Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced. Prove that $QO$ is perpendicular to $BC$.
Problem
Source: APMO 2000
Tags: ratio, geometry, circumcircle, trigonometry, angle bisector, perpendicular bisector, Hi
yetti
02.04.2006 09:23
Let the line PN meet the (extended) triangle side CA at R. The cross-ratio of 4 points is preserved in a central projection of the line BC to the line PR, hence, $\frac{NB}{NC} \cdot \frac{MC}{MB} = \frac{NP}{NR} \cdot \frac{QR}{QP}$ M is the midpoint of BC, N is the midpoint of PR, hence, $\frac{NB}{NC} = \frac{QR}{QP}$ The angle bisector AN meets the circumcircle of the triangle $\triangle ABC$ at the midpoint K of the arc BC opposite to the vertex A. Since $PR \perp AN, PO \perp AB$, it follows that $\angle OPR = \angle KAB = \angle KAC = \angle KBC$ and the isosceles triangles $\triangle ROP \sim \triangle BKC$ together with the points $Q, N \in RP$ resp. $N, M \in BC$ on their bases are similar. Hence, the angles $\angle NOQ = \angle NKM$ are equal, which means that the lines $OQ \parallel KM$ are parallel and $KM \perp BC$ is the perpendicular bisector of BC.
mr.danh
11.02.2010 02:55
mr.danh wrote:
Let $ L$ be the second intersection of $ AN$ and the circumcircle of triangle $ ABC$. Let $ D$ on $ AB$ and $ E$ on $ AC$ such that $ LD\perp AB$, $ LE\perp AC$. Easy to see that the Simson line $ (DMN)$ is perpendicular to $ AN$. So $ DM\parallel PQ$. Apply Thales's theorem to pairs of parallel lines $ DM\parallel PQ$, $ LD\parallel OP$, we have that $ \frac {AO}{AL} = \frac {AP}{AD} = \frac {AQ}{AM}$.
Hence $ OQ\parallel LM$. And $ OQ\perp BC$ (since $ LM\perp BC$).
Posted
jayme
22.10.2010 08:51
Dear Mathlinkers, let X be the second point of intersection of AD with the circumcircle of ABC. The problem can be solved without calculation by considering the circle passing through M, N and X. Sincerely Jean-Louis
Zhero
18.12.2010 05:15
Let $PN$ hit $AC$ at $R$. Note that there exists a circle $\omega$ tangent to $AP$ and $AR$. Let $Q'$ be the intersection of $PR$ and the line through $O$ perpendicular to $BC$. $Q'$ lies on the polar of $A$, so $A$ lies on the polar $\ell$ of $Q'$. Let $E$ be the intersection of $PR$ and $\ell$, and let $M'$ be the intersection of $AQ$ and $BC$. Because $E$ lies on the polar of $Q'$, $(P, R; Q', E) = -1$. $\ell$ and $BC$ are both perpendicular to $OQ'$, so they are parallel. Hence, $(P, R; Q', E) = A(P, R; Q', E) = A(B, C; M', \infty) = -1$, i.e., $M'$ and $\infty$ are harmonic conjugates, whence $M' = M$ and $Q' = Q$.
Attachments:
vslmat
21.08.2012 22:10
Another solution: Let $AN$ and $AM$ cut the circumcirle of $ABC$ with centre $H$ at $F$ and $G$, respectively. It is obvious that $H, M, F$ are collinear and $HF\perp BC$. Let $HF$ cut the circumcircle again at $D, DG$ cut $BC$ at $K$. Easy to see that $APOE\sim DBFC$, and since $\angle OAQ =\angle FDK$, line $AQ$ corresponds to $AK, OQ$ corresponds to $FK$. Thus $\angle NOQ = \angle MFK$, but $\angle MFK = \angle MGK$ (since $MKGF$ is cyclic)$= \angle AFD$. $\angle NOQ = \angle AFD$, this means that $OQ\parallel FD$, thus $OQ\perp BC$.
Attachments:
NewAlbionAcademy
03.06.2013 02:03
We do not use isothermal coordinates.
Let $PN \cap AC=R$, let the perpendicular from $O$ to $BC$ meet $PR$ at $Q'$, and let $AQ'$ meet $BC$ at $M'$. It is sufficient to prove that $M'=M$.
Clearly $\triangle APO \cong \triangle ARO$. Then by the Ratio Lemma, we have $\dfrac{PQ}{QR}=\dfrac{PO}{OR}\cdot\dfrac{\sin \angle POQ}{\sin \angle ROQ}=\dfrac{\sin \angle ABC}{\sin \angle ACB}=\dfrac{AC}{AB}$. Then, since $AP=AR$, by the Ratio lemma we have $\dfrac{\sin \angle BAQ}{\sin \angle CAQ}=\dfrac{AC}{AB}$. Finally, the Ratio Lemma gives us that $\dfrac{BM'}{M'C}=\dfrac{AB}{AC}\cdot \dfrac{\sin \angle BAQ}{\sin \angle CAQ}=1$, as desired.
QED
djmathman
21.08.2013 08:34
Here's a solution for those that aren't scared of diving into a bit of algebra. It's not an efficient solution at all (NAA's above beats it with a very similar idea), but hey, it works.
Let $R=AC\cap PQ$ and $X=QO\cap BC$. I first propose a lemma.
LEMMA: Suppose $a,b,c,d$ are angles such that $\dfrac{\sin a}{\sin b}=\dfrac{\sin c}{\sin d}$ and $a+b=c+d$. Then $a=c$ and $b=d$.
Proof. Let $k=a+b=c+d$. Then $b=k-a$ and $d=k-c$. Cross-multiplying and expanding gives \begin{align*}\sin a\sin(k-c)&=\sin c\sin(k-a)\\\sin a(\sin k\cos c-\sin c\cos k)&=\sin c(\sin k\cos a-\sin a\cos k)\\\sin a\sin k\cos c-\sin a\sin c\cos k&=\cos a\sin c\sin k-\sin a\sin c\cos k\\\sin a\cos c&=\sin c\cos a\\\sin(a-c)&=0\\a&=c,\end{align*} as desired. $\blacksquare$
Note that it is easy to see that $\triangle PAR$ is isosceles, which implies that $\angle RAN=\angle PAN=\angle NPO\implies AROP$ is cyclic. Therefore $\angle PAO=\angle PRO$ and $\triangle POR$ is also isosceles.
Applying the Ratio Lemma to $\triangle ABC$ gives \[\dfrac{AB\sin\angle BAM}{AC\sin\angle MAC}=\dfrac{BM}{CM}=1\implies \dfrac{AB}{AC}=\dfrac{\sin\angle QAR}{\sin\angle PAR}.\] Next, applying the Ratio Lemma to $\triangle PAR$ gives $\tfrac{PQ}{QR}=\tfrac{AP\sin\angle PAQ}{AR\sin\angle QAR}=\tfrac{\sin\angle PAQ}{\sin\angle QAR}$ and applying it to $\triangle POR$ gives $\tfrac{PQ}{QR}=\tfrac{PO\sin\angle POQ}{OR\sin\angle QOR}=\tfrac{\sin\angle POQ}{\sin\angle QOR}$. Therefore \[\dfrac{\sin\angle POQ}{\sin\angle QOR}=\dfrac{PQ}{QR}=\dfrac{\sin\angle PAR}{\sin\angle QAR}=\dfrac{AC}{AB}.\] In addition, by LoS we have $\tfrac{AB}{\sin\angle ACB}=\tfrac{AC}{\sin\angle ABC}$, so all in all we have $\tfrac{\sin\angle POQ}{\sin\angle QOR}=\tfrac{\sin\angle ABC}{\sin\angle ACB}$. But note that since $ARPO$ is cyclic we have $\angle POQ+\angle QOR=180^\circ-\angle BAC=\angle ABC+\angle ACB$. Therefore by our Lemma, $\angle ABC=\angle QOP$. It follows that $BXOP$ is cyclic, so $\angle BXO=180^\circ-\angle BPO=90^\circ$ and $BC\perp QO$, as desired. $\blacksquare$
djb86
26.06.2015 21:23
This is similar to some of the previous solutions, but uses homothety to simplify one step. Let $D$ be the other intersection point of $AN$ and the circumcircle of $ABC$. Then $DM\perp BC$ by the South Pole Theorem. Let $E$ and $F$ be the points on $AC$ and $BC$ such that $DE\perp AC$ and $DF\perp BC$, respectively. Note that there is a homothety centred at $A$ sending $O$ to $D$, $P$ to $F$ and (since clearly $OR\perp AC$) $R$ to $E$. Since $FME$ is the Simson line of point $D$, it sends the entire segment $PR$ to $FE$, and thus $Q$ to $M$. Now it is clear that $QO\parallel MD$, so $QO\perp BC$.
PRO2000
27.06.2015 06:08
I will be using homothety. Extend PN to meet AC at J.Let $O*$ be the second intersection of AO with circumcircle of $ABC$. Consider the homothety centred at $A$ that maps $O$ to $O*$. Drop $OX\perp AB$ and $OY\perp AC$.$P$ maps to $X$ and $J$ maps to $Y$.So $PJ$ maps to $XY$. So, $AM\cap PJ$ maps to $AM\cap XY$.Invoking Simpson, easy to see that $X,M,Y$ collinear.So, $Q$ maps to $M$. $OQ\perp BC$ iff $O*M\perp BC$ . But this is trivial. Q.E.D. Please check my solution whether it is correct.
PRO2000
27.06.2015 06:10
jayme wrote: Dear Mathlinkers, let X be the second point of intersection of AD with the circumcircle of ABC. The problem can be solved without calculation by considering the circle passing through M, N and X. Sincerely Jean-Louis Can you please elaborate on your method.It seems very nice.
MSTang
01.10.2016 04:33
You can also use Cartesian coordinates with, say, $N=(0,0)$, $A=(-1,0)$, $P=(0,k)$, and $O=(k^2,0)$. Since $AN$ is the angle bisector, it's actually easy to find the equation of $AC$.
palisade
01.08.2019 06:44
Dynamic points: We will vary the point $B$ on $AP$ ($A,P,O,N$ will all remain fixed). Let $X=BN\cap QO$ and $X'=BN\cap (ON)$. We have $B\mapsto C\mapsto M \mapsto Q$ is a projective transformation, and so lines $BN$ and $OQ$ each have degree 1, therefore their intersection $X$ has degree 2. Since $(ON)$ is fixed, $X'$ has degree 1. Now to prove $X=X'$ we only must check 4 points for $B$. These four points will be $AB_{\infty}, P, A, K$ where $K\in AB$ with $KN\parallel AC$. For each of these points the result is trivial, thus $X=X'$ and this implies $XO\bot XQ$.
BOBTHEGR8
16.01.2020 14:44
shobber wrote: Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced. Prove that $QO$ is perpendicular to $BC$. Proof- Let $R$ denote intersection of $PN$ with $AC$. Let $\omega$ denote circumcircle of triangle $APR$. Note by symmetry about $AN$ that $O\in\omega$ and is the $A-antipode$. Let $OQ\cap\omega=S,AQ\cap\omega=T$. Now $H(A,O,P,R)\xRightarrow{\text{Q}}H(T,S,R,P)\xRightarrow{\text{A}}H(AT,AS, AP,AR)\implies H(AS,AM, AB,AC)$ and hence $AS\parallel BC$ and we know $AS\perp OS$ So $OQ\perp BC$ Hence proved.
Stormersyle
11.04.2020 01:01
Extend $AO$ to meet $(ABC)$ at $L$. Let $D$ be the altitude from $L$ to $AB$, $E$ be the altitude from $L$ to $AC$, and $T$ be the altitude from $O$ to $AC$. Note that by Simson line wrt $\triangle{ABC}$, we have $D, M, E$ collinear, because $LM\perp BC$. Also, it is easy to see $APOT$ is cyclic, so by Simson line wrt $\triangle{APT}$, we have $P, N, T$ collinear; hence, $Q\in PT$. Finally, let $h$ be the homothety about $A$ mapping $L$ to $O$. Because $PO||LD, OT||LE$, we have that $h(D)=P$ and $h(E)=T$; hence, $h(M)$ lies on both $AM$ and $PT$, so $h(M)=Q$. Thus, by similar triangles we see $QO||ML$, meaning that $QO\perp BC$, as desired.
dgreenb801
11.04.2020 06:46
Here is my solution: Let the perpendicular from $O$ to $BC$ meet $PR$ at $Q'$. We wish to show $Q'=Q$. Let the parallel through $Q'$ to $BC$ meet $AB$ and $AC$ at $S$ and $T$, respectively. Then since $OQ' \perp TS$, and $\angle OPA=\angle ORA=90$, both $PSQ'O$ and $RTOQ'$ are cyclic. Then $\angle Q'OS= \angle Q'PS = \angle Q'PA = \angle Q'RA = \angle Q'OT$. Thus, since $Q'O \perp ST, Q'S=Q'T$, in which case $Q'$ is on the median from $A$ and $Q'=Q$.
mathlogician
20.10.2020 04:51
Redefine $Q$ as $T$, let $Q = PN \cap AC,$ note that $OQ \perp AC$. now let $X$ be arc midpoint of $BC$, and let $E$ and $F$ be the foot of altitude from $X$ to $AB$ and $AC$. Now by simson line we know that $EMF$ is collinear and $EMF \parallel PQ,$ so there's a homothety now at A sending $P \to E, Q \to F, T \to M, O \to X$ implying the desired result.
AwesomeYRY
18.05.2021 01:45
Oops it looks like I missed the homothety, and kinda rederived homothety? Let $D$ be the arc midpoint of $BC$. Let the foot from $D$ to $AB$ be $X$, and the foot to $AC$ be $Y$. Let $R=XY\cap AN$. Then, by Simson's theorem $XMY$ are collinear, call this line $L_1$. We will attempt to show that $L_1\parallel PNQ$. It suffices to note that \[\angle ARX = 180 - \angle XAR - \angle AXR = 180-\angle DAC-\angle BXM = 180-\angle DBM - \angle BDM = 90\]Thus, $AR\perp XY$, but since we also have $AR\perp PNQ$, then $XY\parallel PNQ$. Thus, we may now create a series of ratios since $PQ\parallel XM$ \[\frac{AQ}{AM}=\frac{AP}{AX}=\frac{AO}{AD}\]Thus, by SAS-ratio similarity, we have $\triangle AQO\sim \triangle AMD$. Thus, $QO\parallel MD$ and since $MD\perp BC$, we clearly have $QO\perp BC$ and we are done.
DottedCaculator
10.12.2021 18:50
[asy][asy]
unitsize(0.8cm);
pair A, B, C, B1, C1, M, N, P, Q, O;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
M=(4,0);
N=(3.5,0);
P=(29/4,3sqrt(5)/8);
Q=(127/32,3sqrt(5)/64);
O=(127/32,-15sqrt(5)/16);
B1=(-31/16,-93sqrt(5)/32);
C1=(221/16,-93sqrt(5)/32);
draw(A--B1--C1--A--(-1/4,-3sqrt(5)/8)--P--A--B--C--P--O--A--M--Q--N--P--N);
draw(circle(O,63sqrt(5)/32));
label("$A$", A, dir(90));
label("$B$", B, W);
label("$C$", C, NE);
label("$B'$", B1, SW);
label("$C'$", C1, SE);
label("$M$", M, S);
label("$N$", N, SSW);
label("$O$", O, S);
label("$P$", P, NE);
label("$Q$", Q, NNE);
[/asy][/asy]
Consider the circle centered at $O$ passing through $P$. Then, this circle is tangent to $AB$ and $AC$. Let $B'C'$ be the line such that $BC\parallel B'C'$ and $B'C'$ is tangent to the circle. Then, since $PN\perp AO$, this means that the reflection of $P$ over $N$ lies on $AB$, $PN$, and the circle. Since $AM$ is a median of $AB'C'$, this means that $PN$, $AM$, and the line through $O$ perpendicular to $B'C'$ are concurrent, so this means that $OQ\perp B'C'$, which means $QO\perp BC$.
guptaamitu1
14.02.2022 19:48
Let $\ell \equiv \overline{PQ}$ and $R = \ell \cap \overline{AC}$. Loot at the isosceles $\triangle APR$ (with $AP = AR$). By symmetry we have $\overline{OR} \perp \overline{AC}$. Let $O' = \overline{AO} \cap \odot(ABC)$ and $\ell'$ be the Simson line of $O'$ wrt $\triangle ABC$ ; $\ell'$ intersect $\overline{AB},\overline{AC}$ at $P',R'$, respectively. As $\overline{O'M} \perp \overline{BC}$ so $M \in \ell$. [asy][asy] size(200); pair A=dir(145),B=dir(-150),C=dir(-30),M=1/2*(B+C),Op=dir(-90),Pp=foot(Op,A,B),Rp=foot(Op,A,C),N=extension(A,Op,B,C),P=extension(A,B,N,N+M-Rp),R=extension(N,P,A,C),Q=extension(N,P,A,M),O=extension(Q,foot(Q,B,C),A,N); draw(circumcircle(A,B,C),red); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(-60)); dot("$O'$",Op,dir(Op)); dot("$P'$",Pp,dir(Pp)); dot("$R'$",Rp,dir(-90)); dot("$N$",N,dir(N)); dot("$R$",R,dir(90)); dot("$P$",P,dir(-170)); dot("$Q$",Q,dir(90)); dot("$O$",O,dir(-140)); draw(Pp--A--C--B^^Op--A--M,red); draw(P--O--R^^Q--O,blue); draw(Pp--Op--Rp^^Op--M,blue); draw(P--R^^Pp--Rp,green); [/asy][/asy] Consider the homothety $\mathbb H$ at $A$ sending $O \to O'$. Note $\mathbb H(P) = P'$ and $\mathbb H (Q) = Q'$ as $\overline{OP} \parallel \overline{O'P'}$ and $\overline{OR} \parallel \overline{O'R'}$. Thus $\mathbb H(\ell) = \ell'$. It follows $\mathbb H(Q) = M$ as $$\mathbb H (Q) = \mathbb H ( \ell \cap \overline{AM}) = \mathbb H(\ell) \cap \mathbb H(\overline{AM}) = \ell' \cap \overline{AM} = M$$Now since $\overline{O'M} \perp \overline{BC}$ and $\overline{OQ} \parallel \overline{O'M}$. Hence $\overline{OQ} \perp \overline{BC}$, as desired. $\blacksquare$
bluedragon17
15.03.2022 17:08
Let the circumcircle of $\triangle{ABC}$ be $\Omega$, let AO meet $\Omega$ at D.
Let E,G be the foot of the perpendiculars from D to AB and AC respectively. Now note that E,M,G are collinear by Simson's Line .Let F be the intersection of PN with AC. Note that since AN is the angle bisector of $\angle{BAC}$, quad APOF is a cyclic kite.
Now notice that triangles POF and EDG are similar and thus we have,
$\frac{PO}{DE} = \frac{PF}{EG} = \frac{OF}{DG}$
Also, since we have $OP \parallel DE$ and $OF \parallel DG$,
we have $\frac{AF}{AG} = \frac{OF}{DG}$
and
$\frac{AP}{AE} = \frac{OP}{DE}$
Therefore,
$\frac{AP}{AE} = \frac{AF}{AG} = \frac{PF}{DG}$
$\Rightarrow PF \parallel DG \Rightarrow PQ \parallel EM $.
Now using Thales theorem on the triangles AEM and AMD we will get $QO \parallel LM$.
Assassino9931
12.12.2022 07:04
Very efficient coordinate bash here: https://www.math.hkust.edu.hk/excalibur/v5_n3.pdf
lifeismathematics
19.03.2023 13:44
one of the finest use of phantom points! consider $PN \cap AC=R$ and perpendicular to $BC$ from $O$ to hit $BC$ at $X$ such that $OX \cap PR=Q'$ and $AQ' \cap BC=M'$ Join $OR$ and $OC$ for later use. now beforehand solving this problem we propose a lemma: Lemma:- In a triangle $ABC$ denote $AX$ to be a cevian, s.t. $\angle{BAX}=\alpha$ and $\angle{XAC}=\beta$ then $\frac{BX}{XC}=\frac{AB}{AC}\cdot \frac{\sin{\alpha}}{\sin{\beta}}$ Proof:- denote $\angle{AXB}=\theta$ then from sine rule we get $\frac{BX}{\sin{\alpha}}=\frac{AB}{\sin{\theta}}$ and $\frac{CX}{\sin{\beta}}=\frac{AC}{\sin{\theta}}$ , dividing these we get $\frac{BX}{CX}=\frac{AB}{AC}\cdot \frac{\sin{\alpha}}{\sin{\beta}}$ $\square$ mark $\angle{BAQ'}=x$ and $\angle{Q'AR}=y$ now we notice that since $AN$ is angle bisector and $AN\perp PR$ we get $\triangle{APN}\cong \triangle{ARN}$ hence we get : $\angle{APN}=\angle{ARN}$ and hence $\angle{OPN}=90-\angle{APN}$ and as $N$ is mid-point of $PR$ we get $\triangle{OPN} \cong \triangle{ORN}$, giving us $\angle{ORN}=90-\angle{APN}$ hence we get $OR \perp AC$ and therefore points $C,R,X,O$ are concyclic which gives $\angle{XOR}=C$ also points $B,P,O,X$ are concyclic which gives us $\angle{XOP}=B$ from Lemma we get $\frac{PQ'}{Q'R}=\frac{AP}{AR}\cdot \frac{\sin{x}}{\sin{y}}$ and since $AP=AR$ we get $\frac{PQ'}{Q'R}=\frac{\sin{x}}{\sin{y}}$ also similarly we have in $\triangle{POR}$ applying Lemma we get $\frac{PQ'}{Q'R}=\frac{OP}{OR}\cdot \frac{\sin{B}}{\sin{C}}$ and since $OP=OR$ we get $\frac{\sin{x}}{\sin{y}}=\frac{\sin{B}}{\sin{C}}$ and now we again apply Lemma in $\triangle{ABC}$ we get$ \frac{AM'}{M'C}=\frac{AB}{AC}\cdot \frac{\sin{x}}{\sin{y}}=\frac{AB}{AC}\frac{\sin{B}}{\sin{C}}=\frac{2R}{2R}=1$ hence we get $AM'=M'C$ and hence $M'$ is midpoint of $BC$ which gives $AM'$ to be the median of $\triangle{ABC}$ and hence the problem statement follows as $M=M'$ and $Q=Q'$ $\blacksquare$
Attachments:
huashiliao2020
30.08.2023 05:19
Solved with a hint to construct DE//BC The circle centered at O through P is tangent to AB,AC, since it lies on the angle bisector. Let DE be the unique line //BC with DE tangent to this circle, with R the foot from O onto AD; since PR perp. NO (look at kite APOR, PR diagonal) along with the well known fact that AM (median in ADE), the line perp. to DE through O are concurrent, this finishes because Q is indeed the concurrence point. $\blacksquare$
asdf334
29.10.2023 19:41
wait what Let $L$ be the midpoint of arc $BAC$ and let $K$ be the midpoint of arc $BC$. Notice that $ANML$ is cyclic so we get $APQO\sim LCNK$. This means \[\measuredangle(OQ,KN)=\measuredangle(PQ,CN)\implies \measuredangle QOK=\measuredangle CNP.\] Subtract $\measuredangle BNA$: \[\measuredangle(OQ,BC)=\measuredangle(OQ,KN)-\measuredangle(KN,BC)=\measuredangle QOK-\measuredangle BNA\]\[\measuredangle QOK-\measuredangle BNA=\measuredangle CNP-\measuredangle BNA=\measuredangle BNP-\measuredangle BNA=90^{\circ}\]done!
asdf334
29.10.2023 20:02
we can also solve as follows: Let $R$ be on $(APQ)$ where $AR\parallel BC$. Let $T=PN\cap AC$. \[-1=(\infty_{BC},M;B,C)\stackrel{A}{=}(R,AQ\cap (APO);P,T)\]so $\frac{RP}{RT}=\frac{QP}{QT}$ and $\overline{RQO}$ bisects $\angle PRT$.
shendrew7
31.12.2023 23:08
Let $AN \cap (ABC) = E$. By Simson, the line connecting the projections from $E$ to $AB$ and $AC$ passes through $M$. Hence the homothety at $A$ sending this line to $PN$ and $E$ to $O$ also sends $M$ to $N$. Thus \[QO \parallel ME \perp BC. \quad \blacksquare\]
EquationTracker
06.01.2024 13:06
By $(PQR)$, we denote the circumcircle of $PQR$.
Simson line theorem : Click here to learn more.
Let $D$ be the circumcenter of $\Delta ABC$. Let $E$ be the concurrency point of $DM$, $AN$ and $(ABC)$. $F$ and $G$ are the perpendicular foot from $E$ to $AB$ and $AC$ respectively. $H$ is the intersection of $FG$ and $AE$. Let $\frac{\angle A}{2}=\alpha$.
[asy][asy]
import graph; size(8cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -6.141728024042078, xmax = 3.174575507137483, ymin = -5.81445529676934, ymax = 3.2614274981217113; /* image dimensions */
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draw((-1.480249684031839,-1.4582986225585077)--(-2.468833572833071,-1.892791333831124), linewidth(0.8));
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draw((-1.480249684031839,-1.4582986225585077)--(-1.494317431054321,-4.110068936296458), linewidth(0.8));
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draw((-1.494317431054321,-4.110068936296458)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8));
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draw((-1.1261681919167021,-4.947704176298286)--(2.66,-1.92), linewidth(0.8));
draw((-5.398473910768285,-3.784828063474779)--(-5.211598643768283,-3.0982645825399944), linewidth(0.8));
draw((-5.398473910768285,-3.784828063474779)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8)); /* special point */
draw((0.9060966758985711,-1.0139049292406515)--(-1.1261681919167021,-4.947704176298286), linewidth(0.8));
draw((0.9060966758985711,-1.0139049292406515)--(-5.398473910768285,-3.784828063474779), linewidth(0.8)); /* special point */
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label("$A$", (-4.443756574004514,1.578482344102179), NE * labelscalefactor);
label("$B$", (-5.345334335086407,-1.8925920360631068), NE * labelscalefactor);
label("$C$", (2.7087603305785044,-1.9677235161532645), NE * labelscalefactor);
label("$M$", (-1.0778662659654463,-2.238196844477832), NE * labelscalefactor);
label("$N$", (-3.1162584522915163,-1.7122764838467284), NE * labelscalefactor);
dot((-5.211598643768283,-3.0982645825399944),linewidth(3pt) + dotstyle);
label("$P$", (-5.630833959429006,-3.0946957175056298), NE * labelscalefactor);
label("$Q$", (-2.0944703230653709,-1.5418031555221607), NE * labelscalefactor);
label("$O$", (-1.9193388429752132,-4.4719308790383105), NE * labelscalefactor);
dot((-1.1057141289867563,-1.0921133140036046),linewidth(3pt) + dotstyle);
label("$D$", (-1.363365890308046,-0.9008564988730251), NE * labelscalefactor);
label("$E$", (-1.228129226145762,-5.318587528174298), NE * labelscalefactor);
dot((-5.398473910768285,-3.784828063474779),linewidth(3pt) + dotstyle);
label("$F$", (-5.8562283996994795,-3.966220886551459), NE * labelscalefactor);
label("$G$", (0.8304733283245604,-0.8858302028549936), NE * labelscalefactor);
label("$H$", (-2.595522163786633,-2.9242223891810623), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy][/asy]
By Simson line theorem, we know that $F,M,G$ are collinear. As, $\angle FAE=\angle EAG$, $\triangle AFE\cong\triangle AEG\implies AF=AG$. Thus, $AE\perp FG\implies PQ\parallel FM$. Also, $OP\parallel FE$. Therefore, $\triangle OPQ$ and $\triangle EFM$ are homothetic with center $A$, which implies $QO\parallel ME$. But $ME\perp BC$. So, $QO\perp BC$ as desired. And we are done.