[bary bash solution to be added oop] Comments - I bary bashed this problem in contest LoL (wasn't too bad! OnLy 15 PaGeS)

Problem was proposed by Anant Mudgal. Here are the two official solutions: Official Solution 1. (by Anant Mudgal) We shall only deal with the cases when $\triangle ABC$ is acute and $AB<AC$. Other cases are analogous. All angles and similarities are directed. Let $\overline{AA_1}$ be a diameter of $\Gamma$. Let $\overline{AD}, \overline{BE},\overline{CF}$ be the altitudes in $\triangle ABC$. Let $\overline{AK}$ meet $\Gamma$ at $M \ne A$. Reflect $M$ in $\overline{BC}$ to obtain $G$. Claim 1. $Q$ lies on $\overline{MD}$. Proof. Observe that $\angle AQK=90^{\circ}$ so $Q$ lies on ray $\overrightarrow{A_1K}$. Thus $$\angle KQM=\angle A_1QM=\angle A_1AM=\angle KAD=\angle KQD$$proving the claim. Claim 2. Let $\overline{YZ}$ meet $\overline{BC}$ at $P$. Then $\frac{PY}{PZ}=\frac{BH}{HC}$. Proof. Note that $$\angle YQZ=\angle YAZ=\angle CAB,$$and $$\angle QYZ=\angle QAZ=\angle QAB=\angle QCB,$$so $\triangle QYZ \sim \triangle QCB$. Consider point $R$ such that $\triangle QBR \sim \triangle QZP$. Since $\angle QZY=\angle QBC,$ we get $R$ lies on $\overline{BC}$. By $\triangle QYZ \sim \triangle QCB$ and $\triangle QZP \sim \triangle QBR$, we get $\frac{PY}{PZ}=\frac{RB}{RC}$. Let $\overline{AH}$ meet $\Gamma$ at $A_2 \ne A$. Observe that $QPBZ$ is cyclic, so $$\angle QRK=\angle QPZ=\angle QPB=\angle QMK$$hence $QRMK$ is cyclic. By power of point, $$\overline{DR} \cdot \overline{DK}=\overline{DQ} \cdot \overline{DM}=\overline{DA} \cdot \overline{DA_2}$$so $ARA_2K$ is cyclic. So $$\angle AA_2R=\angle AKB=\angle ACM=180^{\circ}-\angle AA_2M,$$so $R$ lies on $\overline{MA_2}$. Since $\overline{MA_2}$ and $\overline{HG}$ are reflections in $\overline{BC}$, we conclude $R$ lies on $\overline{HG}$. By the external angle bisector theorem, $\frac{RB}{RC}=\frac{BH}{HC}$, proving the claim. Claim 3.$T$ lies on $\overline{HG}$. Proof. Let $U$ and $V$ be projections of $T$ on $\overline{BE}$ and $\overline{CF}$ respectively. Observe that $\frac{TU}{EY}=\frac{BT}{BY}$ and $\frac{TV}{FZ}=\frac{CT}{CZ}$, so $$\frac{TU}{TV}=\frac{BT}{BY} \cdot \frac{CZ}{CT} \cdot \frac{EY}{FZ}.$$Since $\angle HEY=\angle HFZ=90^{\circ}$ and $AZHY$ is cyclic, we conclude $\triangle HEY \sim \triangle HFZ$. Consequently, we have $\frac{EY}{FZ}=\frac{HE}{HF}=\frac{CH}{BH}$ so $$\frac{TU}{TV}=\frac{BT}{BY} \cdot \frac{CZ}{CT} \cdot \frac{PZ}{PY},$$proving $TU=TV$. The last assertion follows from Menelaus's theorem in $\triangle TYZ$. Thus, we are done! Official Solution 2. (by Pranjal Srivastava) The following solution is a beautiful non harmonic cross ratio chase based solution! Let $\ell$ be the angle bisector of $\angle BHC$. Now, observe that $\ell \parallel AK$. Thus, it suffices to prove that $T$ lies on the exterior angle bisector of $\angle BHC$. Let $A'$ be the antipode of $A$ in $(ABC)$. Clearly, $A', A$ and $Q$ are collinear. Let $O_1$ be the circumcenter $OKH$. Let $R=KZ\cap CH$ and similarly $S=KY\cap BH$ $U=AK\cap (AQH)$, $V=\ell' \cap (AQH)$ where $\ell'$ is the exterior angle bisector of $\angle BHC$. Thus, we want $T\in \ell'$. Claim 1. $\triangle KQC\sim \triangle ZQH$ Proof. \[\angle KQC=\angle A'AC=\angle A'AC=\angle HAB=\angle HAZ=\angle HQZ\] Also, \[\angle QCK=\angle QCB=\angle QAB=\angle QAZ=\angle QHZ\] Claim 2. $Q$ is the miquel point of $ZHCK$ Proof. Follows immediately from Claim 1. Claim 3. $KC=KR=KY$ and $KB=KZ=KS$ Proof. Observe that by Claim 2, $R,S\in (AQH)$. Now, \[\angle KRC=\angle ZRH = \angle ZAH=90^\circ -B=\angle KCH\]Thus, $KC=KR$. Similarly, $KB=KS$. Now, \[\angle BKZ=\angle KCR+\angle KRC=180^{\circ}-2B\]Thus, $KB=KZ$ and similarly, $KY=KC$. Claim 4. $RYUV$ is an isosceles trapezoid. Proof. \[\angle VUR = \angle VHR= \frac{A}{2} =\angle UAY=\angle UVY\] Claim 5. The reflection around $O_1K$ swaps $(R,Y)$, $(V,U)$, and $(S,Z)$. Let this reflection be $\Phi$. Proof. Immediate from previous claims. We now define a few more points for convenience in the upcoming arguments. $L=CH\cap BY$ $F=CH\cap AB$ $N=AR\cap BC$ $P=HV\cap BR$ \[(BT;LY)\stackrel{C}{=}(BZ;FA)\stackrel{R}{=}(BK;CN)\stackrel{A}{=}(ZU;YR)\stackrel{\Phi}{=}(SV;RY)\stackrel{H}{=}(B,HV\cap BY;L,Y)\implies T=HV\cap BY\] Thus, we are done!

Remarks

Very beautiful problem, but I solved this in 20 minutes. I guess I'm very lucky. The key claim of my solution is the following. Claim: If $X=YZ\cap BC$, then $HX$ externally bisects $\angle YHZ$. Proof. Let $D = AH\cap BC$, and let $AK$ meet $\odot(ABC)$ again at $M$. Clearly, $ADQK$ is cyclic, so by Reim's on $\odot(MMAQ)$ and $\odot(DKAQ)$, we get that $M,D,Q$ are collinear. Combining this with the spiral similarity $\triangle QBZ\sim\triangle QYC$, we get $$\frac{BZ}{CY} = \frac{QB}{QC} = \frac{BD}{DC} = \frac{\cot B}{\cot C}.$$Thus, by Menelaus' theorem on $\triangle AYZ$, $$\frac{YX}{XZ}\cdot \frac{ZB}{BA}\cdot\frac{AY}{YC} = 1 \implies \frac{XZ}{XY} = \frac{BZ}{CY}\cdot\frac{AC}{AB} = \frac{\cot B}{\cot C}\frac{\sin B}{\sin C} = \frac{\cos B}{\cos C},$$and the result follows from law of sine in $\triangle HYZ$. $\blacksquare$ We now finish by DDIT. We have an involution swapping $(HB, HZ)$, $(HC, HY)$, and $(HX, HT)$. However, notice that $\measuredangle BHC = \measuredangle YHZ$, so this involution is actually isogonality. Since we know $HX$ externally bisects $\angle YHZ$, it follows that $HT$ externally bisects $\angle BHC$, done.

Doing the other DDIT allows you to generalize the problem to the following: In triangle $ABC$, with orthocenter $H$ and circumcircle $\Gamma$, $K$ is an arbitrary point on $\overline{BC}$. Point $Q$ lies on $\Gamma$ such that $\overline{AQ} \perp \overline{QK}$. Circumcircle of $\triangle AQH$ meets $\overline{AC}$ at $Y$ and $\overline{AB}$ at $Z$. Let $\overline{BY}$ and $\overline{CZ}$ meet at $T$. Prove that $\overline{TH}$ is perpendicular to the isogonal of $\overline{AK}$ in $\angle BAC$. Do DDIT on $\overline{AB},\overline{BT},\overline{CT},\overline{AC}$ from $H$, and then project this involution onto $(BHC)$. The point is that since $\angle YHZ=\angle BHC$, hopefully we can turn the problem into an angle condition. It suffices to show $Y'=\overline{HY}\cap (BHC),Z'=\overline{HZ}\cap (BHC),$ and $K$ are collinear by the involution in $(BHC)$, and because of $Y'Z'=BC$, we want $\overline{Y'Z'}$ to be the reflection of $\overline{BC}$ in the line through $K$ and the center of $(BHC)$, and angle chasing means we want to show $\angle KQH=\angle KOM$, where $M$ is the midpoint of $BC$ and $O$ is the center of $(ABC)$. Let $A'$ be the antipode of $A$ in $(ABC)$, so $Q-K-A'$, then we want to show $\angle HQA'=\angle KOM$. Let $O_1$ be the center of $(HQA')$, which means $\overline{OO_1}\perp\overline{KA'}$ and $\overline{MO_1}\perp\overline{HA'}$. Then we want to show $\angle A'O_1M=\angle KOM$. However $\triangle OMO_1\sim \triangle KMA'$, so by spiral similarity $\angle A'O_1M\sim \angle KOM$, as desired.

Wait, what? (Title is a scam) Let $AK \cap (AQH)=W$, $AH \cap BC=D$ and let $HW \cap BC=X$ and let $YZ \cap BC=X'$, we start by proving $X=X'$. Note that $Q$ is miquel point of $BZYC$ and $DHWK$ so $\angle QXB=\angle QHA=\angle QZA=\angle QYA$ this implies that $(QZB) \cap (QYC)=X$ but then remember that $(QZB) \cap (QYC)=X'$ so $X=X'$. Now to finish note $\angle YHZ=180-\angle BAC=\angle BHC$ so angle bisectors of $\angle ZHB, \angle YHC$ are the same line and now by DDIT we have that $(HY, HC), (HZ, HB), (HT, HX)$ are pairs of Involution but by the previous results we have that as $H,X,W$ are colinear $HX$ is external bisector of $\angle YHZ$ and that this involution is in fact a isogonality so using this we get that $HT$ is external bisector of $\angle BHC$ but as the angle bisectors of $\angle BHC$ and $\angle BAC$ are parallel (easy to see with angle chase ) we have $HT \perp AK$ as desired, thus we are done .

Pranav1056 wrote: In triangle $ABC$, with orthocenter $H$ and circumcircle $\Gamma$, the bisector of angle $BAC$ meets $\overline{BC}$ at $K$. Point $Q$ lies on $\Gamma$ such that $\overline{AQ} \perp \overline{QK}$. Circumcircle of $\triangle AQH$ meets $\overline{AC}$ at $Y$ and $\overline{AB}$ at $Z$. Let $\overline{BY}$ and $\overline{CZ}$ meet at $T$. Prove that $\overline{TH} \perp \overline{KA}$ The problem is just a corollary of the two following statements: $1$. Given triangle $ABC$ and a point $D$ lying on line $BC$. $E,F$ are the points respectively on line $AB$, $AC$ such that $DE=DB$, $DF=DC$. Then $(AEF)$ pass through the orthocenter of $\triangle$$ABC$. $\textbf{Proof}\,(1)$: Denote the midpoint of $BE$ is $M $. The altitude of $\triangle$$ABC$ intersects $BC$ and $(ABC)$ at $K$, $G$. It's quite straightforward that $DM \perp AB$, so there exists a spiral similarity which transform $M\to K$, $B\to G$, so $E$ will be transformed into the reflection of $G$ through $K$, which is $H$. So proof's done. $2$. Given an arbitrary quadrilateral $ABCD$, $P$ is the intersection of $AC$ and $BD$; $M$, $N$ are midpoints of segments $AD$, $BC$ respectively. $Q$ is a point satisfying $\triangle AQB \backsim \triangle DQC$. Then $PQ\parallel MN$. $\textbf{Proof}\,(2)$: Let $AB$ meets $CD$ at $I$. It's obvious that $MN$ pass through the midpoint of $IP$ (Newton-Gauss line), so it suffices to prove that $J$- the midpoint of $IQ$, also lies on $MN$. Denote $\{E\}=\{DQ\cap AB\}$, $\{F\}=\{AQ\cap CD\}$. Due to Newton-Gauss line, $J$, $M$ and the midpoint of $EF$ ($G$) are collinear. But now $\triangle AQE \backsim \triangle DQF$, so $\dfrac{\overline{AE}}{\overline{EB}}=\dfrac{\overline{DF}}{\overline{FC}}$, resulting in $\overline{G,M,N}$ (spiral similarity angle argument). Proof's complete.

Here's a proof using a mix of synthetic observations + coordinates to finish: Let $D$ be the foot of perpendicular from $A$ to $BC$, $U$ be the foot of perpendicular from $K$ to $AC$, $V$ be the foot of perpendicular from $K$ to $AB$. Let $\omega$ be the circle with diameter $AK$, then $\omega$ intersects $\Gamma$ at $A,Q$. Thus, we see from the Forgotten Coaxiality Lemma, that the set of points $P$ on $\omega$ satisfy that $\text{Pow}_\Gamma(P)/\text{Pow}_{(AQH)}(P)$ is a fixed constant. But for $P = D$, we see that $\text{Pow}_\Gamma(D) = \text{Pow}_{(AQH)}(D)$, therefore the ratio must be 1. Hence, $U$ is the midpoint of $CY$ and $V$ is the midpoint of $BZ$. So we may simply ignore $Q$ from now. Now, we use coordinates: $A = (0,0), B = (b, \lambda b), C = (c, -\lambda c)$. A quick calculation reveals $$K = \left(\frac{2bc}{b+c}, 0 \right), \ H = \left(\frac{(1-\lambda^2)(b+c)}2, \frac{(1-\lambda^2)(c-b)}{2 \lambda} \right).$$ Now, let $V = (v, \lambda v)$, then clearly $U = (v, -\lambda v)$ (symmetry across angle bisector). Also $KU$ has slope $\frac 1 \lambda$, so we have $$\frac{-\lambda v}{v - \frac{2bc}{b+c}} = \frac 1 \lambda \implies (1+ \lambda^2)v = \frac{2bc}{b+c} \implies v = \frac{2bc}{(b+c)(1+ \lambda^2)}.$$ So, if $Z = (l, \lambda l)$, then we know that $$l = 2v - b = b \left(\frac{4c}{(b+c)(1+\lambda^2)} - 1 \right) = \frac{(3-\lambda^2)bc - (1+ \lambda^2)b^2}{(1+\lambda^2)(b+c)}.$$By symmetry with $b \leftrightarrow c, \lambda \rightarrow -\lambda $, we get $Y = (m, -\lambda m)$ with $$m = \frac{(3-\lambda^2)bc - (1+ \lambda^2)c^2}{(1+\lambda^2)(b+c)}.$$ Now, we will need the ratio $(m+b)/(b-m)$ for later so let us compute it here: $$\frac mb = \frac{(3-\lambda^2)bc - (1+ \lambda^2)c^2}{(1+\lambda^2)b^2 + (1+\lambda^2)bc} \implies \frac{(m+b)}{(b-m)} = \frac{(1+\lambda^2)(b^2-c^2) + 4bc}{(1+\lambda^2)(b^2+c^2)+ (2\lambda^2 - 2)bc}$$ Now, the main point: We will find the point of $BY$ with $x-$coordinate the same as $x-$coordinate of $H$, and then all we need to observe is that the $y-$coordinate is symmetric with $b \leftrightarrow c, \lambda \rightarrow -\lambda$. If this happens then that point will automatically lie on $CZ$ by this symmetry, finishing the proof. Now, we calculate the $y-$coordinate, it satisfies: $$\frac{y - \lambda b}{\frac 12((1-\lambda^2)c - (1+\lambda^2)b)} = \frac{- \lambda(m+b)}{(m-b)} = \lambda \frac{(1+\lambda^2)(b^2-c^2) + 4bc}{(1+\lambda^2)(b^2+c^2)+ (2\lambda^2 - 2)bc}.$$So, we get $$y = \lambda b + \lambda \frac{((1+\lambda^2)(b^2-c^2) + 4bc)((1-\lambda^2)c - (1+\lambda^2)b))}{2(1+\lambda^2)(b^2+c^2)+ (4\lambda^2 - 4)bc} $$The denominator is symmetric wrt $b \leftrightarrow c, \lambda \rightarrow -\lambda$, so it suffices to see it for the numerator. The numerator is of the form $\lambda(a_0b^3 + a_1b^2c + a_2bc^2 + a_3c^3)$ so it suffices to see that $a_0 \rightarrow -a_3, a_1 \rightarrow -a_2$ under $\lambda \rightarrow -\lambda$. And indeed a fairly simple computation shows that $a_0 = (1- \lambda^4) = -a_3$, and $a_2 = (\lambda^4 + 3) = -a_1$. So we are done.

@Pranav1056

This is truly one of the problems of all time, I present a solution using DDIT and several other properties related to this config.

Solution Key Claim: If $YZ \cap BC = X$, $HX$ externally bisects $YHZ$

Proof

BY DDIT, on $BYZC$, there exists an involution swapping $B,Y$;$C,Z$;$T,X$ however by Claim this is an isogonality, therefore $TH$ externally bisects $BHC$. Done Other Properties Define $E$ and $F$ to be the feet of altitude from $B$ and $C$ respectively, $M,N$ be the feet of perpendicular from $K$ to $AB,AC$.Let $I$ be the miquel point of $FYEZ$. Property 1: $(AFY),(AZE),(AQDKMN)$ concur at $I$

Proof

Let $J = FY \cap EZ$, $L = EF \cap BC$ Property 2: $AI,EF,MN,BC$ concur at $L$

Proof

Property 3:(miquel) $JZFI,JIYE$ is cyclic

Proof

Property 4: $J-T-H$

Proof

Remark: If we find a solution without using orignal problem to Property 4 then we will find another solution.

Quote: We now finish by DDIT. Can anyone please tell what DDIT means?

guruguha9 wrote: Can anyone please tell what DDIT means? It is an abbreviation for "Dual of Desargues' Involution". This article from markbcc168 is about DIT and DDIT

Seungjun_Lee wrote: guruguha9 wrote: Can anyone please tell what DDIT means? It is an abbreviation for "Dual of Desargues' Involution". This article from markbcc168 is about DIT and DDIT Oh ok thank you for the PDFs