Geometry Finale: Incircles and concurrency

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Geometry Finale: Incircles and concurrency

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IMO Shortlist 2019

mixtilinear incircle

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Source: IMO 2019 Problem 6

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#200 h Dec 29, 2023, 8:35 AM

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Define the linear functions $f(\bullet) := \text{Pow}(\bullet, (PBF)) - \text{Pow}(\bullet, (PCE))$ and $g(\bullet) := \text{Pow}(\bullet, (BDIF)) - \text{Pow}(\bullet, (CDIE)).$ Let $L=\overline{DI} \cap \overline{PQ}$ . Note that $f(A)=g(A)$ and $f(L)=g(L)$ . We want to show that $\overline{AL}$ is the external angle bisector $\ell$ of $\angle BAC$ , so it suffices to find some point on $\ell$ that has an equal $f$ - and $g$ -value. Select $K=\ell \cap \overline{BC}$ . Observe that $KB/KC=AB/AC$ .

Let $(PBF)$ intersect $BC$ at $B_1$ and $(PCE)$ intersect $BC$ at $C_1$ . Since we need to show that $f(K)=g(K)$ , we can reduce this (by PoP) to $\frac{DB_1}{DC_1} = \frac{KC}{KB} = \frac{AC}{AB}.$ Fortunately, this is something very much in our hands.

Claim: In fact, we have $\frac{DB_1}{DC_1} = \frac{AC}{AB}.$ Proof. First, we need to understand what $P$ is with respect to standard configs. Do everything with respect to $\triangle DEF$ , so that $R$ has a simple definition. Redefine $P$ as the intersection of the $R$ -symmedian in $\triangle REF$ with $(REF)$ . Thus, we have $\frac{PE}{PF} = \frac{RF}{RE} = \frac{\cos(\angle F)}{\cos(\angle E)}.$ Second, redefine $B_1$ and $C_1$ with respect to $P$ and $\triangle DEF$ . Note that the center of the spiral similarity that maps $BF \mapsto FE$ is $P$ . Thus, we readily have $\triangle B_1PC_1 \sim \triangle EPF$ , so taking similarity ratios we have $\frac{PB_1}{PC_1} = \frac{PE}{PF} = \frac{\cos(\angle F)}{\cos(\angle E)} = \frac{\sin(\angle C/2)}{\sin(\angle B/2)}.$ Finally, we apply LoS on $\triangle DPB_1$ and $\triangle DPC_1$ to involve $D$ , from which we find that it suffices to prove that $\angle DPB_1 = 90^{\circ} + \angle B/2.$ This is easy angle chasing, in that we write $\angle DPB_1 = 360^{\circ} - (\angle B_1PF + \angle DPF) = 360^{\circ} - ((180^{\circ}-\angle B)+(90^{\circ}+\angle B/2)) = 90^{\circ}+\angle B/2,$ as desired.
:yoda:

:yoda:

Remark: I think the hardest part is actually the ratio computation at the very end. If you want to force linearity of PoP onto the problem, then that is not hard, since you just need to bring another pair of circles whose radical axis is $DI$ . Then the rest of the solution flows until the point at which you have to do the ratio computation, which, rather than requiring bash, needs some intuition with American configs.

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#201 h Jan 16, 2024, 4:15 PM

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let $DI$ intersect $\omega$ at $Y$ , let $PR$ intersect $EF$ at $H$ , let $DR$ intersect $EF$ at $G$ , and let $DI$ and $PQ$ intersect at $X$
because $D$ is a point of tangency, $YDB=YRD$ , $YR||EF$ , $YFE=REF$ , and $YF=RE$
because $E$ and $F$ are points of tangency, $AEF$ is isoceles, and we can prove $AY=AR$ , $AI||DR$ , $AR||YI$ , and that $AYIR$ is a rhombus
then, we prove that $AYIP$ is and isoceles trapezoid, and that $AYX$ and $PID$ are congruent, and from that we can prove that $AX \perp AI$

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8857 posts

#202 h Mar 13, 2024, 3:59 AM

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chat is this real

We will complex bash with $(DEF)$ the unit circle, where $x, y, z$ represent $D, E, F$ . Note that $a=\frac{2yz}{y+z}$ and cyclic permutations. Let $X$ correspond to $kx$ for some real constant $k$ . As $\frac{x-a}a$ must be pure imaginary,
$\begin{align*} \frac{kx-\frac{2yz}{y+z}}{\frac{2yz}{y+z}} + \frac{\frac kx-\frac 2{y+z}}{\frac 2{y+z}} &= 0 \\ \iff \frac{kx(y+z)-2yz}{2yz}+\frac{k(y+z)-2x}{2x} &= 0 \\ \iff kx^2(y+z)+kyz(y+z)-4xyz &= 0 \\ \iff \frac{4xyz}{(y+z)(x^2+yz)} &= k. \end{align*}$ So $X$ is located at $\frac{4x^2yz}{(y+z)(x^2+yz)}$ . Now, compute $p = \frac{-\frac{yz}x-\frac{2yz}{y+z}}{-\frac{yz}x \cdot \frac 2{y+z}-1} = \frac{yz(y+z)+2xyz}{2yz+x(y+z)}.$ It follows that the circumcenter $o_B$ of $(BFP)$ is located at
$\begin{align*} o_B &= z -\frac{(b-z)(p-z)(\overline b - \overline p)}{(b-z)(\overline{p-z}) - (\overline{b-z})(p-z)} \\ &= z + \frac{-\left(\frac{2xz}{x+z}-z\right)\left(\frac{yz(y+z)+2xyz}{2yz+x(y+z)} - z\right)\left(\frac 2{x+z} - \frac{xy+xz+2yz}{yz(2x+y+z)}\right)}{\left(\frac{2xz}{x+z}-z\right)\left(\frac{xy+xz+2yz}{yz(2x+y+z)} - \frac 1z\right)-\left(\frac 2{x+z}-\frac 1z\right)\left(\frac{yz(y+z)+2xyz}{2yz+x(y+z)}+z\right)} \\ &= z + \frac{\frac{z(x+y)(z-x)(y-z)(x^2y+x^2z-xyz+xz^2-2y^2z)}{y(x+z)^2(2x+y+z)(xy+xz+2yz)}}{\frac{(x+y)^2(x-z)(y-z)^2}{y(x+z)(2x+y+z)(xy+xz+2yz)}} \\ &= z + \frac{z(x^2y+x^2z-xyz+xz^2-2y^2z)}{(x+y)(x+z)(y-z)} \\ &= \frac{yz(z-x)(2x+y+z)}{(x+y)(x+z)(z-y)}. \end{align*}$ Intermediate steps are omitted so the reader does not need to suffer the factorization process like I did. Then we have $o_B-o_C = \frac{yz(z-x)(2x+y+z)}{(x+y)(x+z)(z-y)} -\frac{yz(y-x)(2x+y+z)}{(x+y)(x+z)(y-z)} = -\frac{yz(2x+y+z)(y+z-2x)}{(x+y)(x+z)(y-z)}.$ It will suffice to show that $\frac{o_B-o_C}{p-x}$ is pure imaginary. In particular, $p-x = \frac{yz(y+z)-2xyz}{2yz-x(y+z)} - \frac{4x^2yz}{(y+z)(x^2+yz)} = \frac{-yz(2x+y+z)(x^2y+x^2z-4xyz+y^2z + yz^2)}{(y+z)(xy+xz-2yz)(x^2+yz)}.$ So it suffices that the ratio $\frac{(2x+y+z)(y+z)(x^2+yz)(xy+xz+2yz)}{(x+y)(x+z)(y-z)(x^2y+x^2z+4xyz+y^2z+yz^2)} \in i\mathbb R$ which is clear as the expression equals the negative of its own conjugate.

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CT17

#204 h May 17, 2024, 2:47 PM • 1 Y

Y by GeoKing

Observe that

$\measuredangle BQC = \measuredangle BQP + \measuredangle PQC = \measuredangle BFP + \measuredangle PEC = \measuredangle BIC$
so $Q$ lies on $(BIC)$ . Moreover, if $PQ$ intersects $(BIC)$ again at $T$ , then

$\measuredangle BIT = \measuredangle BQT = \measuredangle BQP = \measuredangle BFP = \measuredangle FRP$
so quadrilaterals $RFPE$ and $IBTC$ are similar. Since $P$ lies on the $R-$ symmedian of $\triangle IFE$ , it follows that $T$ lies on the $I-$ symmedian of $\triangle IBC$ , which passes through the midpoint $M$ of $\widehat{BAC}$ . It suffices to show that $X$ lies on $PT$ . Now, the spiral similarity centered at the Sharkydevil point $S$ mapping $EF$ to $BC$ maps $P$ to $T$ and $A$ to $M$ . Hence, it suffices to show that $X$ lies on $(APS)$ . Inverting about the incircle, $S$ gets sent to the foot form $D$ to $EF$ , $A$ gets sent to the midpoint of $EF$ , $X$ gets sent to the foot from $A^*$ to $DI$ , and $P$ remains as the $D-$ Queue point of $\triangle DEF$ , so all four inverted points lie on the circle with diameter $DA^*$ , as desired.

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#206 h May 18, 2024, 6:32 PM

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$[asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(31, 191, 184); pen sec; sec=RGB(5, 113, 108); pen tri; tri=RGB(25, 120, 165); pen qua; qua=RGB(3, 17, 99); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair O, A, B, C, I, D, E, F, R, P, Q, K, D_prime; O=(0, 0); A=dir(125); B=dir(220); C=dir(320); I=incenter(A, B, C); D=foot(I,B,C); E=foot(I,A,C); F=foot(I,A,B); R=2*foot(I,D,foot(D,E,F))-D; P=2*foot(I,A,R)-R; Q=intersectionpoints(circle(B,F,P), circle(C,E,P))[1]; K=extension(D,I,P,Q); D_prime=2I-D; fill(D--E--F--cycle,sfil); filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); filldraw(circumcircle(B, P, F), tfil, tri); filldraw(circumcircle(C, E, P), tfil, tri); filldraw(circumcircle(B, I, C), sfil, sec); filldraw(circumcircle(A, B, C), fil,pri); filldraw(incircle(A, B, C), fil, pri); draw(R--D,tri+dashed); draw(P--K,tri+dashed); draw(D--K,tri+dashed); draw(A--K,tri+dashed); dot("$A$",A,dir(110)); dot("$B$",B,dir(200)); dot("$C$",C,dir(-15)); dot("$K$",K,dir(-5)); dot("$I$",I,dir(-30)); dot("$D$",D,S); dot("$E$",E,dir(75)); dot("$F$",F,dir(130)); dot("$R$",R,NW); dot("$P$",P,dir(120)); dot("$Q$",Q,dir(-20)); dot("$D'$",D_prime,dir(45)); [/asy]$
Claim: $Q$ lies on $(BIC)$ .
Proof. Angle chasing we have,
$\begin{align*} \angle BQC = \angle BFP + \angle PEC &= 360 - \angle PFA - \angle PEA\\ & = 360 - (\angle PFE + \angle PEF) - (180 - \angle A)\\ &= 90 + \angle A/2 \end{align*}$ which proves the claim. $\square$

Let $D'$ be the antipode of $D$ in the incircle, and let $T$ be the mixintillinear touch point.

Claim: $\overline{AP}$ and $\overline{AD'}$ are isogonal.
Proof. Invert about the incircle, so that $A$ maps to the midpoint of $\overline{EF}$ and $T$ maps to the midpoint of $\overline{DH}$ where $H$ is the orthocenter of $\triangle DEF$ , namely $T^*$ lies on $\overline{DR}$ . Then it suffices to show that $A^*$ , $T^*$ , $I$ and $R$ are concyclic. This follows as upon reflection about $\overline{BC}$ we observe $R$ maps to $H$ , and it is easy to show say by complex numbers that $A^*H \parallel T^*O$ , which proves the claim. $\square$

From now on much of our work will deal with the inverted figure.

$[asy] import geometry; size(7cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(31, 191, 184); pen sec; sec=RGB(5, 113, 108); pen tri; tri=RGB(25, 120, 165); pen qua; qua=RGB(3, 17, 99); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair I, D, E, F, A, B, C, X, Y, P, Q, D_prime, K; I=(0, 0); D=dir(110); E=dir(220); F=dir(320); A=((E+F)/2); B=((D+F)/2); C=((D+E)/2); X=((D+A)/2); Y=intersectionpoints(circle(D,E,F),line(D, A))[0]; P=intersectionpoints(circle(X,abs(X-D)), circle(D, E, F))[1]; Q=intersectionpoints(circle(P, E, C), circle(B, P, F))[0]; D_prime=2I-D; K=foot(A,D,D_prime); filldraw((path)(D--E--F--cycle), white+0.1*pri, pri); filldraw(circumcircle(D, E, F), sfil, pri); filldraw(circumcircle(P,A,D), sfil, tri); filldraw(circumcircle(P,Q,Y), sfil, tri); draw(circumcircle(P,Q,C), dashed+sec); clipdraw(circumcircle(P,Q,B), dashed+sec); draw(D--Y,tri); draw(Q--B,tri); draw(D--D_prime,tri); draw(A--K,tri); draw(D_prime--P,tri+dashed); dot("$A^*$",A,dir(-120)); dot("$B^*$",B,dir(40)); dot("$C^*$",C,dir(-110)); dot("$I$",I,dir(20)); dot("$D$",D,N); dot("$E$",E,dir(-100)); dot("$F$",F,dir(-40)); dot("$P$",P,dir(130)); dot("$Q^*$",Q,dir(W)); dot("$X$",X,dir(240)); dot("$Y$",Y,dir(-40)); dot("$D'$",D_prime,dir(-50)); dot("$K$",K,dir(-15)); [/asy]$

First however, we provide a characterization of $P$ .

Claim: $\angle DPA^* = 90$ .
Proof. Upon inverting back to our original figure this is equivalent to showing $D'PIA$ is cyclic. This follows as,
$\begin{align*} \angle AD'I = \angle ARD' + \angle RD'D &= 180 - \angle D'RP + \angle RD'D \\ &= 180 - (180 - \angle D'DP) + (90 - \angle D'DR)\\ &= \angle D'DP - \angle D'DR + 90 \\ &= \frac{1}{2}\angle RIP + 90\\ &= 180 - \angle IPA \end{align*}$ which proves the claim. $\square$

Now define $K$ as the intersection of the external angle bisector of $\angle BAC$ and $\overline{DI}$ , let $X$ be the center of $(PDA^*)$ and $Y = \overline{DA^*} \cap (DEF)$ .

Note immediately that $X$ is the midpoint of $\overline{B^*C^*}$ by homothety.

Claim: $X$ and $Y$ both lie on $(PQ^*I)$ .
Proof. First note that $Y \in (PQ^*I)$ as,
$\begin{align*} \angle IYP = 90 - \frac{1}{2}\angle PIY &= 90 - \frac{1}{2}(180 - \angle PID - \angle YID')\\ &= \angle PD'D + \angle YDD'\\ &= \angle PFB + \angle XDD'\\ &= \angle PQB + \angle XPI \\ &= \angle PQB + \angle XQI = \angle PQI \end{align*}$ where we have $\angle XDD' = \angle XPI$ as both $X$ and $I$ lie on the perpendicular bisector of $\overline{PD}$ because $\overline{PD}$ is a chord in circles centered at $X$ and $I$ . Now $X \in (PQ^*I)$ because,
$\begin{align*} \angle PQX = \angle PFB = \angle PYD = \angle PYX \end{align*}$ so we have proven the claim. $\square$

Now we may delete both $P$ and $Q^*$ from the picture.

Claim: $K^ *$ lies on $(XIY)$ .
Proof. Note that,
$\begin{align*} \angle IKX = \angle DKX = \angle KDA^* = \angle IYD \end{align*}$ and thus the claim is proven. $\square$

Inverting back the claim follows as $K \in (PQ^*I)$ implies $K$ , $P$ and $Q$ are collinear.

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#207 h Jun 10, 2024, 6:06 PM • 1 Y

Y by GeoKing

We will split the solution into 2 parts.

Part 1) We will prove that lines $PQ,IR,EF$ are concurrent.

(Sketch of) proof:
Let $K = IR \cap EF$ , $B_1=(BPF) \cap EF$ , $C_1 = (CPE) \cap EF$ . Also let $\angle BAC=\alpha, AB=c$ and so on. Define $f(X)=P(X,(BPF))-P(X,(CPE))$ , there $P(X, \omega)$ is power of point $X$ onto circle $\omega$ . Note that since $f$ is linear function, we have $f(K)=\frac{1}{EF} \left( KE \cdot f(F) + KF \cdot f(E) \right)=KF \cdot EB_1-KE \cdot FC_1$ . So we want to prove that $\frac{EB_1}{FC_1}=\frac{KE}{KF}$ . Note that $\frac{KE}{KF}=\frac{AC}{AB}$ .

Let $Q'=BC_1 \cap CB_1$ . By easy angle chasing $Q'=Q$ . So $\Delta PBB_1 \sim \Delta PC_1C$ and $\Delta PBC_1 \sim \Delta PB_1C$ . Now we can find $AB \cdot EB_1 = \frac{AB \cdot FB \cdot CB_1}{BC_1}$ and similar equality with segment $CB_1$ . After this good approach will be using trig-Ceva to triangle $ABC$ and point $P$ and not hard bashing.

Part 2): If $X=PK \cap DI$ , then $AX \perp AI$ .

Sketch of proof: If $U=AP \cap EF, V=DI \cap EF$ , then $FREP$ is harmonic quadrilateral, so $-1=^(A,U;R,P)=^K(AI \cap DI, V; I, X)$ and it ie enough to prove that $AI$ is bisector of $\angle KAV$ or $FK=VE$ , which is too not hard (now we can delete all points but $D,E,F,R,I,K,V$ ).

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DistortedDragon1o4

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DistortedDragon1o4

#208 h Jun 30, 2024, 6:22 AM • 1 Y

Y by GeoKing

Our main idea is incircle inversion and the angle chasing.

Define $O$ as the intersection of the external bisector of $\angle CAB$ and $DI$ .
Define $M = (\omega \cap DI) - D$

On inverting about $\omega$ we have that
$O \iff O'$
$Q \iff Q'$
$A \iff A'$
$B \iff B'$
$C \iff C'$

$\therefore PQ \iff (Q'PI)$
$(BPQF) \iff (B'PQ'F)$
$(CPQE) \iff (C'PQ'E)$

It is easy to see that $\angle CQB = 90^\circ + \frac{\alpha}{2}$
Which implies $Q\in (CIB)$

$(CIQB) \iff C'B'Q'$

Define $S = ((BIC)\cap ID) - I$
$S\iff S'$
$S' = B'C'\cap ID$

Now we claim that $Q\in PO$
Proof:

We have $\angle DMP = \angle DFP \equiv \angle B'FP = \angle B'Q'P$

Define $W = ((Q'PQ)\cap B'C') - Q'$
Let $W' \iff W$

$\because$ $(Q'PWQ)$ is orthogonal $W' = ((Q'PWQ)\cap (CIQB)) - Q$

$\angle IAP = \angle DRP = \angle B'FP = \angle B'Q'P \equiv \angle WQ'P = \angle WW'P$
$\implies W', P, I, A$ are concyclic

$\because P, I, M, A$ are concyclic, $W', P, I, M, A$ are concyclic

Also $(W'PIA) \iff PWM$

$\angle S'MP\equiv\angle DMP = \angle B'Q'P\equiv\angle S'Q'P$
$\therefore S', M ,Q', P$ are concyclic
$\therefore S, M, Q, P$ are concyclic

Claim inside claim: $WS \parallel A'D$
Proof:
$\angle ERD = \angle EFD = 90^\circ - \frac{\gamma}{2} = \angle CID\equiv\angle CIS$
$\angle RDE = 90^\circ - \angle DEF = \frac{\beta}{2} = \angle IBC = \angle ISC$
$\Delta ERD \sim \Delta CIS$

Define $T = DR\cap EF$ and $T' = DR\cap B'C'$

$\therefore \frac{MD}{DS} = \frac{2\cdot ID}{DS} = \frac{2\cdot RT}{TD} = \frac{RT}{TT'}$

Now $\frac{RT}{TT'} = \frac{MK}{A'W}$
$\therefore A'D\parallel WS$

We know that $A, O, D, P$ are concyclic (this is obvious)
$\therefore A', O', D, P$ are concyclic

$\angle OPM = \angle MPO' = \angle A'DO' = \angle WSM\equiv\angle QSM = \angle QPM$

$\therefore Q\in PO$

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#209 h Aug 25, 2024, 4:49 PM • 1 Y

Y by dolphinday

this is honestly not that bad

Let $J$ be $IR\cap EF$ , $M$ be the midpoint of $EF$ , and $J'$ as $DI\cap EF$ . Let $D'$ be on $\omega$ such that $DD' \parallel EF$ (so the antipode of $R$ ), $G$ as $AD'\cap \omega$ , $K=D'P\cap EF$ , $N$ the midpoint of $AK$ . Define $T$ to be $DI$ intersect the $A$ external bisector.
We first note a few properties of the configuration.

Note that $-1=(PR;EF)\overset{D'}{=}(KJ;EF)$ and $-1=(GD';EF)\overset{R}{=}(RG\cap EF,J;E,F)$ . Thus $R,G,K$ collinear.
As $RG\perp AD'$ we get $R$ is the orthocenter of $AKD'$ .
Note that $IJ$ and $IJ'$ are symmetric about line $AI$ . It is well known $AJ'$ is a median so $AJ$ is a symmedian. This implies that $AK$ is tangent to $(ABC)$ by projecting $(KJ;EF)$ onto $(ABC)$ .
As $AN$ is tangent to $(ABC)$ , $AM$ is the bisector, and $NA=NM$ from $\angle AMK=90 ^{\circ}$ , we conclude that $NM \parallel BC$ .

We are now ready to tackle the problem. Let $Q'$ and $T'$ denote the inverses of $Q$ and $T$ about $\omega$ .

Claim: Line $PQ$ inverts to the nine point circle of $AKD'$ .
Proof. It would suffice to show $Q$ goes to the midpoint of $KD'$ , since $PQ$ passes through the midpoint of $RD'$ and the foot from $A$ to $KD'$ . Let $B',C'$ be the midpoints of $DF$ , $DE$ , so $Q'PB'F$ and $Q'PC'E$ are cyclic. Then, $\measuredangle PQB'=\measuredangle PFB'=\measuredangle PFD=\measuredangle PED=\measuredangle PEC'=\measuredangle PQC'$ so $Q',B',B'$ collinear. Also, $\measuredangle D'PE=\measuredangle FED=\measuredangle B'C'D=\measuredangle Q'C'E=\measuredangle Q'P'E$ so $D',P,Q'$ collinear. A homothety at $D'$ with scale factor $2$ now sends line $B'C'$ to $EF$ , so $Q'$ goes to $K$ . $\blacksquare$

Thus it suffices to show $T'$ lies on the nine point circle of $AKD'$ . But as $MT'\parallel BC$ as both are perpendicular to $DI$ , we have $\angle NT'I=90^{\circ}$ as desired.

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Eka01

#210 h Aug 28, 2024, 5:02 AM • 1 Y

Y by Sammy27

It is well known that $P$ is the $D$ queue point in $\Delta DEF$ . We now invert with respect to the incircle and restate in terms of $\Delta DEF$ .

Quote:

In $\Delta ABC$ with center $O$ and $A$ queue point $P$ , midpoint of $AC$ is $M$ and midpoint of $AB$ is $N$ . $(PBN) \cap (PCM)=Q \neq P$ . If $K$ is midpoint of $BC$ and $D \in AO$ such that $KD \perp AO$ , then prove that $(PQOD)$ is cyclic.

First notice by angle chasing that $Q$ lies on the $A$ midline of $\Delta ABC$ and that $APKD$ is a cyclic quadrilateral with diameter $AK$ . Notice that the center of this quadrilateral lies on the $A$ midline of $BC$ . Let the center be called $E$ . Angle chase some more to show that $(PEOD)$ are cyclic and finally even more angle chasing shows that $(PQODE)$ is cyclic.

Remark

Attachments:

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#211 h Aug 30, 2024, 1:19 PM

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Let $DI$ and the perpendicular from $A$ to $AI$ intersect at $S$ . $DI\cap (ABC)=K$ , let $M$ be the midpoint of $EF$ .

Claim: $P,M,K$ are collinear.
Proof: Since $PM$ is median on $\triangle PEF$ and $PA$ is $P-$ symmedian, we have
$\measuredangle KPE=\measuredangle KDE=\frac{\measuredangle C}{2}=\measuredangle FPA=\measuredangle MPE$ Hence $P,M,K$ are collinear. $\square$

Invert the diagram from $I$ with radius $ID$ . $A$ swaps with $M$ and $B,C$ swap with the midpoints of $DF,DE$ respectively. $S^*$ is the foot of the altitude from $M$ to $ID$ .

New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$ . $D,E,F$ are the midpoints of $BC,CA,AB$ respectively and $AO$ intersects $(ABC)$ at $S$ for second time. $SD$ intersects $(ABC)$ at $P$ and $Q$ is the intersection of $(PBF)$ and $(PEC)$ . If $H$ is the altitude from $D$ to $AS,$ then $P,Q,O,H$ are concyclic.

Let $T$ be the point on $(ABC)$ where $AT\parallel BC$ .

Claim: $Q$ lies on $EF$ .
Proof:
$\measuredangle PQF=\measuredangle PBF=\measuredangle PBA=\measuredangle PCA=\measuredangle PCE=\measuredangle PQE$ Thus, $Q,E,F$ are collinear. $\square$

Claim: $Q,P,T$ are collinear.
Proof: Let $T'=PQ\cap (ABC)$ .
$\measuredangle T'QF=\measuredangle PQF=\measuredangle PBA=\measuredangle PT'A$ This yields $AT'\parallel BC\iff T'=T$ . $\square$

Claim: $QF$ is the angle bisector of $\measuredangle PQH$ .
Proof: Let the perpendicular from $H$ to $BC$ intersect $TP$ and $EF$ at $X,Y$ .
$-1\overset{?}{=}(Q(XY)_{\infty},\overline{QFE};\overline{QPT},QH)\overset{XY}{=}(XY_{\infty},Y;X,H)$ $-1\overset{?}{=}(XY_{\infty},Y;X,H)\iff YX\overset{?}{=}YH$ .
Let the reflection of $H,D,T$ with respect to $EF$ be $H',D',T'$ . Note that $D'$ lie on $AT$ and $T'$ lie on $BC$ . Also $S,T,T'$ are collinear.
Since $\measuredangle SHD=90=\measuredangle ST'D,$ we get that $D,H,T',S$ are concyclic. Also $DD'H'H$ and $TT'HH'$ are isosceles trapezoids. By $DHT'\sim D'H'T$ we have
$\measuredangle PTA=\measuredangle PSA=\measuredangle DSH=\measuredangle DT'H=\measuredangle H'TD'=\measuredangle H'TA$ This yields $T,H',P$ are collinear. Hence $YX=YH$ . $\square$

Claim: $P,Q,O,H$ are concyclic.
Proof:
$\measuredangle PQH=2\measuredangle PQF=2\measuredangle PBA=\measuredangle POA=180-\measuredangle HOP$ As desired. $\blacksquare$

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#212 h Nov 16, 2024, 3:02 PM

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Invert at $I$ .

Let $M$ be the midpoint of $\overline{BC}$ . It suffices to show that $QPMIX$ is cyclic. Points $P$ , $M$ , $I$ and $X$ are all feet/midpoints of $\triangle DAD'$ , where $D'$ is the antipode of $D$ WRT $(DEF)$ , meaning that they lie on the nine-point circle of $\triangle DAD'$ . Now we show that $Q$ lies on this circle too. Firstly, point $Q$ lies on line $BC$ since $\angle PQC = \angle PEC = \angle PFB = \angle PQB.$ Now, $\angle PIM = \angle IPD = \angle DD'P = \angle PEC = \angle PQM$ finishes.

This post has been edited 1 time. Last edited by ihatemath123, Nov 16, 2024, 3:04 PM

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#213 h Nov 27, 2024, 3:38 PM • 1 Y

Y by cosdealfa

Solved with emotional support from cosdealfa. I had a horrible time solving this problem because I though it was going to be harder than it actually was.

Let $T$ be the $A$ -mixtilinear touch point, let $S$ be the foot from the $A$ -excenter to $BC$ , let $N$ be the midpoint of major arc $BC$ , let $IM$ meet $(BIC)$ again at $M$ , let $DI$ meet $AN$ at $K$ and let $H$ be the orthocenter of $\triangle DEF$ .

We need to prove that $K$ lies on $PQ$ . In fact, we will show that $M$ also lies on $PQ$ .

It is well known that $\overline{N-I-T}$ and we will also prove $\overline{A-R-T}$ . Now note that $AT$ and $AS$ are isogonal so it suffices for $\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle BAT)}{\sin(\angle CAT)}=\frac{\sin(\angle CAS)}{\sin(\angle BAS)}$ By ratio lemma $\frac{\sin(\angle CAS)}{\sin(\angle BAS)}=\frac{CS}{BS}\cdot\frac{AB}{AC}=\frac{p-b}{p-c}\cdot\frac{AB}{AC}$ By trig ceva $\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle AFR)}{\sin(\angle RFE)}\cdot\frac{\sin(\angle FER)}{\sin(\angle REA)}=\left(\frac{\sin(\angle FDR)}{\sin(\angle EDR)}\right)^2=\left(\frac{\sin C/2}{\sin B/2}\right)^2$ One can easily check that these are equal so we get $\overline{A-R-T}$ .

Now note that $\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle EDF= \measuredangle BIC$ So $Q$ lies on $(BIC)$ .

Invert around the incircle and denote $X$ 's inverse by $X'$ . Clearly, $A'$ , $B'$ and $C'$ are the midpoints of $EF$ , $DF$ and $DE$ . $T'$ is the midpoint of $DH$ , $M'$ is the midpoint of $B'C'$ , $K'$ and $N'$ are the intersections of $DI$ and $DN$ with the circle with diameter $IA'$ and $Q$ lies on $(PB'F)$ , $(PC'E)$ and $B'C'$ .

Also note that $IT'PR$ is cyclic. Now look at the problem from the perspective of $\triangle DEF$ . Clearly, $T'C'A'B'$ is the nine-point circle. The problem translates to $Q'IPK'$ being cyclic.

Claim: $M'$ lies on $(IPK')$ .

Proof: We firstly prove that $PA'\parallel IM'$ . To do this, redefine $P$ to be the intersection of the circle with diameter $DA'$ (which is centered at $M'$ ) with $(DEF)$ . By orthocenter config, we know that $H$ lies on $PA'$ and since $T'IA'G$ is a parallelogram, we get that $PA'\parallel IM'$ . Since $\measuredangle DPH=90^\circ$ we have $PT'=T'H=IA'$ so $PT'IA'$ is an isosceles trapezoid, which suffices.

From the above, we also know that $PT'=TD$ so $M'T'$ is actually the perpendicular bisector of $DP$ . This implies $\measuredangle DIM'=\measuredangle M'IP= \measuredangle K'IN'$ And since $M'P=M'K'$ we get that $PM'IK'$ is cyclic (well known lemma). $\square$

We are finally ready to finish the problem.

Claim: $Q'$ lies on $(IPK'M')$ .

Proof: Note that $\measuredangle C'Q'P=\measuredangle DEP$ and $\measuredangle PDK'=\measuredangle PDE+\measuredangle EDK'=\measuredangle PDE+90^\circ - \measuredangle DFE=90^\circ-\measuredangle PED$ Therefore, using that $M'$ is the circumcenter of $\triangle DPK'$ we get $\measuredangle M'K'P=90^\circ-\measuredangle PDK'=\measuredangle PED=\measuredangle C'Q'P. \ \ \ \blacksquare$

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OronSH

#214 h Dec 22, 2024, 6:30 PM • 7 Y

Y by megarnie, ihatemath123, YaoAOPS, anantmudgal09, GrantStar, TestX01, AndreiVila

Let $M$ be the midpoint of $EF$ , let $Z$ be the reflection of $D$ over $AI$ , let $S$ be the reflection of $D$ over $I$ , let $P_B,P_C$ be the feet from $B,C$ onto $AI$ , let $Q_B,Q_C$ be the intersections of $BI,CI$ with $EF$ , let $R_B,R_C$ be the second intersections of $(BPF),(CPE)$ with $EF$ .

We claim $MQ_B=MR_B$ and $MQ_C=MR_C$ .

First, since $PR$ is a symmedian of $\triangle PEF$ we have $P,M,S$ are collinear. Next, $P_B\in DE$ by Iran lemma, and by symmetry $P_B\in ZF$ . Thus arc chasing implies $\measuredangle FPM=\measuredangle FP_BM$ ,so $FPP_BM$ is cyclic and $\angle FPP_B=90^\circ$ .

Let $X_C=FP\cap CP_C$ . Then $P,P_C$ lie on the circle with diameter $P_BX_C$ . We show $Z$ lies on this circle as well. First $P_C\in DF$ by Iran lemma, so by symmetry $P_C\in ZE$ and thus $\measuredangle PZP_C=\measuredangle PZE=\measuredangle PFE=\measuredangle PX_CP_C$ as desired.

Now observe $FR_C\cdot FE=FP\cdot FX_C=FP_B\cdot FZ=EP_B\cdot ED=EQ_C\cdot EF,$ since $Q_C\in(BDP_BIF)$ by Iran lemma. This implies $MQ_C=MR_C$ , and similarly we get $MQ_B=MR_B$ .

Next notice $\measuredangle PQC=\measuredangle PEC=\measuredangle PFE=\measuredangle PFR_B=\measuredangle PQR_B$ , so $Q\in CR_B$ . Similarly $Q\in BR_C$ .

Now apply DIT on points $B,Q,I,C$ and line $EF$ . Since $BQ\cap EF=R_C,CI\cap EF=Q_C$ and $BI\cap EF=Q_B,CQ\cap EF=R_B$ we get the involution is reflection over $M$ . Thus the conic through $B,Q,I,C,E$ must also pass through $F$ .

Let $\mathcal I,\mathcal J$ be the circular points at infinity. Construct the cubic $\mathcal C$ through $A,B,C,D,E,F,\mathcal I,\mathcal J,P$ . By Cayley-Bacharach with $((BDIF\mathcal I\mathcal J)\cup\overline{AEC})\cap((CDIE\mathcal I\mathcal J)\cup\overline{AFB})$ we see $I\in\mathcal C$ . Next since $B+F+P+\mathcal I+\mathcal J=P+\mathcal I+\mathcal J-A=C+E+P+\mathcal I+\mathcal J$ we get $Q\in\mathcal C$ .

Additionally, since $B+F+\mathcal I+\mathcal J+P+Q=0=B+F+\mathcal I+\mathcal J+D+I,$ we get $X=PQ\cap DI\in\mathcal C$ . We also get $B+D+I+F+\mathcal I+\mathcal J=0=B+F+A+D+I+X,$ so $A+X=\mathcal I+\mathcal J$ .

Finally, note $\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle FEP+\measuredangle PFE=\measuredangle FPE=\measuredangle FDE=\measuredangle BIC,$ so $B+Q+I+C+\mathcal I+\mathcal J=0$ . Since $B,Q,I,C,E,F$ are coconic, $B+Q+I+C+E+F=0$ , so $E+F=\mathcal I+\mathcal J=A+X.$ Thus $EF$ and $AX$ intersect on the line at infinity, as desired.

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#215 h Jan 9, 2025, 7:29 PM • 1 Y

Y by anantmudgal09

This is the best problem of all time, perfect for my 1700th post. Solved in 2.5 hours, no hints.

We identify that $P$ is the $D$ -queue point of $\triangle DEF.$ Let $D'$ be the point diametrically opposite $D$ on $\omega.$ Let $M$ be the midpoint of segment $EF,$ and let $DM$ meet $\omega$ again at point $K.$ We start with a preliminary claim:

Claim 1: Lines $PK$ and $DI$ meet on the line through $A$ perpendicular to $AI.$
Proof: By reciprocating poles/polars, this amounts to showing that if the tangents at $P$ and $K$ to $\omega$ meet at point $L,$ then $LM \parallel BC.$ Let $\ell$ be the line through $M$ parallel to $BC,$ and let $L$ intersect $\omega$ at two points $V_1, V_2.$ It is clear that $(D,D';V_1, V_2) = -1.$ Projecting this harmonic bundle through $M$ gives $(K,P;V_1,V_2) = -1,$ so $V_1 V_2 = \ell$ passes through $L,$ as claimed.

Thus it suffices to show that $P,Q,K$ are collinear.

Now, we will reflect a number of objects across the line $AI.$

The point $D$ is sent to the point $X$ such that $DX \parallel EF.$
The point $P$ gets sent to the second intersection of $AD'$ and $\omega,$ say $P'.$
Some simple angle-chasing yields that the point $B$ gets sent to the second intersection of $(BIC)$ with $AC,$ say $B'.$ Similarly define $C'.$
Obviously, points $E$ and $F$ are sent to each other.
By symmedian properties, if $AD$ intersects $\omega$ again at point $K',$ then $K$ is sent to $K'.$
We must show that if $(B'EP')$ intersects $(C'FP')$ at $Q',$ then $P', Q', K'$ are collinear.

We will explicitly construct a $Q'$ which satisfies that $B'EP'Q'$ and $C'FP'Q'$ are cyclic and $P', Q', K'$ are collinear. Let $I_A$ be the $A$ -excenter, so that $B,I,C,B',C',I_A$ are concyclic, and let $N$ be the midpoint of $BC.$ Redefine $Q'$ as the intersection of $I_A N$ with $(BIC).$ We will show that this $Q'$ satisfies the desired properties.

Claim 2: $DHP'Q'I$ is cyclic.
Proof: We show that they lie on the circle with diameter $IH.$ Clearly $D$ lies on this circle, as does $Q'$ since $\angle IQ'H = \angle IQ' I_A = 90^\circ.$ Finally, say by the incircle diameter lemma, we see that $NP'$ is tangent to $\omega,$ so $\angle IP'N = 90^\circ.$ This proves our claim.

Claim 3: $P'Q'EB'$ is cyclic.
Proof: Invert about $\omega.$ By symmetry, $\overline{B'XC'}$ is tangent to $\omega,$ so $B'$ is sent to the midpoint of $EX.$ The circle $(BIC)$ is sent to the $D$ -midline of $\triangle DEF$ , and by Claim 2, the inverse of $Q'$ lies on $DP'.$ This if $DP' \cap EF = Y,$ then $Q'$ is sent to the midpoint of $DY.$ Therefore, our inverted claim is as follows:

Let $\triangle ABC$ be a triangle, and let $A'$ be on $(ABC) = \omega$ such that $AA' \parallel BC.$ Let $Q$ be the $A$ -queue point of $BC,$ and let $Y = A'Q \cap BC.$ Let $D, E$ be the midpoints of $A'Y, AB,$ respectively, Prove that $DEBQ$ is cyclic.

Since $DE \parallel AA',$ the claim thus follows by Reim's Theorem.

Similarly, $C'FQ'P'$ is cyclic. This equates our two definitions of $Q'$ . Finally:

Claim 4: $P',Q', K'$ are collinear.
Proof: Note that $\triangle DEF \sim \triangle I_A BC,$ so symmedian properties give $\angle EDA = \angle CI_A Q'.$ But $\angle CI_A Q' = 180^\circ - \angle CB'Q' = \angle EB'Q' = \angle EP'Q'$ and $\angle EDA = \angle EDK' = \angle EP'K'.$ Therefore, $\angle EP'Q' = \angle EP'K',$ and we are done!

This post has been edited 4 times. Last edited by EpicBird08, Jan 11, 2025, 5:34 PM

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#216 h Apr 4, 2025, 1:17 PM

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Here's a solution that I don't think was posted before.
Invert from $D$ with radius $\sqrt{DE.DF}$ . $X'$ is the image of $X$ under this inversion. After the inversion, let the $D$ symmedian intersect $(DEF)$ at $L$ . Let the foot of the perpendicular from $D$ to $EF$ be $D_1$ and the antipode of $D$ in $(DEF)$ be $D_2$ . Then by definition, $LD_2 \cap EF = P'$ . We must show that if $(DQ'P')\cap DD_1 = G \Rightarrow \angle GA'I' = 90$ where $A'$ is the Humpty point of $D$ in $DEF$ and $I'$ is the reflection of $D$ over $EF$ .
It is well-known that $A'$ is the reflection of $L$ over $EF$ . Let $DL \cap EF=K$ and $P'A' \cap DD_1=G_1$ then by angle chasing we have $\angle G_1P'K = \angle LP'K = \angle KDG_1$ So $PKDG_1$ is cyclic. Also by PoP we have $D_1K.KP' = KL.KD = KA'.KI'$ Which gives $I'D_1A'P'$ is cyclic and so $\angle I'A'G_1 = 90$ .
Therefore it's enough to prove that $DKP'Q'$ is cyclic. Let $\omega_1 = (C'P'F)$ and $\omega_2=(B'P'E)$ . Let $\omega_1 \cap B'C' = X$ and $\omega_2 \cap B'C' = Y$ . We'll prove $\frac{P_{\omega_1}^D}{P_{\omega_2}^D}=\frac{P_{\omega_1}^K}{P_{\omega_2}^K}$ which would imply the result.
Notice that $EF \parallel B'C'$ and $DF=FC'=P'X$ which gives $DXP'F$ is a parallelogram and so $DX=FP'$ . Similarly $DY = EP'$ . Let $f(Z) = \frac{ZF}{ZE}$ . And so $\frac{P_{\omega_1}^D}{P_{\omega_2}^D}=-\frac{DX.DC'}{DY.DB'} = -f(P')f(D_1) = -f(L)f(D_2)f(D_1)=-f(K)\frac{f(D_2)f(D_1)}{f(D)}$ And $\frac{P_{\omega_1}^K}{P_{\omega_2}^K} = -\frac{KF.KP'}{KE.KP'}=-f(K)$ so we must prove that $f(D_1)f(D_2)=f(D)$ and this is true since
$\frac{f(D_1)}{f(D)} = \frac{D_1F}{DF}.\frac{DE}{D_1E} = \frac{\cos \angle F}{\cos \angle E} = \frac{ED_2}{DD_2}.\frac{DD_2}{FD_2} = \frac{1}{f(D_2)}$ And we are done.

This post has been edited 2 times. Last edited by Parsia--, Apr 4, 2025, 1:20 PM

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