Collinearity with orthocenter

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Source: Baltic Way 2017 Problem 11

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#1 h Nov 11, 2017, 1:55 PM • 2 Y

Y by Adventure10, Mango247

Let $H$ and $I$ be the orthocenter and incenter, respectively, of an acute-angled triangle $ABC$ . The circumcircle of the triangle $BCI$ intersects the segment $AB$ at the point $P$ different from $B$ . Let $K$ be the projection of $H$ onto $AI$ and $Q$ the reflection of $P$ in $K$ . Show that $B$ , $H$ and $Q$ are collinear.

Proposed by Mads Christensen, Denmark

This post has been edited 4 times. Last edited by math163, Aug 15, 2018, 6:37 PM

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#2 h Nov 11, 2017, 2:27 PM • 2 Y

Y by Adventure10, Mango247

WLOG $\angle B > \angle C$ . It suffices to show that $d(K,BH) = \frac{1}{2} d(P,BH)$ .

Trivially, we have $AP=AC$ by simple angle chase, so $d(P,BH) = BP \cdot \sin \angle ABH = (b-c) \cos A = 2R (\sin B- \sin C) \cos A$ .
Now, doing some angle chasing and calculation, we find $d(K, BH) = HK \cdot \sin \frac{1}{2} A = AH \cdot \sin (90-C-\frac{1}{2}A) \cdot \sin \frac{1}{2}A = 2R \cos A \cdot \sin (90-C-\frac{1}{2}A) \cdot \sin \frac{1}{2}A$ .

Now, plugging stuff in, it suffices to show that $2 \sin (90-C-\frac{1}{2}A) \sin \frac{1}{2}A = \sin B - \sin C$ , which is easy by the sum of trigonometric functions formula.

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#3 h Jul 29, 2018, 4:23 AM • 3 Y

Y by coldheart361, Adventure10, Mango247

My solution :

Let $BB_0, CC_0$ be the altitudes of $ABC$ .
$M$ be the midpoint of $BC$ .
$M_a$ be the midpoint of arc $BC$ not containing $A$ .
$N$ be the midpoint of $BP$ .
$T$ be the midpoint of $AH$ .

We will prove that $NK$ is parallel to $BH$ .

Step 0. $M_a$ is the circumcenter of $BIC$ .

This is well-known.

Step 1. $MK = MM_a$ .

Firstly, $AB_0C_0H$ is cyclic with circumcenter $T$ and diameter $AH$ . It is also clear that $K$ lies on circle with diameter $AH$ , so $A,B_0,C_0,H,K$ are concyclic. Since $AK$ is the angle bisector of $\angle AB_0C_0$ , then $K$ is the mid of arc $B_0C_0$ of circle $AB_0C_0$ , so $KB_0 = KC_0$ . But we have $MB_ 0 = MC_ 0$ and $TB_0 = TC_0$ . So, $M,K,T$ are collinear. Because $AH$ and $MM_a$ are parallel, $\triangle{AHK} \sim \triangle{M_aMK}$ so $MK = MA_a$ .

Step 2. $MN \perp AM_a$

Clearly, $N,M$ are the projection of $M_a$ on $AB,BC$ . Thus $MN$ will be the Simson line of $M_a$ . So it passes the projection of $M_a$ the $AC$ , say $Q$ . But $NQ$ must be perpendicular to $AM_a$ since $M_a$ lies on angle bisector of $A$ . So, step 2 is proved.

Step 3. $NK$ paralel to $BH$ .

From step 1,2 we see that $NI = NM_a$ . So
$\angle KPM_a = 2\angle MPM_a = 2\angle M_aBC = \angle BAC.$ So, $\angle APK = 90^{\circ} - \angle BAC = \angle ABH$ . And step 3 is proved.

Step 4. $B,H,Q$ collinear.

Since $N$ is the mid of $BP$ and $NK$ parallel to $BH$ then the reflection of $P$ w.r.t $K$ lie on $BH$ . And we are done.

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#4 h Apr 9, 2020, 8:57 AM • 1 Y

Y by MeineMeinung

Nice solution Meine, I will give mine
With angle chasing, we can get $AI{\perp}PC$ meaning $\bigtriangleup{PAC}$ is isosceles.
Suppose the intersection of $HB$ and $CP$ is $X$ , the intersection of $AI$ and $CP$ is $D$ and the reflection of $H$ to $K$ is $H'$ . $\angle{HKA}=\angle{PDA}=90^{\circ}$ so we have $HH'{\parallel}PC$
since $PD=DC$ and $HK=KH'$ we have $PH'=CH$ . with some more angle chasing, we get $\angle{HXC}=\angle{HCX}$ resulting $HX=HC=PH'$ , since $XP{\parallel}HH'$ we have $XPH'H$ is a parallelogram. so $HH'=XP=2.HK$ so the intersection of $XH$ and $PK$ , we call $T$ , makes $\bigtriangleup{TPX}$ $\simeq$ $\bigtriangleup{TKH}$ , since $XP=2.HK$ , we have $K$ is the midpoint of $PT$ , also meaning that $T=Q$

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#5 h Nov 7, 2020, 10:48 PM

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For obvious reasons, $\angle{ICB}=\angle{ICA}$ , thus $AP=AC$ , hence $AI\perp PC\implies KH\parallel BC$ .
Also, $PK=KC=KQ\implies \angle PCQ=90^{\circ}$ , thus $KH\perp QC$ . Let $R=KH\cap AC$ .
$KH$ bisects the angle $\angle CKQ$ , since $KH$ is perpendicular bisector of $CQ$ . Also, this implies that $\triangle QHC$ is isosceles. Hence, $R$ lies on $KH$ , thus $QRC$ is isosceles.
Now, $\angle{BAC}=2\angle{RCQ}=\angle{ARQ}\implies QR\parallel AB\perp CH$ . Hence, $R$ is the orthocentre of $\triangle HQC$ , thus $AC\perp HQ$ and because $BH\perp AC$ , we conclude that $B$ , $H$ and $Q$ are collinear.

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#6 h Aug 13, 2021, 9:22 PM

Y by

Other write-up with asy

It is obvious that $AP=AC$ , this is due to trivial angle chase.
Let $R$ be the reflection of $P$ over $A$ . Thus, we have $AP=AR=AC$ , hence $\measuredangle RCP=90^\circ$ . Hence, $K$ is the circumcentre of $\triangle PCQ$ .
As $KH\perp AI\perp PC$ , we get that $KH$ is the perpendicular bisector of $CQ$ . Let $D$ be the intersection of $HQ$ and $AC$ , we claim that $AKHD$ is cyclic, from which the desired collinearity follows. Indeed,
$\measuredangle DHK=\measuredangle KHC=\measuredangle PCH=\measuredangle CPA-90^\circ=\measuredangle ACP-90^\circ=\measuredangle DAK.$ We are done.

$[asy] size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,I,H,K,P,R,Q,D; O=(0,0);A=dir(60);B=dir(200);C=dir(340);I=incenter(A,B,C);H=orthocenter(A,B,C);path w=circle(A,abs(A-C));K=foot(H,A,I);P=intersectionpoints(w,B--2A-B)[1];R=intersectionpoints(w,B--2A-B)[0];Q=extension(P,K,R,C);D=extension(H,Q,A,C); draw(A--B--C--cycle,deep);draw(A--foot(A,P,C),deep);draw(P--C,deep);draw(P--Q,deep);draw(A--R--C,deep);draw(circumcircle(P,Q,C),med);draw(K--C,deep);draw(K--foot(K,C,Q),light);draw(circumcircle(A,K,H),light+dashed);draw(H--Q,med);draw(H--C,med); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$H$",H,dir(H)); dot("$K$",K,dir(K)); dot("$P$",P,dir(P)); dot("$R$",R,dir(R)); dot("$Q$",Q,dir(Q)); dot("$D$",D,dir(D)); [/asy]$

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111 posts

#7 h Apr 7, 2025, 7:07 AM

Y by

This problem is easier than you think.

Let $BH$ and perpendicular line to $CP$ at $C$ intersect at $Q$ . Let $K=PQ \cap AI$ . It's clear that $K$ is the midpoint of $PQ$ . Let $D=PC \cap BQ$ , Then it's suffices to prove that $H$ is the midpoint of $DQ$ , and it's just some easy angle chase.

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