Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$.
Problem
Source: 2019 Junior Balkan MO
Tags: geometry, JBMO, perpendicular bisector, circumcircle, geometry solved, Balkan, Junior Balkan
22.06.2019 14:36
This problem was proposed by Pavle Martinović and Aleksa Milojević from Serbia.
22.06.2019 15:03
Just note that if $R$ is the reflection of the orthocenter across the $M$ and $D$ is the intersection of $HM$ and $AN$, it suffices to prove that $RADB$ is cyclic (since R lies on the circumcircle), i.e. $\angle{ADM}=90$ So it remains to prove that $ADZM$ is cyclic, i.e. $\angle{DAH}=\angle{HMZ}$,which is equivalent to $\angle{ANQ}=\angle{HMB}$ (since $AH$ and $NQ$are parallel). Now notice that the triangles $HBC$ and $AQP$ are similar. Since $M$ and $N$ are the respective midpoints of the sides $PQ$ and $BC$, we have that $HBM$ and $AQN$ are similar. Thereby $\angle{ANQ}=\angle{HMB}$ and the proof is completed.
22.06.2019 16:13
$\triangle HBC \sim \triangle AQP$. There is a spiral similarity with rotation on $90^\circ$ mapping them to each other. So $MH \perp NA$ as corresponding lines in this triangles. If ray $MH$ meet circumcircle at $T$, $\angle MTA = 90^\circ$, so $T, A, N$ are collinear, and this finishes the proof.
22.06.2019 16:28
Let $X=AN \cap BC$ and $Y=HM \cap AN$ and let $D,E,F$ be the feet of the prependiculaar from $A,B,C$ to the respective sides. Then $(P,Q;N,P_{\infty})=-1$ and projecting from A to $BC$ we have $(B,C;X,D)=-1$ which means $X=EF \cap BC$. And hence $Y$ is the miquel point of the complete quadrilateral $(EFBC)$ and so $Y$ lies on the circumcircle of (ABC).
22.06.2019 17:33
Let $T = AN \cap \odot(ABC)$, $H_A = AH \cap \odot(ABC)$, then \[ -1 = (P,Q; N, \infty) \stackrel{A}{=} (B,C; T,H_A)\]On the other hand, let $A'$ be the anti-pole of $A$, then $A',M,H$ are collinear. Let $T' = A'M \cap \odot (ABC)$, then \[ -1 = (B,C; M, \infty) \stackrel{A'}{=} (B,C; T', H_A) \]Therefore $T'$ coincides with $T$.
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23.06.2019 06:07
Very nice! It is well known that reflecting $H$ over $M$ (call it $H'$) lies on $(ABC),$ and $AH \cap (ABC) = A',$ where $BC \parallel A'H'.$ Then $\angle AA'H' = 90^{\circ},$ and it suffices to show $\angle ADH = 90^{\circ}.$ Call foot from $A$ to $BC$ be $A_H.$ It is well known that $\angle HBC = 90^{\circ} - \angle C,$ and $\angle HCB = 90^{\circ} - \angle B,$ but this means $\angle HBC = \angle AQP,$ and $\angle HCB = \angle APQ,$ and $\triangle HBC \sim \triangle AQP.$ Then, since $M$ and $N$ are the midpoints of sides $BC$ and $QP,$ respectively, we also must have $\triangle HMB \sim \triangle ANQ.$ Thus, $\angle HMB = \angle ANQ = \angle DAH,$ because $NM \parallel AH.$ Thus, $\triangle DAH \sim \triangle A_H M H,$ and so $\angle ADH = \angle M A_H H = 90^{\circ},$ as desired.
19.11.2019 16:58
Lukaluce wrote: Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$. Lemma : The antipode of $A$ ,the midpoint of $BC$ and the orthocentre of $\triangle ABC$ are collinear , we can prove this simple lemma in many ways ... But my synthetic proof is the same as the official one , so I 'll provide a solution using Complex Numbers . My solution : Let $AN\cap{HM}=X$ Let's call the $A$-antipode $A'$ . We let $(ABC)$ be the unit circle. Notice that $a'=-a$ , $m=\frac{b+c}{2}$ , $h=a+b+c$ Then $\frac{h-m}{h-a'}=\frac{h-m}{h+a}=\frac12$ ($A'$ is actually the reflection of $H$ over $M$) But $\frac{\bar{h}-\bar{m}}{\bar{h}+\bar{a}}=\frac12$ Thus $X-H-M-A'$ ,and $AA'$ is the diameter of $(ABC)$ Hence $\angle AXA'=90^{\circ}\rightarrow{\boxed{X\in(ABC)}}$. $\text{Q.E.D}$
19.11.2019 22:24
Lukaluce wrote: Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$. Solution:- Let $HM\cap\odot(ABC)=K$ which is actually the $A-\text{Queue Point}$ of $\triangle ABC$ and $AH\cap \odot(ABC)=R$. It is well known that $BKCR$ is a harmonic quadrilateral. Let $AN\cap\odot(ABC)=K'$. So it suffices to prove that $BK'CR$ is also a harmonic quadrilateral. Note that $$-1=(P,Q;N,\infty_{PQ})\overset{A}{=}(B,C;K',R)\implies \text{BK'CR is a Harmonic quadrilateral}$$Hence, $K'\equiv K$. Hence, $AN,HM$ meet on $\odot (ABC)$.
26.11.2019 12:14
it was an easy problem.
26.11.2019 14:10
@bove whatever
17.04.2020 20:54
17.04.2020 23:01
In my opinion, this was one of the easier geometries on #3 (although I did not solve this on the contest)
17.04.2020 23:34
Flash_Sloth wrote: Let $T = AN \cap \odot(ABC)$, $H_A = AH \cap \odot(ABC)$, then \[ -1 = (P,Q; N, \infty) \stackrel{A}{=} (B,C; T,H_A)\]On the other hand, let $A'$ be the anti-pole of $A$, then $A',M,H$ are collinear. Let $T' = DM \cap \odot (ABC)$, then \[ -1 = (B,C; M, \infty) \stackrel{A'}{=} (B,C; T', H_A) \]Therefore $T'$ coincides with $T$. I think you made some typing error. It should be : $T' = A'M \cap \odot (ABC)$
20.05.2020 09:11
Let $D$ be foot of altitude from $A$. In fact the result is also true for arbitrary perpendicular line to $BC$, which intersects $BC$ at point, which lies between $D$ and $C$. Let $E$, $F$ be foots of altitudes from $B$, $C$. Let $EF \cap BC =X_A$ Claim : $AN \cap BC = X_A$ Proof: Note that: $$-1=(P,Q;N, \infty) \overset{A}{=}(B,C;AN \cap BC, D)$$But it is well - known that $(B,C; X_A,D)=-1$. Therefore $AN \cap BC = X_A$ It is well - known that $AN \cap \odot (ABC) = Q_A$ ( A - Queue point ) and that $M, H, Q_A$ are collinear, therefore we are done.
22.05.2020 16:09
Let $E=BH\cap AC, F=CH\cap AB, X=(ABC)\cap (AH)$, also let $XA$ meet $PQ$ at a point $N'$ It is well known that $X$ is the A-Queue point and lies on $MH$ We attempt to prove that the points $X,A,N$ are collinear. Claim 1: $FEXH$ is a harmonic quadrilateral. $proof$: By the well-known 3 tangents lemma, we see that $ME$ and $MF$ are both tangent to $(AEHFX)$ and since $M,H,X$ are collinear it follows that $FEXH$ is harmonic Now, $$-1=(F,E;X,H)\overset{A}{=}(P,Q;N',P_{\infty})$$Hence $N'$ is the midpoint of $PQ$ which implies $N'=N$ and we are done.$\blacksquare$
12.06.2020 14:12
Assume that $H'$ be the reflection of the orthocenter across the $M$ and $F$ be the intersection of $HM$ and $AN$. Then it suffices to prove that $A, F, B, H'$ is cyclic. (Because points $A, B, H'$ lie on the circumcircle of $ABC$) We know that $AH'$ is the diameter of the circumcircle of $ABC$. Since $\angle PAQ=180^\circ-\angle BAC=\angle BH'C ; \angle AQP=\angle MQC=90^\circ-\angle MCQ=\angle MCH'=\angle BCH'$, we have $\triangle AQP \sim \triangle H'CB$. Also, $AN$ and $H'M$ are both medians. Then we have $\angle MH'B=\angle NAP$ and also $\angle NAP=\angle FAB$. Then $\angle FH'B=\angle FAB$ so $A, F, B, H'$ is cyclic.
13.09.2020 22:52
Let $E$ be the reflection of $H$ on side $BC$. $\angle PAQ=\angle BEC$, $\angle APQ=90-\angle B=\angle EBC$. So $\triangle AQP \sim \triangle ECB$ and in particular $\triangle ANQ \sim \triangle EMC$. $\angle ANQ=\angle EMC=\angle BMH$.So, $\angle ADE=\angle NDM=180-\angle ANQ-\angle HMQ=180-\angle BMH-\angle HMQ=90=\angle ABE$. Hence $D$ lies on $(ABC)$.
24.03.2021 00:12
There is a spiral similarity sending $\triangle BMH$ to $\triangle QNA$, so $\angle BMH=\angle QNA$ and $\angle HMC=90-\angle BMH=90-\angle QNA$, thus $HM\perp NA$ and so $AN$ is a tangent to the circumcircle of $ABC$ at which it meets $MA$, as desired. EDIT: Pretty much the same as post 4 oopsies.
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24.06.2021 17:48
We can also prove that $\triangle BMH$ and $\triangle APQ$ are similar by saying that their sides are perpendicular to each other
26.07.2021 07:02
15.12.2021 12:14
Let HM meet Circumcircle of ABC at S and D such A,D are on the same side of BH. we know AS is diameter of it. with some easy angle chasing we can prove triangles BHC and QAP are similar so we need to prove ∠DHB = ∠DAQ. ∠DAQ = 180-∠CBD and ∠DHB = ∠BHM. so we need to prove ∠BDH = ∠MBH. ∠MBH = ∠HAC and ∠BDH = ∠BAS. ∠BDH = ∠BAS = ∠BAH + ∠HAS = 90 - ∠CBA + ∠HAS = 90 - ∠CSA + ∠HAS = ∠SAC + ∠HAS = ∠HAC = ∠CBH = ∠MBH so ∠DHB = ∠DAQ and we have N,A,D are collinear and we're Done.
15.12.2021 21:56
Let $H_a$ be the reflection of $H$ over $BC$. By the Orthocenter Reflection Lemma, we know $H_a$ lies on $(ABC)$. It's well-known that ray $MH$ meets $(ABC)$ again at the $A$-Queue point, which we denote by $Q_a$. Now, Brazil 2011/5 implies $BH_aCQ_a$ is harmonic. Redefine $N$ as the intersection between $AQ_a$ and $PQ$. It's easy to see $$-1 = (B, C; H_a, Q_a) \overset{A}{=} (P, Q; \infty_{PQ}, N)$$which clearly finishes. $\blacksquare$ Remark: Radical Axes and Ceva-Menelaus can be used to show $BH_aCQ_a$ is harmonic.
06.01.2022 04:57
Let $(AH) \cap (ABC)=Q_A$ It is well known that: $Q_A-H-M-A'$ Then: $\angle AQ_AA'=90=\angle AQ_AM$ Now, $\triangle AQP \sim \triangle HBC$ $\angle HMB=\angle ANQ=90-\angle Q_AMN=\angle Q_ANM$ $\angle AQ_AA'=\angle NQ_AM =\angle AQ_AM=90 \implies Q_A ; A ; N$ are collinear.$\blacksquare$
07.04.2023 11:19
Flash_Sloth wrote: Let $T = AN \cap \odot(ABC)$, $H_A = AH \cap \odot(ABC)$, then \[ -1 = (P,Q; N, \infty) \stackrel{A}{=} (B,C; T,H_A)\]On the other hand, let $A'$ be the anti-pole of $A$, then $A',M,H$ are collinear. Let $T' = A'M \cap \odot (ABC)$, then \[ -1 = (B,C; M, \infty) \stackrel{A'}{=} (B,C; T', H_A) \]Therefore $T'$ coincides with $T$. Can someone explain why last one is $H_A$ in projecting A?
11.04.2023 00:30
wow. first time I used projective without knowing beforehand that it is projective Let $AN$ intersect the circumcircle again at $P$, and let the altitude from $A$ intersect the circumcircle of $\triangle ABC$ at $R$. Projecting $(PQ;N\infty)$ from $A$ onto $(ABC)$ gives that $PBRC$ is a harmonic quadrilateral. Thus, $\angle PMB=\angle RMB.$ Note that by Reflecting the Orthocenter, $\angle HMB=\angle RMB$. Hence, $M,H,P$ are collinear, so $HM$ and $AN$ meet at $P$ and we are done.
07.08.2023 01:55
Define $X$ as the intersection of $\odot(A,AM)$ with $\odot (ABC)$, with $X \in \overset{\large\frown}{BA}$ and redefine $N:=(XA) \cap (PQ)$ and $D:=(XA) \cap (BC)$ and let $E$,$F$,$G$ be the foot of the altitudes through $A$,$B$,$C$ Since $DE \dot DM = DX \dot DA = DB\dot DC $, then: $$-1= (BC;DE) \stackrel{A}{=} (QP;N\infty) $$Thus, $N$ is the midpoint of $PQ$. Now since $(BC;DE)$ then $D\in (FG)$, and Brokard on $\odot (BGFC)$ finishes the problem.
20.01.2024 20:36
Do not flood with harmonics and deep configs pls, this is a junior olympiad! We begin with the well known fact that the reflection of $H$ with respect to $M$ lies on the circumcircle of $ABC$ and is diametrically opposite $A$ - hence if $HM$ intersects the arc $\widehat{AB}$ at $K$, then $\angle AKM = 90^{\circ}$. Our aim is to show that $A$, $K$, $M$ are collinear, which we shall do via $\angle KAB + \angle BAC + \angle QAN = 180^{\circ}$. Let $AH \cap BC = A_1$ - then $AKA_1M$ is cyclic (with two right angles), so $\angle KAB = \angle KAH - \angle BAH = \angle BMH + \angle ABC - 90^{\circ}$. On the other hand, $\angle APQ = 90^{\circ} - \angle ABC = \angle BCH$ and $\angle AQP = \angle MQC = 90^{\circ} - \angle ACB = \angle HBC$, thus $\triangle BHC \sim \triangle QAP$. Now the corresponding medians $HM$ and $AN$ yield $\angle BMH = \angle ANQ$. Hence \[ \angle KAB = \angle ANQ + \angle ABC - 90^{\circ}\]and so the desired $\angle KAB + \angle BAC + \angle QAN = 180^{\circ}$ is equivalent to showing that \[ \angle ANQ + \angle QAN = 90^{\circ} + \angle ACB.\]This is true, since both sides are equal to $\angle AQM$ and the proof is completed.
17.04.2024 20:33
Take $(ABC)$ unit center. We will use complex bash. Let $E$ be the altitude from $A$ to $MH$ which is on $(ABC)$. \[e=\frac{a(\overline{(a+b+c)}+\overline{{a}})+\overline{a}(2a+b+c)-a.(\overline{a+b+c})(a+b+c).\overline{a}}{2(\overline{2a+b+c})}=\frac{2abc+b^2c+bc^2}{2bc+ab+ac}\]$P$ is the intersection of $OM$ and $AB$ thus \[p=\frac{(\overline{a}b-a\overline{b})(\frac{b+c}{2})}{(\overline{a}-\overline{b})(\frac{b+c}{2})-(a-b)(\frac{1}{2b}-\frac{1}{2c})}=\frac{(b^2-a^2)(b+c)}{(b-a)(b+c)+(b-a)(a+\frac{ab}{c})}=\frac{(b+a)(b+c)}{a+b+c+\frac{ab}{c}}\]\[q=\frac{(a+c)(b+c)}{a+b+c+\frac{ac}{b}}\]\[n=\frac{1}{2}(\frac{(a+c)(b+c)}{a+b+c+\frac{ac}{b}}+\frac{(a+b)(b+c)}{a+b+c+\frac{ab}{c}})\]\[EA\cap OM=\frac{(\overline{a}e-a\overline{e})(\frac{b+c}{2})}{(\overline{a}-\overline{e})(\frac{b+c}{2})-(a-e)(\frac{1}{2b}+\frac{1}{2c})})=\frac{(e+a)(b+c)}{b+c+\frac{ae}{b}+\frac{ae}{c}}\]\[\frac{1}{2}(\frac{(a+c)(b+c)}{a+b+c+\frac{ac}{b}}+\frac{(a+b)(b+c)}{a+b+c+\frac{ab}{c}})=n\overset{?}{=}EA\cap OM=\frac{(\overline{a}e-a\overline{e})(\frac{b+c}{2})}{(\overline{a}-\overline{e})(\frac{b+c}{2})-(a-e)(\frac{1}{2b}+\frac{1}{2c})})=\frac{(e+a)(b+c)}{b+c+\frac{ae}{b}+\frac{ae}{c}}\]\[\frac{a+c}{a+b+c+\frac{ac}{b}}+\frac{a+b}{a+b+c+\frac{ab}{c}}\overset{?}{=}\frac{2(a+e)}{b+c+\frac{ae}{b}+\frac{ae}{c}}\]\[\frac{b(a+c)}{(b+a)(b+c)}+\frac{c(a+b)}{(c+a)(c+b)}=\frac{ab+ac}{b^2+ab+bc+ca}+\frac{ac+bc}{c^2+ca+ab+bc}\overset{?}{=}\frac{2bc(a+e)}{(bc+ae)(b+c)}\]\[\frac{b(a+c)}{(b+a)(b+c)}+\frac{c(a+b)}{(c+a)(c+b)}=\frac{ab+ac}{b^2+ab+bc+ca}+\frac{ac+bc}{c^2+ca+ab+bc}\overset{?}{=}\frac{2bc(a+e)}{(bc+ae)(b+c)}\]\[\frac{b(a+c)}{a+b}+\frac{c(a+b)}{a+c}\overset{?}{=}\frac{2bc(a+e)}{bc+ae}\]\[2a^3bc+2a^2b^2c+2a^2bc^2+ab^2ce+abc^2e+2b^2c^2e\overset{?}{=}a^2bc^2+a^3ce+2a^2bce+b^3c^2+b^2c^3+a^2b^2c+a^3be+2ab^2c^2\]\[e(ab^2c+abc^2+2b^2c^2-a^3c-a^3b-2a^2bc)\overset{?}{=}2ab^2c^2+b^3c^2+b^2c^3-2a^3bc-a^2b^2c-a^2bc^2\]By replacing $\frac{bc(2a+b+c)}{2bc+ab+ac}$ with $e$, we get \[\frac{2a+b+c}{2bc+ab+ac}.(ab^2c+abc^2+2b^2c^2-a^3c-a^3b-2a^2bc)\overset{?}{=}2abc+b^2c+bc^2-2a^3-a^2b-a^2c\]\[(2a+b+c)(ab^2c+abc^2+2b^2c^2-a^3b-a^3c-2a^2bc)\overset{?}{=}(2bc+ab+ac)(2abc+b^2c+bc^2-2a^3-a^2b-a^2c)\]\[(2a+b+c)(ab^2c+abc^2+2b^2c^2-a^3b-a^3c-2a^2bc)\overset{?}{=}(2bc+ab+ac)(2a+b+c)(bc-a^2)\]\[ab^2c+abc^2+2b^2c^2-a^3b-a^3c-2a^2bc\overset{?}{=}(2bc+ab+ac)(bc-a^2)=2b^2c^2-2a^2bc+ab^2c-a^3b+abc^2-a^3c\]Which is true as desired.$\blacksquare$
17.04.2024 21:13
Let $X$ be the $A$-Queue point of $\triangle ABC$, $H_A = \overline{AH} \cap (ABC)$, and $A'$ be the antipode of $A$ wrt $(ABC)$. Then we find that: $X$, $H$, $M$, and $A'$ are collinear (properties of Queue point config) $\overline{H_AA'} \parallel \overline{BC}$ by a half-homothety at $H$ since $H_A$ and $A'$ are the reflections of $H$ over $\overline{BC}$ and $M$, respectively $AH_A$ is parallel to the perpendicular bisector of $\overline{BC}$ since they're both perpendicular to $\overline{BC}$ Thus $$-1 = (B, C; M, \infty_{BC}) \overset{A'}{=} (B, C; X, H_A) \overset{A}{=} (P, Q; \overline{AX} \cap \overline{PQ}, \infty_{PQ}),$$so $AX$ passes through $N$, which finishes.
14.10.2024 16:59
The condition is equivalent to proving $N-A-Q-X$ are collinear where $Q$ is the $A$ Queue point and $X$ is the $A$ expoint. Let $D$ be the foot of $A$ altitude. We prove that $NA$ passes through $X$ which finishes since $\overline{A-Q-X}$ are known to be collinear. Since $AD \parallel PQ$, we get:- $$-1=(P,Q;N,PQ \cap AD) \stackrel{A}{=} (B,C;AN \cap BC,D)$$Which implies $AN \cap BC=X$ and so we are done.