In triangle $ABC$ the orthocenter is $H$ and the foot of the altitude from $A$ is $D$. Point $P$ satisfies $AP=HP$, and the line $PA$ is tangent to $(ABC)$. Line $PD$ intersects lines $AB, AC$ at points $X,Y$ respectively. Prove that $\angle YHX = \angle BAC$ or $\angle YHX+\angle BAC= 180^\circ$.
Problem
Source: 2023 Israel TST Test 2 P3
Tags: TST, geometry
SerdarBozdag
23.03.2023 21:58
Assume that $PD$ intersects rays $AB, AC$. $D$ is the midpoint of $BB',CC'$ and $HH'$. $M$ is the midpoint of $AH$. $A'$ is the antipode of $A$ in $(ABC)$. $PD$ intersects $(ABC)$ at $U,V$. $O$ is the circumcenter. $\frac{PA}{PM}=\frac{AH'}{AA'}=\frac{DM}{AO} \implies PMD \sim PAO \implies APDO$ is cyclic $\implies D$ is the midpoint of $UV$. Inversion with center $M$ radius $DU=DV$ yields that $U,H,B',V$ are cyclic. From radical axis theorem on $(ABH'C), (UHB'V), (AHB'C)$; $UV, HB', AC$ are concurrent. Note that $\angle DHB'=DHB$ and by symmetry $\angle DHC'=\angle DHC$. Thus $\angle BHC= \angle YHX$.
GuvercinciHoca
24.03.2023 02:06
Denote $K=EF\cap PD, G=EF\cap AD, R=EH\cap PD, Q=FH\cap PD$ and let $M$ be the mid-point of $AH$. We have $(A,H;G,D)-1 \implies DH.DA=DG.DM \implies \frac{DG}{DA}=\frac{DH}{DM}$ which means $HK\parallel PM$ since $PA\parallel EF$. $$-1=(A,H;G,D)=(Y,R;K,D) \text{and} (A,H;G,D)=(X,Q;K,D) \implies \angle RHD= \angle YHD \text{and} \angle XHD= \angle QHD $$Hence, we get $\angle XHB=\angle YHC \implies \angle BHC=\angle YHX$ as desired.
lkjin7932
24.03.2023 04:21
Denote the reflection of $H$ over $BC$ as $E$, and the intersection of $(ABC), (D,DH)$ is $E,F$. $A'$ is the reflection of $A$ over perpendicular bisector of $BC$ and $PH$ meets $AA'$ at $K$. $\angle HFE=\frac{\pi}{2}$, so $A'$ is on $FH$. By easy angle chasing, $HK$ is perpendicular to $A'F$, so $H$ is the orthocenter of $\triangle A'KE$. $A'F \bot DK$ so $K$ lies on $DF$. $A,H,F,K$ lie on a circle which center is $P$. $F$ is the reflection of $H$ over $DP$. $\angle ACF=\angle AEF=\angle ADY=\angle YDF$, so $(D,C,Y,F) cyclic$ $F$ is the miquel point of $ABDYXC$, which kills the problem.
Attachments:
Cookierookie
24.03.2023 11:02
To overcome configuration issues, assume that $X,Y$ lie on the rays $[AB , [AC$. Let the reflection of $B$ wrt. $D$ be $B'$ and define $C'$ similiarly. Delete points $X,Y$ in the problem and define $X=C'H \cap AB$, $Y=B'H \cap AC$. Claim 1: X,D,Y are collinear. Proof: Define $E$, $F$ the other feets of the altitudes. Notice that $$(Y,E;A,C)=(B',B;D,C)=\frac{BC}{B'C} \implies \frac{YA}{YC} = \frac{EA}{EC} \frac{BC}{B'C}$$Similiarly, $$\frac{XB}{XA} = \frac{FB}{FA} \frac{BC'}{BC}$$Multiplying these two toghether and using Menelaus finishes the proof. Now we want to show that $XY$ passes trough $P$, let $PH \cap BC = Q$ and $PA \cap BC = K$, this is equivalent to $$(Q,D;C',B') = (K,D;B,C)=\frac{AB^2}{AC^2} \frac{DC}{DB} \iff \frac{QC'}{QB'}=\frac{AB^2}{AC^2} \frac{DC^2}{DB^2}$$And since $\frac{CH}{BH} = \frac{AB}{AC} \frac{DC}{DB}$, we need to show that $PH$ is tangent to $(B'HC')$. Reflecting among the altitude, it boils down to the tangents to $(BHC)$ at $H$ and $(ABC)$ at $A$ being parallel, which is obvious angle chasing. $\blacksquare$
math_comb01
12.06.2024 16:12
Cute! Let $B',C'$ be the reflections of $B,C$ over $D$, redefine $X,Y$ as $X = \overline{HC' \cap AB}$ and $Y = \overline{HB' \cap AC}$, Claim 1: $X,D,Y$ are collinear.
Follows by DDIT on point $H$ and quadrilateral $BCXY$.
Claim 2: `$P \in \overline{XDY}$
Observe if we reflect $HP$ over $AD$ to meet $BC$ at $Q$ then $HQ \parallel AP$, now redefine $P = \overline{XY \cap AA}$ and then project $(PD;XY)$ onto $BC$ from $A$ and $H$.
swynca
05.11.2024 12:15
Let the intersection of the line $AD$ and $(ABC)$ be $H'$. $\mathcal{H}$ is the homothety which takes $H'$ to $H$ with center $A$. $R$ and $S$ is the image of $B,C$ under $\mathcal{H}$ respectively. Let's define $X' = AR\cap HS$, $Y'=AS \cap HR$ and $Z=AH \cap RS$. Let $P'$ lay on both the line $X'Y'$ and perpendicular bisector of $AH$. Obviously $A,R,S,H$ are cyclic.
$P'$ is on dual of Z wrt $(ARS)$. If we take $O'$ as the center of $(ARS)$ O'P is perpendicular to line $AH$. So the line $AH$ is dual of $P'$. Therefore the line $PA$ is tangent to $(ARS)$ and also to $(ABC)$. Hence $P'=P$.
$\frac{DH}{DA}=\frac{DH'}{DA}=\frac{ZH}{ZA} \implies (D,Z;H,A)=-1 \implies D$ is on dual of $Z$ which is the line $X'Y'$. That means $X'$ is the intersection of $PD$ and $AB$. Thus $X'=X$ and similarly, $Y'=Y$.
Finally,we have $\measuredangle XHY= \measuredangle RHS = \measuredangle RAS= \measuredangle BAC \blacksquare$
Also a different solution sketch by bin_sherlo
We consider $\sqrt{AH.AD}$ inversion $+\sqrt{bc}$ inversion $+$ reflection over angle bisector of $A$. In our new problem, $ABC$ is a triangle with circumcenter $O$ and $F$ is taken on the parallel line to $BC$ through $A$ with $\measuredangle AOF=90$. $D$ is the antipode of $A$, $AO\cap (BOC)=E,H$ lies on $(ABC)$ with $AH=AC$. If we prove that $HD,AC,EF$ are concurrent, then $RS\parallel BC$ and $A,H,R,S$ are concyclic. Let $AC\cap HD=W$. By angle chasing, $E,H,B$ are collinear. Pascal at $DACBHD$ gives $E,W,DD\cap BC$ are collinear. Let $DD\cap BC=I$. If $AD\cap BC=J$, triangles $AOF$ and $JDI$ are homothetic so $F,I,E$ are collinear. This yields the concurrency of $AC,HD,EF$.