This problem deserves a bump!

This is one of my best problems. Thanks to Telv Cohl for posting and recommending it to Taiwan TST. I am happy that my problem appears in Taiwan TST. My proof is based on a more general lemma, hopefully the problem will be discussed passionately, I send here the figure. Thanks again Telv.

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This is very nice problem! Here is my solution. Lemma. Given a parallelogram $ABCD$ with an arbitrary point $P$ on $AB$. Let $J$ and $L$ be the incenters of triangles $PAD$ and $PBC$, respectively. Let $H$ and $K$ be orthocenters of triangles $PLB$ and $PJA$, respectively. Construct a point $Q$ such that $KQ \perp PL$ and $HQ \perp PJ$. Then $PQ \perp AB$. Proof of the lemma. By some easy angle chasing, we can prove that $\triangle JKP \stackrel{+}{\sim} \triangle JAD$, then $\frac{PK}{AD} = \frac{JK}{JA}$. Similarly, $\frac{PH}{BC} = \frac{LH}{LB}$. Note that $AD = BC$, we get $$ \frac{PK}{PH} = \frac{PK}{AD} \cdot \frac{BC}{PH} = \frac{JK}{JA} \cdot \frac{LB}{LH}. $$Redefine the point $Q$ such that $KQ \perp PL$ and $PQ \perp AB$. By some easy angle chasing, we see that $AJ \perp BL$, that means $PK \parallel LB$. Combine with $PQ \parallel LH$ and $QK \parallel BH$, it follows that $\triangle PKQ \stackrel{+}{\sim} \triangle LBH$ and $\frac{PK}{PQ} = \frac{LB}{LH}$. Hence, $$ \frac{PK}{PH} = \frac{JK}{JA} \cdot \frac{PK}{PQ} \Longrightarrow \frac{PQ}{PH} = \frac{JK}{JA}. $$Note that $PQ \parallel JK$ and $PH \parallel JA$, then $\triangle PQH \stackrel{+}{\sim} \triangle JKA$. Hence, $HQ \parallel KA$ and it follows that $HQ \perp PJ$. The proof is done. $\qquad \square$ Back to the main problem. Note that $PL \parallel CI$ and $PJ \parallel DI$, it follows from the lemma that $\triangle PHK$ and $\triangle ICD$ are orthologic. Moreover, we see that $OC \perp PK$ and $OD \perp PH$. Therefore, we get $OI \perp HK$. $\qquad \square$

Just been sniped by the exact same solution Here is my solution, not a elegant one. To start, it suffice to prove that $\triangle ICD$ is orthologic to the triangle $\triangle PHK$, in which case the three perpendiculars : $C$ to $PK$, which is $CO$; $D$ to $PH$, which is $DO$ and $I$ to $KH$ are concurrent, implying $IO \perp KH$. Now it suffices to prove that the three perpendiculars $H$ to $ID$, $K$ to $IC$ and $P$ to $BC$ are concurrent. Note that $ID \parallel PJ$, $IC \parallel PL$, we can remove the dependency on $I$ and prove the following statement: let $\ell_P$ be the line from $P$ perpendicular to $AB$, $\ell_H$ be the line from $H$ perpendicular to $JP$, $\ell_K$ be the line from $K$ perpendicular to $PL$. Let $X = \ell_H \cap \ell_P$, then $\triangle XPH \sim \triangle JKA$, hence $XP = PH \cdot JK / AJ$. Similarly if $X' = \ell_K \cap \ell_P$, then $X'P = PK \cdot LH /LB$. Therefore it suffices to prove that \[ \frac{PH \cdot BL}{LH} = \frac{PK \cdot AJ}{JK} \]Now use trigonometric with the fact $H$ is the orthocenter of $\triangle PBL$, we have \[ \frac{PH \cdot BL}{LH \cdot PB} = \frac{ \sin{2 \angle PBH} }{ \sin {2 \angle BHP} } = \frac{ \sin{ \angle PBC} }{ \sin{ \angle BCP} } = \frac{BC}{PB} \]Hence $\frac{PH \cdot BL}{LH} = BC = AD = \frac{PK \cdot AJ}{JK}$. PS. Is it well known that $P,J,I,L$ are concyclic? PS 2. Is there any handout available on orthologic triangles?

Flash_Sloth wrote: Just been sniped by the exact same solution This happens to me all the time. Quote: PS 2. Is there any handout available on orthologic triangles?

Flash_Sloth wrote: PS. Is it well known that $P,J,I,L$ are concyclic? I believe that it is well known and easy to prove by isogonal conjugate, and it only requires $ABCD$ to have incircle. It was once a contest problem, but I forget where and when. @below: Thanks!

@above Bulgaria 2018 Edit: throw some wobbly diagram here to the main problem also:

Here's a simple solution: Note that $PK\perp AJ = CO$, $PH\perp BL = DO$, the original statement is equivalent to showing that the perpendicular lines from $P$, $H$, $K$ to $CD$, $DI$, $IC$ concurrent. Let $X$ be the intersection of the perpendicular lines from $H$, $K$ to $DI$, $IC$, $P'$ the reflection point of $P$ with respect to $O$, $I'$ the incenter of $\triangle P'AB$, then $AK$, $BH$ intersect at the orthocenter $H'$ of $\triangle AI'B$. Since $HH'KX$ is a parallelogram, we only need to prove that the midpoint of the projection points from $H$, $K$ to $AB$ is the midpoint of the projection point from $H'$ to $AB$ and $P$, and these points are exactly the tangent points $Q$, $R$, $S$ of the incircles of $\triangle PBC$, $\triangle PDA$, $\triangle P'AB$ and $AB$. Let $\overline{PA} = a$, $\overline{PB} = b$, $\overline{PC} = c$, $\overline{PD} = d$, we see that (oriented length) $$PQ = \frac{c - a}{2},\quad PR = \frac{b - d}{2}, $$$$PS = PB - SB = b - \frac{(a + b) + d - c}{2} = PQ + PR,$$this completes the proof.

Thank you very much for your contribution. I send here the official proof based on a general lemma.

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THVSH wrote: This is very nice problem! Here is my solution. Lemma. Given a parallelogram $ABCD$ with an arbitrary point $P$ on $AB$. Let $J$ and $L$ be the incenters of triangles $PAD$ and $PBC$, respectively. Let $H$ and $K$ be orthocenters of triangles $PLB$ and $PJA$, respectively. Construct a point $Q$ such that $KQ \perp PL$ and $HQ \perp PJ$. Then $PQ \perp AB$. Proof of the lemma. By some easy angle chasing, we can prove that $\triangle JKP \stackrel{+}{\sim} \triangle JAD$, then $\frac{PK}{AD} = \frac{JK}{JA}$. Similarly, $\frac{PH}{BC} = \frac{LH}{LB}$. Note that $AD = BC$, we get $$ \frac{PK}{PH} = \frac{PK}{AD} \cdot \frac{BC}{PH} = \frac{JK}{JA} \cdot \frac{LB}{LH}. $$Redefine the point $Q$ such that $KQ \perp PL$ and $PQ \perp AB$. By some easy angle chasing, we see that $AJ \perp BL$, that means $PK \parallel LB$. Combine with $PQ \parallel LH$ and $QK \parallel BH$, it follows that $\triangle PKQ \stackrel{+}{\sim} \triangle LBH$ and $\frac{PK}{PQ} = \frac{LB}{LH}$. Hence, $$ \frac{PK}{PH} = \frac{JK}{JA} \cdot \frac{PK}{PQ} \Longrightarrow \frac{PQ}{PH} = \frac{JK}{JA}. $$Note that $PQ \parallel JK$ and $PH \parallel JA$, then $\triangle PQH \stackrel{+}{\sim} \triangle JKA$. Hence, $HQ \parallel KA$ and it follows that $HQ \perp PJ$. The proof is done. $\qquad \square$ Back to the main problem. Note that $PL \parallel CI$ and $PJ \parallel DI$, it follows from the lemma that $\triangle PHK$ and $\triangle ICD$ are orthologic. Moreover, we see that $OC \perp PK$ and $OD \perp PH$. Therefore, we get $OI \perp HK$. $\qquad \square$ Good proof. I propose here a general lemma and the generalization of original problem General lemma. Let $ABCD$ be a parallelogram. $P$ is an arbitrary point such that lines $PD$, $PC$ meet side $AB$ at $M$, $N$, respectively. Let $L$, $J$, and $G$, be incenters of triangles $MAD$, $NBC$, and $PMN$, respectively. Let $H$ and $K$ be orthocenters of triangles $JNB$ an $LMA$, respectively. $Q$ is the point such that $QK\perp GJ$ and $QH\perp GL$. Then $GQ\perp AB$. Generalization of Taiwan TST 2021. Let $ABCD$ be a parallelogram. $P$ is an arbitrary point such that lines $PD$, $PC$ meet side $AB$ at $M$, $N$, respectively. Let $L$, $J$, $I$, and $G$, be incenters of triangles $MAD$, $NBC$, $PCD$, and $PMN$, respectively. Let $H$ and $K$ be orthocenters of triangles $JNB$ an $LMA$, respectively. Let $O$ be a point such that $OD\perp GH$ and $OC\perp GK$. Prove that $OI\perp HK$.

I give here my proof of general lemma and the problem. Thanks again to Telv Cohl for always encouraging me

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Teacher Nguyen Minh Ha contributes a similar problem with rhombus. Problem (Nguyen Minh Ha). Let $ ABCD $ be a rhombus with center $ O. $ $ P $ is a point lying on the side $ AB. $ Let $ J$ and $ L $ be the incenters of triangles $ PAD, $ and $PBC, $ respectively. Let $Q$ be tangent point of incircle triangle $PCD$ and side $CD$. Prove that $OQ\perp JL.$ And I give the general problem of this Problem (Buratinogigle). Let $ABCD$ be a rhombus with center $O$. $P$ is an arbitrary point such that lines $PA$, $PD$ meet side $BC$ at $M$, $N$, respectively. Let $K$, $L$ be incenters of triangles $MAB$, $NCD$, respectively. Let $Q$ be tangent point of incircle triangle $PAD$ and side $AD$. Prove that $OQ\perp KL$. Hope for your interest.

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@above Nice.

buratinogigle wrote: Teacher Nguyen Minh Ha contributes a similar problem with rhombus. And I give the general problem of this Problem (Buratinogigle). Let $ABCD$ be a rhombus with center $O$. $P$ is an arbitrary point such that lines $PA$, $PD$ meet side $BC$ at $M$, $N$, respectively. Let $K$, $L$ be incenters of triangles $MAB$, $NCD$, respectively. Let $Q$ be tangent point of incircle triangle $PAD$ and side $AD$. Prove that $OQ\perp KL$. Hope for your interest. My proof.

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I didn't solve this problem during the test last year. Here's the solution with some trigonometric calculations that I came up with recently. Since $PK\perp CO$, in order to prove $HK\perp OI$, it suffices to show $\angle PKH=\angle COI$. We prove this by showing $\tan \angle PKH=\tan \angle COI$. Notice that $\angle KPH=90^\circ$, so $\tan \angle PKH=\dfrac{PH}{PK}$. Similarly, because $\angle COD=90^\circ$, $\tan COI=\dfrac{\sin \angle COI}{\sin \angle IOD}$. It is easy to see $\triangle PLH \stackrel{+}{\sim} \triangle CLB$ and $\triangle PJK\stackrel{+}{\sim} \triangle DJA$. Hence, \[\dfrac{PH}{PK}=\dfrac{PH}{BC}\cdot \dfrac{DA}{PK}=\dfrac{PL}{CL}\cdot \dfrac{DJ}{PJ}=\dfrac{\sin \angle PCL}{\sin \angle CPL}\cdot \dfrac{\sin \angle DPJ}{\sin \angle PDJ}. \]Also, $\angle PCL=\angle OCI$, $\angle CPL = \angle ICD$, $\angle DPJ=\angle CDI$, and $\angle PDJ=\angle IDO$. By Ceva's theorem, we have \[\tan \angle PKH=\dfrac{\sin \angle PCL}{\sin \angle CPL}\cdot \dfrac{\sin \angle DPJ}{\sin \angle PDJ}=\dfrac{\sin \angle OCI}{\sin \angle ICD}\cdot \dfrac{\sin \angle CDI}{\sin \angle IDO}=\dfrac{\sin \angle COI}{\sin \angle IOD}=\tan COI.\]It completes the proof.

very nice and creative problem! I will post my sol soon.

Let the line through $P$ parallel to $AD$ meet $DC$ at $E$. $U,V$ be the $P$ excenters of $\Delta PDE, \Delta PCE$.We can observe that $U,V$ are just translations of $K,H$ along vector $\overrightarrow {AD}$.So it suffices to show $OI \perp UV$. $DU \cap CV=R$ is the $P$ excenter of $\Delta PDC$. Let $M$ be the midpoint of $UV$ , observe $\angle UEV=90^\circ \implies M$ is circumcenter of $\Delta UEV$. $C',Q,D'$ be the feet of $U,R,V$ respectively on $DC$. (here we are using $S_{\Delta}$ notation for semiperimeter of $\Delta$)Observe that $Q,E$ lies inside segment $C'D'$ and $S_{\Delta PDC}=S_{\Delta PDE}+S_{\Delta PCE}-PE \implies (S_{\Delta PDC}-CP)-(S_{\Delta PCE}-CP)=S_{\Delta PDE}-PE\implies |CQ-CD'|=C'E \implies QD'=C'E$. Thus $QE,C'D'$ share same perpendicular bisector. Perpendicular bisector of $C'D'$ is the midline of trapezoid $C'UVD'$ and hence pass through $M$, hence $ME=MQ \implies Q \in (UEV)$. Let $RQ$ meet $(UEQV)$ at $F$. Note that $F$ is antipode of $E$ and thus $EVFU$ is a rectangle. Now $RF \perp CD$ ,$UF \parallel EV \perp DO$, $VF \parallel EU \perp CO \implies \Delta RUV ,\Delta ODC$ are orthologic with $F$ being one of the orthology centers. Since $CI \perp RV$ and $DI \perp RU$ we obtain that $I$ is the second orthology center and hence $OI \perp UV$.

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