Let $ABC$ be an acute-angled triangle with $AB<AC$, incentre $I$, and let $M$ be the midpoint of major arc $BAC$. Suppose the perpendicular line from $A$ to segment $BC$ meets lines $BI$, $CI$, and $MI$ at points $P$, $Q$, and $K$ respectively. Prove that the $A-$median line in $\triangle AIK$ passes through the circumcentre of $\triangle PIQ$. Proposed by Pranjal Srivastava and Rohan Goyal
Problem
Source: India IMOTC 2024 Day 3 Problem 1
Tags: geometry, incenter, excenter
Rijul saini
31.05.2024 08:37
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Solution AObserve that $\angle PIQ=90-\frac{\angle A}{2}$, $\angle IQP=90-\frac{\angle C}{2}$ and $\angle IPQ=90-\frac{\angle B}{2}$. Thus, if $\triangle DEF$ is the orthic triangle of $\Delta IPQ$, then it is similar to $\Delta ABC$.
Let $\ell$ be the $I$ midline in $\triangle IPQ$ and $O$ be the circumcenter. Now, let $X=\ell \cap AO$ and $K'$ be the reflection of $I$ across $X$. Then, we just want that $K', K$ coincide or equivalently that $I, M, X$ are collinear.
Now, with respect to triangle $IPQ$, $AI$ is tangent to the circumcircle of $IPQ$ as $\angle AIQ=90-\frac{\angle B}{2}$ as required.
Let $H$ be the orthocenter of $\triangle IPQ$ and $N$ be the midpoint of $PQ$. Then, we have that $\angle AIM=\angle NHD$ so we just want $\angle NHD+\angle AIX= 180^\circ$ But $\angle AIX=90^\circ +\angle OIX$ and $\angle NHD=90-\angle HND$. Thus, we just want that $\angle HND=\angle OIX$
Taking homothety with dilation factor $+2$ from $I$, we have $X$ going to $K'$ which is now the intersection of $PQ$ and the line through the antipode of $I$ in $IPQ$ and the point $R$ on $(IPQ)$ such that $AR$ is tangent to $(IPQ)$.
Now let $HN \cap (IPQ)=S_1, S_2$ where $S_2$ is the antipode of $I$ and let $H'$ be the reflection of $H$ in $PQ$.
Now, we just want $\angle HND=\angle HS_2H'=\angle S_1IH$. Thus, we want $\angle S_1IH=\angle OIK'$ or that $IK'$ and $IS_1$ are isogonal in $\angle QIP$.
Now, performing $\sqrt{bc}$ and reflection in $\triangle IPQ$, we get that $S_1$ and $K'$ interchange. Thus, $IK'$ and $IS_1$ are isogonal in $\angle QIP$ as required. Thus, we are done.
Remark.The use of inversion in the last step is not strictly necessary: it's just used for convenience to prove an isogonality that might be part of the Olympiad lore. The key insight lies in introducing the points $D,E,F$ and completely shifting the result to a statement about $\Delta IPQ$ instead of $(ABC)$. We can restate the result for which we use the invertive step as follows:
Let $\Delta ABC$ be a triangle with midpoint $M$ of $BC$, orthocenter $H$, $A'$ antipode of $A$ in $(ABC)$ and point $T$ on $(ABC)$ such that the $A$ and $T$ tangents to $(ABC)$ concur on $BC$. Now, $X=HA'\cap (ABC)$ and $Y=TA'\cap (BC)$ then $AX$ and $AY$ are isogonal.
A non-invertive and easy way of proving the isogonality would be by proving that triangles $HXA$ and $A'YA$ are indeed similar. This can be done via angles and lengths with sine rule.
Solution BLet $O$ be the circumcenter of $\triangle PIQ$,
$S$ be the point on circumcircle of $\triangle PIQ$ so that $IS \perp AJ$, $R$ be the antipode of $I$ in the circumcircle of $\triangle PIQ$.
I will use the following:
Lemma 1 Let $T$ be a point on the line through $I$ perpendicular to $AI$. Then $(IB, IC; IT, IM) = -1$.
Proof: Let $I_B, I_C$ be the $B,C$-excenters. Then we can project onto $I_BI_C$ and reduce to showing that $M$ is the midpoint of $I_BI_C$ which is known.
Now, we begin investigating the problem at hand:
Lemma 2 $IO \perp AI$.
Proof: This is direct: $\angle AIP = 90^\circ - B/2$ and $\angle PIO = 90^\circ - \angle PQO = 90^\circ - (90^\circ - B/2) = B/2$.
Proposition 3 $S,R,K$ are collinear.
Proof: Let the circumcircle of $\triangle PIQ$ be denoted by $\omega$. From Lemma 2, $AI$ is tangent to $\omega$. Therefore, $IS$ is the $I$-symmedian of $\triangle PIQ$. Hence, $(P,Q; S,I) = -1$.
Projecting from $R$, we get $(RP, RQ; RS, RI) = -1$, hence if $RI \cap PQ = X$ and $RS \cap PQ = K'$, then $(P,Q; X,K') = -1$.
On the other hand, we know $(IB, IC; IR, IM) = -1$ from Lemma 1, so projecting onto $PQ$ we get $(P,Q; X,K) = -1$. Thus $K= K'$ which finishes the proof.
Proposition 4 $AO$ bisects $IK$.
Proof: $RS \perp SI$ since $RI$ is the diameter, hence $RK \parallel AO$. But $O$ is the midpoint of $RI$, hence $AO$ bisects $IK$ by midpoint theorem.
This finishes the solution. $\blacksquare$
Solution CLet $J$ be the $A$-excentre of $\triangle ABC$ and $L$ the point on $BC$ with line $LJ$ is tangent to the $A$-excircle. Let $N$ be the midpoint of arc $BC$ not containing $A$ of the circumcircle of $ABC$. Let $S$ be the point on $BC$ with $\angle ANS=90^{\circ}$. Let $T$ be the point on $BC$ with $TN \perp MI$.
Claim. $S$ is the midpoint of $LT$.
Proof. Note that by the Incenter-Excenter Lemma, $N$ is the midpoint of $IJ$ so $JL \perp AI$ and by standard mixtilinear facts, $TI \perp AI$. Thus, $L, S, T$ are obtained by drawing perpendicular lines to line $AI$ from the points $J, N, I$ respectively and intersecting with line $BC$. Since $N$ is the midpoint of $JI$, then $S$ is the midpoint of $LT$ as desired. $\spadesuit$
Now observe that $\triangle IPQ \sim \triangle JBC$ so the spiral similarity mapping $IPQ$ to $JBC$ consists of a $90^{\circ}$ rotation as $PQ \perp BC$. Since $AI$ is tangent to $(PIQ)$ and $A$ lies on $PQ$, the spiral similarity sends $A$ to $L$ and the line passing through $A$ and the circumcentre of $PIQ$ to the line $LN$. So it suffices to show that the line through $A$ perpendicular to $LN$ bisects $IK$.
Thus we want cross ratio of lines $AP, AI,$ line through $A$ parallel to $IK$ and line through $A$ perpendicular to $LN$ form a harmonic bundle. Taking directions perpendicular to this pencil from $N$, we get the four lines $N\infty_{BC}$ (as $BC \perp AP$), $NS$ (as $NS \perp AI$), $NL$ and the line $NT$ as $NT \perp KI$. Projecting this pencil at $N$ to $BC$ we get that this cross ratio equals since $(\infty_{BC}S; LT)=-1$ as $S$ is the midpoint of $LT$. This concludes the proof.
Solution DLet $D,E,F$ be the foot of perpendiculars from $I$ to $BC,CA,AB$ respectively. Let $N$ be the midpoint of $BC$. Suppose the foot of perpendicular from $N$ onto $EF$ is $L$. We use that $I,L,M$ are collinear. This can be proven in many ways: one sketch of the argument goes as follows: If $X$ (resp. $Y$) is the foot of perpendicular of $B$ (resp. $C$) onto $CI$ (resp. $BI$), then $X,Y$ lie on $EF$, and $N$ is the midpoint of $XY$. On the other hand, if $I_B, I_C$ are the $B-$, $C-$excenters, then $M$ is the midpoint of $I_BI_C$. The result then follows on noting that $XY \parallel I_BI_C$, and considering a homothety at $I$.
Now toss the diagram on the complex plane. Let the incircle of $\triangle ABC$ be the unit circle, so $I$ is the origin, and let $D = 1$, $E=e$, $F = f$ without loss of generality.
Now, it is well known that the intersection of tangents to the unit circle at points $x,y$ is $\displaystyle \frac{2xy}{x+y}$. Therefore, $\displaystyle A =a = \frac{2ef}{e+f}, B = b = \frac{2f}{f+1}, C =c = \frac{2e}{e+1}$.
Thus, the midpoint of $BC$ is $N = \displaystyle \frac{e}{e+1} + \frac{f}{f+1}$.
Hence, the foot of perpendicular from $N$ to $EF$ is $$L =l = \frac 12 \left(e+f + \frac{e}{e+1} + \frac{f}{f+1} - ef \left( \frac{1}{e+1} + \frac{1}{f+1} \right)\right) = \frac {(e+f)(e+f+2)}{2(e+1)(f+1)}.$$
We know $D=1$, therefore the line perpendicular to $BC$ from $A$ is the set of points $z$ with $\mathrm{Im}(z) = \mathrm{Im}(A)$, i.e. $\displaystyle z - \Bar{z} = \frac{2(ef-1)}{e+f}$.
Now, $P$ lies on $BI$, therefore $p = \lambda b$ for some $\lambda \in \mathbb R$. Note that $\Bar{b} = \frac 1f \cdot b$, therefore, $\Bar{p} = \frac 1f \cdot p$. Thus, $$p - \Bar{p} = \frac{(f-1)}{f} \cdot p \implies p = \frac{2f(ef-1)}{(e+f)(f-1)}\text{. Similarly, } q = \frac{2e(ef-1)}{(e+f)(e-1)}.$$
With the same idea, one checks that $\displaystyle \Bar{l} = \frac{(e+f)(2ef+e+f)}{2ef(e+1)(f+1)} = l \cdot \frac{(2ef+e+f)}{ef(e+f+2)}$, therefore, $$k-\Bar{k} = \frac{(ef-1)(e+f)}{ef(e+f+2)} \cdot k \implies k = \frac{2ef(e+f+2)}{(e+f)^2}.$$
\noindent We know that the circumcenter of a circle passing through origin, $p,q$ is given by $\displaystyle \frac{pq(\Bar{p}- \Bar{q})}{\Bar{p}q - p\Bar{q}}$. Note that $\displaystyle p = \frac{2(ef-1)}{(e+f)} \cdot p_1$, and $\displaystyle q = \frac{2(ef-1)}{(e+f)} \cdot q_1$ where $p_1, q_1$ are $\displaystyle \frac{f}{f-1}, \ \frac{e}{e-1}$ respectively. Thus the circumcenter of $\triangle IPQ$ is $\displaystyle \frac{2(ef-1)}{(e+f)} \cdot t$ where $t$ is the circumcenter of $\triangle IP_1Q_1$. This simplifies calculation, and we get
$$ \text{The circumcenter of $\triangle IPQ$} =
\frac{2(ef-1)}{(e+f)} \cdot \frac{\frac{ef}{(e-1)(f-1)} \left(\frac{1}{1-f} - \frac{1}{1-e} \right)}{\frac{e}{(1-f)(e-1)} - \frac{f}{(1-e)(f-1)}} = \frac{2ef(ef-1)}{(e+f)(e-1)(f-1)}.$$
Thus, we need to show that $\displaystyle \frac{2ef}{(e+f)}, \ \frac{2ef(ef-1)}{(e+f)(e-1)(f-1)}, \ \frac{ef(e+f+2)}{(e+f)^2} $ are collinear. Multiplying by $\displaystyle \frac{(e+f)}{2ef}$, we reduce to proving $\displaystyle 1, \ \frac{ef-1}{(e-1)(f-1)}, \ \frac{e+f+2}{2(e+f)}$ are collinear for which we consider:
$$\frac{\frac{ef-1}{(e-1)(f-1)}-1}{\frac{e+f+2}{2(e+f)} - 1} = \frac{\frac{e+f-2}{(e-1)(f-1)}}{\frac{2-e-f}{2(e+f)}} = - \frac{2(e+f)}{(e-1)(f-1)} $$The above quantity may be easily checked to equal the conjugate, so we are done.
math_comb01
31.05.2024 09:03
Let $I^{*}$ denote the antipode of $I$ in $(IPQ)$, then by angle chase $AI$ is tangent to $(IPQ)$ and $II^{*}$ is tangent to $(BIC)$; now using the well known fact $(IP;IQ;II^{*};IK)=-1$ now restate in terms of $IPQ$ and ratio lemma on circle finishes after homothety of 2 finishes
WLOGQED1729
31.05.2024 11:05
This problem reminds me of ISL 2017 G3 which I solved it by trig bash. Unsurprisingly, this problem can also be solved by trig bash too.
Let $\angle ABC=2x$ and $\angle ACB=2y$. Note that $x>y$ since $AB<AC$.
We divide the problem into two parts
Part(i) Compute $\frac{sin\angle QAO}{sin \angle OAI}$
By spamming sine laws on triangles $QAO,OAI$ and simple angle chasing yields
$$\frac{sin \angle QAO}{sin \angle OAI}=\frac{sin \angle QAO}{QO}\div \frac{sin \angle OAI}{OI}=\frac{sin \angle AQO}{AO}\div \frac{sin \angle AIO}{AO}=\frac{sin(90^\circ +x+y)}{sin(90^\circ)}=cos(x+y)$$
Part(ii) Compute $\frac{sin \angle KAD}{sin \angle DAI}$ where $D$ is a midpoint of side $KI$
Let $\angle AIM=\theta $. By spamming sine laws on triangles $KAD,DAI,IAC,CAM$ yields
$$\frac{sin \angle KAD}{sin \angle DAI}= \frac{sin \angle KAD}{AD}\div \frac{sin \angle DAI}{DI} = \frac{sin \angle AKD}{AD}\div \frac{sin \angle AID}{AD}= \frac{sin \angle AKI}{sin \angle AIK}=\frac{sin(\theta -(x-y))}{sin \theta}=cos(x-y)-cot(\theta)sin(x-y)$$and $$cot(\theta) = \frac{AI}{AM}=\frac{AI}{AC} \cdot \frac{AC}{AM}=\frac{sin(y)}{cos(x)} \cdot \frac{sin(2x)}{sin(x-y)} =\frac{2sin(y)sin(x)}{sin(x-y)}$$So, $$\frac{sin \angle KAD}{sin \angle DAI}=cos(x-y)-2sin(x)sin(y)=cos(x+y)$$By part (i) and (ii), note that these two values are equal, we can conclude that $O$ lies on $AN$, as desired.
Attachments:
egxa
31.05.2024 15:40
Define $I_a$ as excenter and $N=AI\cap(ABC)$. See that $\triangle IPQ \sim \triangle I_aBC$. Doing complex bash taking unit circle as $(ABC)$ is easier than #2 complex solution.
MathLuis
31.05.2024 18:39
By trivial angles we see that $AI$ is tangent to $(PQI)$, now let $E$ the antipode of $I$ in $(IPQ)$ and let $F \ne I$ a point on $(IPQ)$ such that $AI=AF$ (so that $AF$ is tangent to $(IPQ)$), now let $I'$ the reflection of $I$ over $A$. Let $I_B, I_C$ the $B,C$ excenters of $\triangle ABC$ respectively. Note that $-1=(I, F; P, Q) \overset{K}{=} (MI \cap (IPQ), KF \cap (IPQ); Q, P) \overset{I}{=} (M, (KF \cap (IPQ))I \cap AM: I_C, I_B)$ But $M$ is known to be the midpoint of $I_CI_B$ and $EI \parallel I_BI_C$ is also true, therefore $KF \cap (IPQ)=E$, now if we extend this line $KF$ to $(A,AI)$, since we have $\angle IFE=90$, we must in fact have $I', F, E, K$ colinear, now take homothety centered at $I$ with scale $0.5$ and notice its equivalent to the problem, thus we are done .
motannoir
02.06.2024 10:30
I like this problem because it is one of the few problems i introduces a lot of points ! Solution:Let $I_A,I_B,I_C$ the excenters;$O_1$ the circumcenter of $(IPQ)$; $T$ the midpoint of $KI$,$T'$ the ,midpoint of $MI$;$S$ the midpoint of $PQ$,$N$ is the midpoint of the little arc. By angle chasing we can see that : 1)$AI$ is tangent to $(IPQ)$; 2)$IQP \sim I_AI_CI_B$ From these 2 observations we can see that $O_1SIA$ is cyclic,so $\angle O_1IS=\angle O_1AS$.It is enough to prove that $\angle O_1IS=\angle TAK$.From similarity we have that $$\angle O_1IS=\angle O_1IQ-\angle QIS=90-\angle QPI-\angle QIS=\frac{B}{2}-\angle I_CI_AM$$. Observe that $IKA \sim IMN$ so $T \sim T'$ so $\angle TAK=\angle T'NM$. So all we have to prove is that $\angle I_CI_AM+\angle T'NM=\frac{B}{2}$. But $T'N\parallel I_AM$ (midlines) so we have that : $$\angle I_CI_AM+\angle T'NM=\angle I_CI_AA+\angle AI_AM+\angle T'NM=\angle I_CI_AA+\angle INT'+\angle T'NM=\angle I_CI_AA+\angle INM=\frac{C}{2}+\angle ABM=\frac{B}{2}$$so the problem is solved $\blacksquare$
sami1618
01.08.2024 05:46
Let $I_a$, $I_b$, and $I_c$ be the respective excenters, let $AI$, $BI$, $CI$, and $BC$ meet the perpendicular bisector of $BC$ at $N$, $P'$, $Q'$, and $S$, and let $H$ be the $I_a$-Humpty point of $I_aI_bI_c$. Claim: $H$ lies on $(IP'Q')$ Since we have that $I_aCHB$ is a harmonic quadrilateral we get that $SHP'B$ is concyclic as $\angle SHB=\angle CBI_a=\angle SP'Q$. Thus $H$ is the Miquel point of $CSBIP'Q'$, finishing the claim. Let $I_aM$ meet $(IP'Q')$ again at $R$ which is the $I$-antipode as $IH\perp I_aM$. To finish the problem notice that $IMP'Q'N$ is similar to $IKPQA$ and homothety with ratio $\tfrac{1}{2}$ at $I$ gives the desired result.
Attachments:
SomeonesPenguin
05.08.2024 17:20
Trig bash for storage.
Let $O$ be the circumcenter of $\triangle IPQ$. By ratio lemma, we have to prove that $\frac{\sin(\angle PAO)}{\sin(\angle OAI)} = \frac{AI}{AK}$.
Denote $\angle BAC = 2a$ and the other similar ones. A quick angle chase gives that $AI$ is tangent to $(IPQ)$. By trig ceva in $\triangle AQI$ and point $O$ we get: $$\frac{\sin(\angle PAO)}{\sin(\angle OAI)}\cdot\frac{\sin(\angle AIO)}{\sin (\angle OIQ)}\cdot \frac{\sin (\angle OQI)}{\sin (\angle OQA)} = 1.$$
But $\angle OIA = 90^\circ$, $\angle OIQ = \angle OQI$ and $\angle OQA = 180^\circ - a$ so we get $\frac{\sin(\angle PAO)}{\sin(\angle OAI)} = \sin(a)$.
Now lets compute $\frac{AI}{AK} = \frac{\sin (\angle AKI)}{\sin (\angle AIM)}$. We have that $\angle KAI = 90^\circ - 2c - a = b-c$. Let $\angle AIM = x$. We have that $\frac{\sin (\angle AKI)}{\sin (\angle AIM)} = \frac{\sin(x-b+c)}{\sin(x)} = \cos(c-b) + \cot(x)\sin(c-b)$. But $\cot(x) = \frac{AI}{AM}$. Apply LOS in $\triangle AIB$ and $\triangle AMB$ to get $\frac{AI}{AM} = -\frac{2\sin(b)\sin(c)}{\sin(c-b)}$. So $\frac{\sin (\angle AKI)}{\sin (\angle AIM)} = \cos(c-b) - 2\sin(b)\sin(c) = \cos(c+b) = \sin(a)$. Done!
Om245
10.09.2024 15:18
I finally have time to submit by contest solution for this problem. This was only geometry problem I fully solved in IMOTC, I have very fun solving. I found little bit different solution with introducing new harmonic cross ratios. At first look this looks like 17G3, but indeed this is very different and amazing problme!! Let $A_0 = AI \cap (ABC)$ Claim: : $AI$ is tangent to $(IPQ)$ Proof. Note that $\measuredangle BPK = 90 - \frac{\measuredangle B}{2}$, $\measuredangle PQC = 90 - \frac{\measuredangle C}{2}$ and $\measuredangle QIP = 90 - \frac{\measuredangle A}{2}$. Also $$\measuredangle AIQ = \measuredangle A_0IC = 90 - \frac{ \measuredangle B}{2} = \measuredangle IPQ$$which give us $AI$ tangent to $(IPQ)$. Define $X= AM \cap BC$ and $G$ is $I-$dumpty point of $\triangle IPQ$. Claim: : $\overline{X-G-I}$ Proof. Let $AN$ be tangent to $(IPQ)$ other then $AI$. It's well known that $\overline{N-G-I}$. If $A'= AI \cap BC$ then following claims are follow from well known results $$(IB,IC;IA',IX) =(IP,IQ;IA,IG)=-1$$Hence $\overline{X-G-I}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.068409913689637, xmax = 27.799478144164162, ymin = -13.781839061307451, ymax = 4.066378941012234; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((5.52,-0.82)--(5.,-6.68)--(18.06,-6.74)--cycle, linewidth(0.8) + ffxfqq); /* draw figures */ draw((5.52,-0.82)--(5.,-6.68), linewidth(0.8) + ffxfqq); draw((5.,-6.68)--(18.06,-6.74), linewidth(0.8) + ffxfqq); draw((18.06,-6.74)--(5.52,-0.82), linewidth(0.8) + ffxfqq); draw(circle((11.541038154299478,-4.307361747480499), 6.958059529302715), linewidth(0.8)); draw((5.52,-0.82)--(5.487250638057658,-7.948444449449819), linewidth(0.8)); draw((11.573004399783407,2.6506243528555724)--(-4.623131731048777,-6.635789593884921), linewidth(0.8)); draw((5.,-6.68)--(-4.623131731048777,-6.635789593884921), linewidth(0.8)); draw(circle((6.314486759525287,-5.0658494289780505), 1.4226958630714122), linewidth(0.8) + red); draw((5.487250638057658,-7.948444449449819)--(11.573004399783407,2.6506243528555724), linewidth(0.8)); draw((5.52,-0.82)--(11.509071908815542,-11.265347847816571), linewidth(0.8)); draw((7.548698020741028,-4.358187022537834)--(-4.623131731048777,-6.635789593884921), linewidth(0.8) + qqwuqq); draw((7.548698020741028,-4.358187022537834)--(5.,-6.68), linewidth(0.8)); draw((5.505871553368096,-3.8952918835443042)--(18.06,-6.74), linewidth(0.8)); /* dots and labels */ dot((5.52,-0.82),dotstyle); label("$A$", (5.161259542112825,-0.5646174894337418), NE * labelscalefactor); dot((5.,-6.68),dotstyle); label("$B$", (4.684247034255729,-7.342170205236651), NE * labelscalefactor); dot((18.06,-6.74),dotstyle); label("$C$", (17.841842042647286,-6.408020710683171), NE * labelscalefactor); dot((7.548698020741028,-4.358187022537834),linewidth(4.pt) + dotstyle); label("$I$", (7.526446560237591,-4.02295817139769), NE * labelscalefactor); dot((11.509071908815542,-11.265347847816571),linewidth(4.pt) + dotstyle); label("$A_0$", (11.561177355862194,-10.840261929522022), NE * labelscalefactor); dot((11.573004399783407,2.6506243528555724),linewidth(4.pt) + dotstyle); label("$M$", (11.720181525147893,2.853972150208781), NE * labelscalefactor); dot((5.487250638057658,-7.948444449449819),linewidth(4.pt) + dotstyle); label("$K$", (5.936404867380606,-8.01793792470087), NE * labelscalefactor); dot((-4.623131731048777,-6.635789593884921),linewidth(4.pt) + dotstyle); label("$X$", (-4.577745826636213,-6.14963893559391), NE * labelscalefactor); dot((5.495150407944361,-6.228927870777472),linewidth(4.pt) + dotstyle); label("$P$", (5.578645486487784,-6.0701368509510605), NE * labelscalefactor); dot((5.505871553368096,-3.8952918835443042),linewidth(4.pt) + dotstyle); label("$Q$", (5.578645486487784,-3.5061946212191684), NE * labelscalefactor); dot((4.907904056900141,-4.852334512692914),linewidth(4.pt) + dotstyle); label("$N$", (4.7836246400592906,-4.321090988808375), NE * labelscalefactor); dot((6.228301038820584,-4.605260767615373),linewidth(4.pt) + dotstyle); label("$G$", (6.31403976943414,-4.440344115772649), NE * labelscalefactor); dot((6.314486759525286,-5.0658494289780505),linewidth(4.pt) + dotstyle); label("$O$", (6.393541854076989,-4.897481102469032), NE * labelscalefactor); dot((6.517974329399343,-6.153315735993827),linewidth(4.pt) + dotstyle); label("$L$", (6.174911121309154,-6.010510287468923), NE * labelscalefactor); dot((8.89020849021879,-6.6978723207820146),linewidth(4.pt) + dotstyle); label("$A'$", (9.116488253094577,-6.4875227953260195), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $L$ be midpoint of $IK$. Claim: : $$(M,L;I,K)=\frac{A_0I}{A_0A}$$Proof. $(M,L;I,K) = \frac{MI}{MK} / \frac{LI}{LK}$ but as $L$ is midpoint we get $\frac{MI}{MK}$. From $AK \parallel MA_0$ we get $\triangle AIK \sim \triangle A_0IM$ $$\frac{MI}{IK} = \frac{A_0I}{AI} \Rightarrow \frac{IK}{MI} = \frac{AI}{A_0I} \Rightarrow \frac{MI}{MK} = \frac{AA_0}{A_0I}$$$$ \frac{MI}{MK} = \frac{A_0I}{A_0A}$$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.50872409290906cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -56.802987153065544, xmax = 105.70573693984352, ymin = -84.39286145450127, ymax = 1.2484731907055522; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((6.806886433199685,-8.072026173044682)--(3.186294770417865,-19.130225512849663)--(37.383746248259556,-20.595392615088233)--cycle, linewidth(0.8) + ffxfqq); draw((29.40745007646128,-56.985950327590295)--(21.07259450977063,-1.4805892995267258)--(-15.402460368442137,-18.3338026414356)--cycle, linewidth(0.4) + qqwuqq); /* draw figures */ draw((6.806886433199685,-8.072026173044682)--(3.186294770417865,-19.130225512849663), linewidth(0.8) + ffxfqq); draw((3.186294770417865,-19.130225512849663)--(37.383746248259556,-20.595392615088233), linewidth(0.8) + ffxfqq); draw((37.383746248259556,-20.595392615088233)--(6.806886433199685,-8.072026173044682), linewidth(0.8) + ffxfqq); draw(circle((20.338091171105066,-18.62412347112488), 17.159261611364997), linewidth(0.8)); draw((21.07259450977063,-1.4805892995267258)--(-15.402460368442137,-18.3338026414356), linewidth(0.4)); draw((3.186294770417865,-19.130225512849663)--(-15.402460368442137,-18.3338026414356), linewidth(0.8)); draw((6.806886433199685,-8.072026173044682)--(29.40745007646128,-56.985950327590295), linewidth(0.8)); draw((21.07259450977063,-1.4805892995267258)--(29.40745007646128,-56.985950327590295), linewidth(0.4) + yqqqyq); draw((22.74318071774964,-12.605735963489321)--(-15.402460368442137,-18.3338026414356), linewidth(0.8) + blue); draw((6.806886433199685,-8.072026173044682)--(19.6035878324395,-35.76765764272303), linewidth(0.8)); draw((6.806886433199685,-8.072026173044682)--(17.686135369787962,-80.52158874888097), linewidth(0.8) + yqqqyq); draw((17.686135369787962,-80.52158874888097)--(21.07259450977063,-1.4805892995267258), linewidth(0.8)); draw((29.40745007646128,-56.985950327590295)--(21.07259450977063,-1.4805892995267258), linewidth(0.4) + qqwuqq); draw((21.07259450977063,-1.4805892995267258)--(-15.402460368442137,-18.3338026414356), linewidth(0.4) + qqwuqq); draw((-15.402460368442137,-18.3338026414356)--(29.40745007646128,-56.985950327590295), linewidth(0.4) + qqwuqq); /* dots and labels */ dot((6.806886433199685,-8.072026173044682),dotstyle); label("$A$", (5.091474077156273,-6.8578892200322885), NE * labelscalefactor); dot((3.186294770417865,-19.130225512849663),dotstyle); label("$B$", (1.6581911737849555,-22.307662285203232), NE * labelscalefactor); dot((37.383746248259556,-20.595392615088233),dotstyle); label("$C$", (36.3724960856505,-18.96974835137), NE * labelscalefactor); dot((9.799725588417735,-14.549364957855785),linewidth(4.pt) + dotstyle); label("$I$", (9.669184614984697,-12.96150327047019), NE * labelscalefactor); dot((19.6035878324395,-35.76765764272303),linewidth(4.pt) + dotstyle); label("$A_0$", (19.87366435556056,-33.751938629774294), NE * labelscalefactor); dot((21.07259450977063,-1.4805892995267258),linewidth(4.pt) + dotstyle); label("$M$", (21.781043746322403,-1.0403820782086615), NE * labelscalefactor); dot((-15.402460368442137,-18.3338026414356),linewidth(4.pt) + dotstyle); label("$X$", (-15.222116434457359,-16.01331029568914), NE * labelscalefactor); dot((29.40745007646128,-56.985950327590295),linewidth(4.pt) + dotstyle); label("$J$", (29.792037187522144,-56.25901544076407), NE * labelscalefactor); dot((22.74318071774964,-12.605735963489321),linewidth(4.pt) + dotstyle); label("$R$", (23.11620931985569,-11.817075636013085), NE * labelscalefactor); dot((7.82409198858344,-14.846032169857061),linewidth(4.pt) + dotstyle); label("$G$", (5.85442583346101,-13.438348118160652), NE * labelscalefactor); dot((8.07908311814851,-16.544126946930962),linewidth(4.pt) + dotstyle); label("$L$", (6.331270681151471,-17.825320716912895), NE * labelscalefactor); dot((12.092612003507117,-19.511810761109412),linewidth(4.pt) + dotstyle); label("$A'$", (13.102467518356015,-18.492903503679543), NE * labelscalefactor); dot((17.686135369787962,-80.52158874888097),linewidth(4.pt) + dotstyle); label("$Z$", (18.061653934336807,-79.71978194713475), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Key Claim: : $A,G,L$ are collinear. Proof. Let $Z = AG \cap MA_0$. Note that $$(M,Z;A_0,\infty{\overline{AK}}) \Rightarrow \frac{MA_0}{M\infty{AK}}/ \frac{ZA_0}{Z\infty{AK}} \Rightarrow \frac{MA_0}{ZA_0} $$ If they are collinear then $$\frac{A_0I}{A_0A} = (M,L;I,K)\stackrel{A}{=}(M,AG \cap MA_0;A_0,\infty{\overline{AK}}) = \frac{MA_0}{ZA_0}$$ Let $J$ be $A$-excenter of $\triangle ABC$. Claim: : $\triangle ZA_0A \sim \triangle MA_0J $ Proof. Observe that in $\sqrt{bc}$ inversion $I \leftrightarrow J$ and $M \leftrightarrow X$ , hence $AI\cdot AJ = AM \cdot AX$. by well known result , we have $I$ as orthocenter of $\triangle JMX$. Hence $MJ \perp IX$ but as $IX \perp AZ$ , we have $AZ \parallel MJ$ Therefore we get $$\frac{MA_0}{JA_0} = \frac{ZA_0}{AA_0} \Rightarrow \frac{MA_0}{IA_0} = \frac{ZA_0}{AA_0}$$
L13832
30.11.2024 12:00
Solved this a few days ago, really cool problem.
[asy][asy]
import graph; size(14cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.747720494268617, xmax = 10.79550249695108, ymin = -5.8578367353619632, ymax = 7.791640248260868; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0);
draw((1,6)--(0,0)--(7,0)--cycle, linewidth(0.5));
draw((2.298740578029825,1.9473254128143442)--(1,2.485281374238571)--(0.5065521266223958,1.7730143606938065)--(1,0.8471270883830366)--cycle, linewidth(0.5) + red);
draw((1,0.8471270883830366)--(2.298740578029825,1.9473254128143442)--(1,2.485281374238571)--cycle, linewidth(0.5) + red);
draw((2.2987405780298245,0)--(3.67570758260774,3.32429241739226)--(0.37791061521389074,2.2674636912833446)--cycle, linewidth(0.5) + red);
draw((4.7012594219701755,-5.54965067985697)--(10.784021952268397,9.135437117483878)--(-3.784021952268396,4.466888149558752)--cycle, linewidth(0.5) + red);
draw((1,6)--(2.298740578029825,1.9473254128143442)--(1,-3.3003964717264056)--cycle, linewidth(0.5) + blue);
/* draw figures */
draw((1,6)--(0,0), linewidth(0.5));
draw((0,0)--(7,0), linewidth(0.5));
draw((7,0)--(1,6), linewidth(0.5));
draw(circle((3.5,2.5), 4.301162633521313), linewidth(0.5));
draw((1,6)--(1,-3.3003964717264056), linewidth(0.5));
draw((3.5,6.801162633521318)--(1,-3.3003964717264056), linewidth(0.5));
draw(circle((1.4215117553751113,1.6662042313108036), 0.9211729077573446), linewidth(0.5));
draw(circle((2.2987405780298245,1.9473254128143438), 1.9473254128143442), linewidth(0.5));
draw((1,6)--(1.4215117553751113,1.6662042313108036), linewidth(0.5) + blue);
draw((1,6)--(3.5,6.801162633521318), linewidth(0.5));
draw((-3.784021952268396,4.466888149558752)--(4.7012594219701755,-5.54965067985697), linewidth(0.5));
draw((10.784021952268397,9.135437117483878)--(4.7012594219701755,-5.54965067985697), linewidth(0.5));
draw((-3.784021952268396,4.466888149558752)--(10.784021952268397,9.135437117483878), linewidth(0.5));
draw((-3.784021952268396,4.466888149558752)--(7,0), linewidth(0.5));
draw((0,0)--(10.784021952268397,9.135437117483878), linewidth(0.5));
draw((4.7012594219701755,-5.54965067985697)--(1,6), linewidth(0.5));
draw((2.2987405780298245,0)--(3.67570758260774,3.32429241739226), linewidth(0.5));
draw((3.67570758260774,3.32429241739226)--(0.37791061521389074,2.2674636912833446), linewidth(0.5));
draw((0.37791061521389074,2.2674636912833446)--(2.2987405780298245,0), linewidth(0.5));
draw((2.2987405780298245,0)--(0.5442829327203977,1.385083049807263), linewidth(0.5));
draw((0.5065521266223958,1.7730143606938065)--(1,6), linewidth(0.5));
draw((0.5065521266223958,1.7730143606938065)--(1,-3.3003964717264056), linewidth(0.5));
draw(circle((1,6), 4.2556901201381745), linewidth(0.5));
draw((0.5442829327203977,1.385083049807263)--(2.298740578029825,1.9473254128143442), linewidth(0.5));
draw(circle((1.6493702890149131,-0.6765355294560309), 2.70302201556467), linewidth(0.5));
draw((1,2.485281374238571)--(0.5065521266223958,1.7730143606938065), linewidth(0.5) + red);
draw((0.5065521266223958,1.7730143606938065)--(1,0.8471270883830366), linewidth(0.5) + red);
draw((1,0.8471270883830366)--(2.298740578029825,1.9473254128143442), linewidth(0.5) + red);
draw((2.298740578029825,1.9473254128143442)--(1,2.485281374238571), linewidth(0.5) + red);
draw((1,2.485281374238571)--(1,0.8471270883830366), linewidth(0.5) + red);
draw((2.2987405780298245,0)--(3.67570758260774,3.32429241739226), linewidth(0.5) + red);
draw((3.67570758260774,3.32429241739226)--(0.37791061521389074,2.2674636912833446), linewidth(0.5) + red);
draw((0.37791061521389074,2.2674636912833446)--(2.2987405780298245,0), linewidth(0.5) + red);
draw((4.7012594219701755,-5.54965067985697)--(10.784021952268397,9.135437117483878), linewidth(0.5) + red);
draw((10.784021952268397,9.135437117483878)--(-3.784021952268396,4.466888149558752), linewidth(0.5) + red);
draw((-3.784021952268396,4.466888149558752)--(4.7012594219701755,-5.54965067985697), linewidth(0.5) + red);
draw((1,6)--(2.298740578029825,1.9473254128143442), linewidth(0.5) + blue);
draw((2.298740578029825,1.9473254128143442)--(1,-3.3003964717264056), linewidth(0.5) + blue);
draw((1,-3.3003964717264056)--(1,6), linewidth(0.5) + blue);
/* dots and labels */
dot((0,0),linewidth(3pt) + dotstyle);
label("$B$", (-0.355879904443651,-0.26653388462407546), NE * labelscalefactor);
dot((7,0),linewidth(3pt) + dotstyle);
label("$C$", (7.120886774427918,-0.20574716365763948), NE * labelscalefactor);
dot((1,6),linewidth(3pt) + dotstyle);
label("$A$", (0.7230843927105794,6.1160718168517), NE * labelscalefactor);
dot((2.298740578029825,1.9473254128143442),linewidth(3pt) + dotstyle);
label("$I$", (2.440309260012383,1.6330511455770482), NE * labelscalefactor);
dot((3.5,6.801162633521318),linewidth(3pt) + dotstyle);
label("$M$", (3.4584868362001777,7.01267595110663), NE * labelscalefactor);
dot((1,0),linewidth(3pt) + dotstyle);
label("$J$", (1.0878047185091926,-0.3729106463153384), NE * labelscalefactor);
dot((1,-3.3003964717264056),linewidth(3pt) + dotstyle);
label("$K$", (0.9206412358514949,-3.336263293429091), NE * labelscalefactor);
dot((1,0.8471270883830366),linewidth(3pt) + dotstyle);
label("$P$", (1.0878047185091926,0.41731672624832905), NE * labelscalefactor);
dot((1,2.485281374238571),linewidth(3pt) + dotstyle);
label("$Q$", (1.0574113580259747,2.149738273791754), NE * labelscalefactor);
dot((1.4215117553751113,1.6662042313108036),linewidth(3pt) + dotstyle);
label("$O$", (1.2245748406836725,1.3595109012280864), NE * labelscalefactor);
dot((3.5,-1.8011626335213178),linewidth(3pt) + dotstyle);
label("$M_{A}$", (3.7928138015155732,-2.166118914825199), NE * labelscalefactor);
dot((2.2987405780298245,0),linewidth(3pt) + dotstyle);
label("$D$", (2.3643258588043383,-0.3729106463153384), NE * labelscalefactor);
dot((3.67570758260774,3.32429241739226),linewidth(3pt) + dotstyle);
label("$E$", (3.7168304003075288,3.4110627338453), NE * labelscalefactor);
dot((0.37791061521389074,2.2674636912833446),linewidth(3pt) + dotstyle);
label("$F$", (0.02403710159657101,2.149738273791754), NE * labelscalefactor);
dot((4.7012594219701755,-5.54965067985697),linewidth(3pt) + dotstyle);
label("$I_A$", (5.01091291785627295,-5.636263293429091), NE * labelscalefactor);
dot((7.791640248260868,9.135437117483878),linewidth(3pt) + dotstyle);
label("$I_C$", (-4.231033366053915,4.490027030999538), NE * labelscalefactor);
dot((0.5442829327203977,1.385083049807263),linewidth(3pt) + dotstyle);
label("$I'$", (0.31277402618713973,1.2379374592952144), NE * labelscalefactor);
dot((0.5065521266223958,1.7730143606938065),linewidth(3pt) + dotstyle);
label("$R$", (0.25198730522070417,1.785017947993138), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy][/asy]
$\rule{40cm}{0.1pt}$
Define $M_A$ to be the arc-midpoint of minor-arc $\widehat{BC}$, $I'$ to be the $I-$antpipode w.r.t $\odot(PIQ)$, $D,E,F$ as the intouch points on $BC,CA,AB$ respectively and $I_A,I_B,I_C$ as the excenters opposite to $A,B,C$ respectively.
Notice that $\angle PQI=90^{\circ}-\frac{\angle C}{2}$ and similarly $\angle PIQ=90^{\circ}-\frac{\angle A}{2}, \angle QPI=90^{\circ}-\frac{B}{2}$, also $\angle AIQ=90^{\circ}-\frac{\angle B}{2}$ and becuase $\angle BI_AC=90^{\circ}-\frac{\angle A }{2}$,$\angle CI_BA=90^{\circ}-\frac{\angle B }{2}$ and $\angle AI_CB=90^{\circ}-\frac{\angle C }{2}$, so we have $\triangle PQI\sim \triangle I_BI_CI_A$ and $AM_A$ tangent to $\odot(PIQ)$.
We introduce a line parallel to $AO$ which passes through $K$ and we show that $I', \odot(PIQ)\cap \odot(A,AI)$ lie on the same line, as what this does is give us $\angle AII'=90^{\circ}\implies II'\parallel I_BI_C$ and hence giving us $AO$ bisecting $IK$.
Since symmedians induce harmonic bundles $-1=(I,R;P,Q)\stackrel{I'}{=}(I'I\cap PQ,I'R\cap PQ; P,Q)$ where $R=\odot(PIQ)\cap \odot(A,AI)$, by projecting this onto $PQ$ we have $(II',IM;IB,IC)$ which is a harmonic bundle, hence $K-I'-R$ and we are finished with the proof.