

Let $S_A$ be the A-sharkydevil point, let $D$ be the intouch point of incircle with $BC$ , also let $A'$ a point such that $AA' \parallel BC$ also $JM \cap BC=T$ and $AJ \cap BC=S$ , also let $L$ midpoint of arc $BAC$ and let $AA' \cap (AS_AI)=U$ . It's clear that $J$ is A-mixtilinear intouchpoint of $\triangle ABC$ from $\sqrt{bc}$ invert and that from easy radax+angle chase we get $A,S_A,T$ colinear.
Claim 1: $Q,I,S$ are colinear.
Proof: From Monge at $\Omega, \omega$ and A-mixtilinear incircle we get that the exsimilicenter of $\omega$ and A-mixtilinear incircle lie in $AJ$ , but also from Monge at $\omega$ . the A-mixtilinear incircle and the incircle of $\triangle ABC$ we get that the exsimilicenter of the incircle and $\omega$ lies on $AJ$ (combined with the first one), but this one also lies in $BC$ due to the common tangent they share, thus the claim is proven as it means $Q$ is such exsimilicenter.
Claim 2: $N$ is center of $(S_ADU)$ and is tangent to $(ABC), BC$ .
Proof: From spiral sim ratios it's easy to see that $S_A,D,M$ are colinear and also $U,I,D$ colinear simply from $\angle AUI=90$ so we have $\angle S_AUD=\angle S_AIM=\angle S_ADB$ so $(S_ADU)$ is tangent to $BC$ so from invert at $(BIC)$ (recall I-E Lemma) we get that it's also tangent to $(ABC)$ as desired, now it's clear that from Reim's we get $S_A,U,L$ colinear so $\angle US_AD=90$ which means the center of $(S_ADU)$ is the midpoint of $UD$ but in fact since $UD=\text{dist}(A, BC)$ we have that the center is our point $N$ as desired.
Claim 3: $NQ, BC, JM$ are concurrent.
Proof: Monge at $(AS_AU), \omega, \Omega$ tells us that the exsimilicenter of $\omega, (AS_AU)$ lies in $AS_A$ , but from common tangent it also lies on $BC$ so it has to be $T$ , thus done.
Finish: From $\sqrt{bc}$ invert note $J,I,L$ colinear and also let $IL \cap BC=V$ then notice that using $\sqrt{bc}$ invert ratios and Ratio Lemma one can easly get $J,D,A'$ colinear thus:
$-1=(A, A'; L, M) \overset{J}{=} (S, D; V, T) \overset{I}{=} (Q, N; LI \cap NQ, T)$ Which implies the desired conclusion as $\angle TJI=90$ thus we are done :cool:

:cool:

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Nice solution! Can you try and prove the claim I posted by the way, that $BN,CQ$ and $BQ,CN$ meet on $JI$

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#8 h Dec 23, 2024, 12:09 PM • 5 Y

Y by bin_sherlo, mxlcv, RM1729, Pranav1056, ehuseyinyigit

solvedy with Siddharth03, Pranav1056, mathsmania, rawlat_vanak while listening to Punjabi Music (Pranav's contribution)

here are two new solutions we discovered

sol 1, due to Siddharth
alternative sol 2
remark

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#9 h Dec 24, 2024, 7:31 AM • 2 Y

Y by trigadd123, abeot

I found this to be quite hard, and my solution is a bit weird. I'm not sure I really understand what's happening even after I solved it. (Comparing with above posts, it seems that I've gone with the sort of low-tech approach that's typical of my geometry these days.)

Preliminaries. Let $D$ be the foot of the angle bisector, $L$ be the midpoint of arc $BAC$ , and $X$ be the foot from $I$ to $\overline{BC}$ . We state the following facts without proof:

$J$ is the point of tangency of $A$ -mixtilinear incircle $\omega_A$ with $\Omega$ . If $E$ and $F$ are the tangency points of $\omega_A$ with
$\overline{AB}$ and $\overline{AC}$ , then the lines $EF$ , $BC$ , and $JM$ concur at a point $Z$ . Furthermore, lines $JA$ and $JX$ are isogonal in $\angle BJC$ .
The circle $\omega$ centered at $Q$ is tangent to $\overline{BC}$ at $D$ . If $T = \overline{AA}\cap \overline{BC}$ , then $T\in (AQD)$ .

$[asy] defaultpen(fontsize(10pt)); size(400); pair A, B, C, I, M, N, U, V, D1, D, E, F, L, O, T, J, Q, X, S, Z, H, H1, JJ, K; A = dir(130); B = dir(200); C = dir(340); I = incenter(A, B, C); L = dir(90); M = dir(270); J = foot(M, I, L); U = (A+B)/2; V = (A+C)/2; N = foot(I, U, V); D = extension(A, I, B, C); D1 = (A+D)/2; O = (0,0); Q = extension(A, O, D, D+dir(90)); E = extension(I, rotate(90,I)*A, A, C); F = extension(I, rotate(90,I)*A, A, B); T = extension(A, rotate(90, A)*O, B, C); S = extension(L, Q, M, T); X = foot(I, B, C); Z = extension(E, F, B, C); H = foot(A, B, C); H1 = (A+H)/2; JJ = extension(Q, Q+J-N, J, M); K = extension(A, J, D, JJ); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--T--B, orange); draw(CP(Q, A), heavyred); draw(U--V^^X--N, orange); draw(J--L, heavycyan); draw(N--J--Q, magenta+dotted); draw(circumcircle(J, E, F), heavycyan); draw(circumcircle(T, A, Q), lightblue); draw(Q--Z, magenta+dashed); draw(E--Z--M, heavygreen); draw(A--K--D^^J--X, dashed+purple); dot("$A$", A, dir(90)); dot("$B$", B, dir(240)); dot("$C$", C, dir(C)); dot("$I$", I, dir(300)); dot("$N$", N, dir(60)); dot("$M$", M, dir(270)); dot("$J$", J, dir(210)); dot("$L$", L, dir(L)); dot("$Q$", Q, dir(15)); dot("$T$", T, dir(200)); dot("$D$", D, dir(315)); dot("$Z$", Z, dir(270)); dot("$E$", E, dir(45)); dot("$F$", F, dir(160)); dot("$D_1$", D1, dir(120)); dot("$X$", X, dir(270)); dot("$J'$", JJ, dir(90)); dot("$K$", K, dir(270)); [/asy]$

The remainder of the proof can be divided into two main parts.

Eliminating $N$ . The key claim is the following, which provides a better understanding of point $N$ :

Claim: Points $Z$ , $N$ , $Q$ are collinear.

Proof. We redefine $Z = \overline{QN}\cap\overline{BC}$ and show that $Z$ lies on $\overline{EF}$ . Let $D_1$ be the midpoint of $AD$ ; then $D_1$ lies on the $A$ -midline and $\overline{QD_1T}$ collinear. Then if $Q_1$ is the intersection of $\overline{QD}$ with the midline, we have
$\frac{TZ}{ZD} = \frac{D_1N}{NQ_1} = \frac{D_1I}{ID},$ so $\overline{ZI}\parallel \overline{TD_1Q}$ . Since $\overline{TD_1Q}$ bisects $\angle ATD$ , it is parallel to $\overline{EF}$ , which means that line $ZI$ must coincide with line $EF$ . $\blacksquare$

Given the previous claim, we would like to finish with angle bisector theorem: it suffices to show $JN/JQ = ZN/ZQ$ . To that end, we take $J'$ to be the intersection of $\overline{JM}$ and the line through $D$ parallel to $\overline{JX}$ . This construction yields homothetic triangles $\triangle NJX\sim \triangle QJ'D$ , so we have
$\frac{QJ'}{NJ} = \frac{QD}{NX} = \frac{ZQ}{ZN}.$ Therefore, we only need to show $QJ = QJ'$ , completely eliminating $N$ .

Finish. Miraculously, we can now solve the problem by introducing the point $K = \overline{AJ}\cap \overline{DJ'}$ . Observe that:

Since $JA$ and $JX$ are isogonal, we have
$\angle JXN = \angle JXB + 90^\circ = \angle JCA + 90^\circ = 180^\circ - \angle JAQ.$ (In the last equality, we use the fact that $\overline{AQ}$ passes through the circumcenter of $\triangle ABC$ .) Hence $K\in (TAQD)$ .
Again by isogonality, we have $\angle AJZ = \angle XJM$ , which implies that $KJ = KJ'$ .
Finally, we have $\angle KQD = \angle KAD = \angle JAM = \angle JLM$ , so $\overline{QK}\parallel \overline{LJ}\perp \overline{ZJJ'M}$ .

Combining the last two bullets yields $QJ = QJ'$ , as desired.

Remark: I solved this problem on a long flight, and I was lucky to stumble into the first claim. I'm not sure I would have noticed it with a hand-drawn diagram.

However, I don't think the problem is easy even after proving $\overline{ZNQ}$ collinear. The key step for me was deciding that $J$ and $N$ were best related through point $X$ , and that $\angle JXN$ was actually workable because of isogonality. When I realized that $\angle JXN + \angle JAQ = 180^\circ$ , introducing $J'$ and $K$ was the natural way to finish.

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#10 h Apr 4, 2025, 5:44 AM

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We will define some new points.

$\bullet$ Let $S_A$ be the $A$ -Sharky devil point.
$\bullet$ Let $\triangle A'B'C'$ be the medial triangle.
$\bullet$ Let $D$ be $A$ -intouch point.
$\bullet$ Let $X=\overline{AM} \cap \overline{BC}$ .
$\bullet$ Let $N_A$ be major arc midpoint of $\widehat{BC}$ .
$\bullet$ Let $D'$ be reflection of $D$ over tangent of $\Omega$ at $M$ .

See that $J$ is $A$ -MIxtilinear intouch point and also $\overline{JN_A}$ is internal angle bisector of $\angle BJC$ .

Claim: $(\overline{JA},\overline{JD})$ ; $(\overline{JX},\overline{JS_A})$ are isogonal pairs in $\angle BJC$ .
Proof: See that it is well known that $\overline{JD}$ , $\overline{A\infty_{BC}}$ , $\Omega$ concur; and so by equal measure of arcs we are done with the first pair.

Also see that $(JDXM)$ is cyclic is well known (just angle chasing anyways) and hence $\angle DJX=\angle DMX=\angle S_AMA=\angle S_AJA$ and we are done with the second pair as well. $\square$

By Shooting lemma $X \in \omega$ and hence if $\sqrt{bc}$ invert at $J$ in $\triangle JBC$ , we get $\omega$ is swapped with the circle tangent to $\Omega$ at $S_A$ and tangent to $\overline{BC}$ at $D$ .

Hence we just need to prove that center of $\omega^*$ lies on $\overline{JN}$ , but we are going to do one better and show it is exactly $N$ .

Now invert at $D$ such that it fixes $\Omega$ and hence $\omega^*$ gets mapped to the tangent line to $(ABC)$ at $M$ and so $N$ maps to $D'$ .

Now by PoP, we just need to prove that $DN$ is $\frac 12$ of length of foot of $A$ to $\overline{BC}$ which is $\frac{bc}{4R}$ .

So see that $\begin{align*} & \frac{bc}{R} \cdot DD'=\frac{bc}{R} \cdot 2A'M = \frac{abc \tan (A/2)}R=\frac{abc}R \left(\frac{1-\cos A}{\sin A} \right)\\ =& 2bc(1-\cos A)=a^2-b^2-c^2+2bc=(a+b-c)(a-b+c)=4DB \cdot DC \end{align*}$ And we are done by PoP.

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#11 h Apr 8, 2025, 8:00 AM

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Let $MJ\cap BC=R,AM$ meet the midbase at $E$ , $AM\cap BC=K,IN\cap BC=D,QE\cap BC=F$ . Let $S$ be $A-$ sharky devil. Note that $FA$ is tangent to $(ABC)$ and $FE$ is the perpendicular bisector of $AK$ .
Claim: $R,N,Q$ are collinear.
Proof: We have $DN=DI.\frac{KE}{KI}=\frac{DI.KF}{KR}=\frac{RD.KQ}{KR}$ thus, $\frac{DN}{DR}=\frac{KQ}{KR}$ which shows the claim.

Claim: If $V$ is the circumcenter of $(ADK)$ , then $V$ lies on $IJ$ .
Proof: Let $AD\cap SK=U,JD\cap (ABC)=D',JK\cap (ABC)=K'$ . Since $J,D,K,M$ are concyclic which can be seen by inversion at $M$ , by Reims $AD'\parallel BC$ and $SK'\parallel BC$ . DDIT at $RSUD$ implies $(\overline{JR},\overline{JU}),(\overline{JS},\overline{JD}),(\overline{JA},\overline{JK})$ is an involution. Projecting onto $(ABC)$ gives $(M,JU\cap (ABC)),(A,S),(D',K')$ is an involution. We see that $AS\cap D'K'$ lies on the perpendicular bisector of $BC$ hence $JU\cap (ABC)=P$ . Brokard at $RSUD$ yields $VU\perp MJ$ and since $\overline{UJPI}$ is the perpendicular from $U$ to $MJ$ , we observe that $V,U,J,P,I$ are collinear.

Notice that $V,E,F$ are collinear since that line is the perpendicular bisector of $AK$ . If $DN\cap EQ=X$ , then $V$ is the midpoint of $XQ$ since it lies on the perpendicular bisector of $DK$ . Also note that $RI\perp AI\perp \overline{VEF}$ . We have
$(JR,JI;JN,JQ)=(R,JI\cap RQ;N,Q)\overset{I}{=}(FQ_{\infty},V;X,Q)=-1$ And since $JR\perp JI$ , we get that $JI$ is the angle bisector of $\measuredangle NJQ$ as desired. $\blacksquare$

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