The Only One Can Beat 9 Points, is 9-Point

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The Only One Can Beat 9 Points, is 9-Point

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Source: 2019 Taiwan TST Round 1

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JustPostTaiwanTST

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JustPostTaiwanTST

#1 h Mar 31, 2020, 2:25 PM

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Given a triangle $\triangle{ABC}$ with orthocenter $H$ . On its circumcenter, choose an arbitrary point $P$ (other than $A,B,C$ ) and let $M$ be the mid-point of $HP$ . Now, we find three points $D,E,F$ on the line $BC, CA, AB$ , respectively, such that $AP \parallel HD, BP \parallel HE, CP \parallel HF$ . Show that $D, E, F, M$ are colinear.

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#2 h Mar 31, 2020, 2:42 PM • 6 Y

Y by AlastorMoody, golema11, amar_04, AmirKhusrau, Cindy.tw, Inconsistent

Let $\mathcal{H}$ be the circumrectangular hyperbola passing through $P$ . Suppose $H_A$ is the orthocenter of $\triangle PBC$ , and define $H_B,H_C$ analogously. Then $H,H_A,H_B,H_C \in \mathcal{H}$ . Also, one can easily see that $APH_AH$ , et al. are parallelograms. In particular, $HH_A \parallel AP$ , and so $D \in HH_A$ (with similar results for $E,F$ ). Then Pascal on $HH_AABCH_C$ gives that line $DF$ passes through $AH_A \cap CH_C$ . But, due to the parallelogram condition, $M$ is the midpoint of segments $AH_A,BH_B$ and $CH_C$ also. Thus, $DF$ passes through $M$ . Similarly, $M \in DE$ , and so $D,E,F,M$ are collinear.

This post has been edited 2 times. Last edited by math_pi_rate, Mar 31, 2020, 3:08 PM
Reason: $D \in HH_A$ not $D \in AH_A$

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#4 h Apr 1, 2020, 2:39 AM • 3 Y

Y by Mango247, Mango247, Mango247

Moving point kills the problem.

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#5 h Apr 3, 2020, 12:00 AM • 1 Y

Y by Mango247

Can be bashed. We get $D=\frac{bc(h+p)}{bc-ap}$ and others analogously. So we need to show collinearity of $D=\frac{ba(h+p)}{ba-cp} , \frac{ac(h+p)}{ac-bp}$ and $\frac{h+p}{2}$ . Divide with $h+p$ and then its trivial.

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Dr_Vex

#6 h Apr 3, 2020, 1:06 AM

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Sugiyem wrote:

Moving point kills the problem.

Solution please!

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#7 h Apr 17, 2020, 6:53 AM • 3 Y

Y by Mango247, Mango247, Mango247

Here is a solution which gives the direction of this line. Let the circumcenter of $\triangle ABC$ be $O$ . We only need to prove that $\angle OMD=90^\circ$ . Denote the midpoint of $BC , AP$ and the antipode of $A$ by $L , N$ and $A'$ respectively. Then it's obvious that $LM \bot AP$ , $ONLM$ is a parallelogram. Let the projection of $O , D , A$ to $LM$ be $U , V , W$ respectively. By $AP//HD$ and $M$ is the midpoint of $HP$ , we can see $WM=VW$ . However, $UW=ON=LM=LW+MV$ , it means that $UL=MV$ , and now it's trivial to see that $L , M$ are both lying on the circle with a diameter $OD$ . So $\angle OMD=90^\circ$ . Q.E.D.

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#8 h Apr 18, 2020, 5:33 AM • 5 Y

Y by AmirKhusrau, Wizard_32, math129, Cindy.tw, k12byda5h

Let $O$ be the circumcenter of $\triangle ABC,$ $V$ be the second intersection of $AH$ with $\odot (O)$ and $J, K$ be the midpoint of $HV, PV,$ respectively, then $JV \stackrel{\parallel}{=} MK$ and $\triangle OVK \stackrel{-}{\sim} \triangle HDJ \stackrel{-}{\sim} \triangle VDJ \Longrightarrow \triangle OVK \stackrel{+}{\sim} \triangle VDJ,$ so $\measuredangle (OV, VD) = \measuredangle (OK,VJ) = \measuredangle (OK,KM) \qquad \text{and}\qquad \frac{OV}{VD} = \frac{OK}{VJ} = \frac{OK}{KM}$ $\Longrightarrow$ $\triangle OVD \stackrel{+}{\sim} \triangle OKM,$ hence we conclude that $\triangle OMD \stackrel{+}{\sim} \triangle OKV$ and $OM \perp MD.$ $\qquad \blacksquare$

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#10 h Jul 2, 2020, 5:07 PM

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WLOG we set $(ABC)$ as a unit circle in the complex plane centered at $O=0$ . $D$ lies on $BC$ iff $d+\overline{d}bc=b+c$ , hence $\overline{d}=\frac{b+c-d}{bc}$ . $AP\parallel HD$ iff $\overline{\big( \frac{d-h}{p-a} \big)}=\frac{d-h}{p-a}$ . Substituting $\overline{d}$ from the previous equation, we obtain $d=(a+b+c+p)\frac{bc}{bc-ap}$ . Similarly $e=(a+b+c+p)\frac{ca}{ca-bp}$ and $f=(a+b+c+p)\frac{ab}{ab-cp}$ . Also, $\overline{d}=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{p})\frac{ap}{ap-bc}$ . $\overline{e}$ and $\overline{f}$ are cyclical. Now we evaluate the determinant
$\begin{vmatrix} d && \overline{d} && 1 \\ e && \overline{e} && 1 \\ f && \overline{f} && 1 \\ \end{vmatrix} = (a+b+c+p)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{p}) \begin{vmatrix} \frac{bc}{bc-ap} && \frac{ap}{ap-bc} && 1\\ \frac{ca}{ca-bp} && \frac{bp}{bp-ca} && 1\\ \frac{ab}{ab-cp} && \frac{cp}{cp-ab} && 1\\ \end{vmatrix} =0$

Also,
$\begin{vmatrix} d && \overline{d} &&1\\ e && \overline{e} &&1\\ m && \overline{m} &&1\\ \end{vmatrix} = (a+b+c+p)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{p}) \begin{vmatrix} \frac{bc}{bc-ap} && \frac{ap}{ap-bc} && 1\\ \frac{ca}{ca-bp} && \frac{bp}{bp-ca} && 1\\ \frac{1}{2} && \frac{1}{2} && 1\\ \end{vmatrix} =0$

Hence $D$ , $E$ , $F$ are collinear and $D$ , $E$ , and $M$ are collinear, which suffices to prove the statement of the problem. $\square$

This post has been edited 2 times. Last edited by snakeaid, Jul 2, 2020, 8:07 PM

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#11 h Aug 12, 2021, 8:01 PM

Y by

dagezjm wrote:

Here is a solution which gives the direction of this line. Let the circumcenter of $\triangle ABC$ be $O$ . We only need to prove that $\angle OMD=90^\circ$ . Denote the midpoint of $BC , AP$ and the antipode of $A$ by $L , N$ and $A'$ respectively. Then it's obvious that $LM \bot AP$ , $ONLM$ is a parallelogram. Let the projection of $O , D , A$ to $LM$ be $U , V , W$ respectively. By $AP//HD$ and $M$ is the midpoint of $HP$ , we can see $WM=VW$ . However, $UW=ON=LM=LW+MV$ , it means that $UL=MV$ , and now it's trivial to see that $L , M$ are both lying on the circle with a diameter $OD$ . So $\angle OMD=90^\circ$ . Q.E.D.

$[asy] size(150); pair A=dir(130),B=dir(-160),C=dir(-20),H=A+B+C,P=dir(-93),M=1/2*(H+P),L=1/2*(B+C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(-120)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); dot(extension(A,P,L,M)); draw(unitcircle,red); draw(A--B--C--A,blue); draw(H--P,brown); draw(L--M^^A--P,green); markscalefactor=0.01; draw(rightanglemark(P,extension(A,P,L,M),L)); markscalefactor=0.008; add(pathticks(P--M,1)); add(pathticks(M--H,1)); [/asy]$
Can someone please tell why is $LM \perp AP$ . Like, this can of course be proven using complex numbers, but how to prove it synthetically ?

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LKira

#12 h Aug 12, 2021, 9:12 PM

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guptaamitu1 wrote:

dagezjm wrote:

Here is a solution which gives the direction of this line. Let the circumcenter of $\triangle ABC$ be $O$ . We only need to prove that $\angle OMD=90^\circ$ . Denote the midpoint of $BC , AP$ and the antipode of $A$ by $L , N$ and $A'$ respectively. Then it's obvious that $LM \bot AP$ , $ONLM$ is a parallelogram. Let the projection of $O , D , A$ to $LM$ be $U , V , W$ respectively. By $AP//HD$ and $M$ is the midpoint of $HP$ , we can see $WM=VW$ . However, $UW=ON=LM=LW+MV$ , it means that $UL=MV$ , and now it's trivial to see that $L , M$ are both lying on the circle with a diameter $OD$ . So $\angle OMD=90^\circ$ . Q.E.D.

$[asy] size(150); pair A=dir(130),B=dir(-160),C=dir(-20),H=A+B+C,P=dir(-93),M=1/2*(H+P),L=1/2*(B+C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(-120)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); dot(extension(A,P,L,M)); draw(unitcircle,red); draw(A--B--C--A,blue); draw(H--P,brown); draw(L--M^^A--P,green); markscalefactor=0.01; draw(rightanglemark(P,extension(A,P,L,M),L)); markscalefactor=0.008; add(pathticks(P--M,1)); add(pathticks(M--H,1)); [/asy]$
Can someone please tell why is $LM \perp AP$ . Like, this can of course be proven using complex numbers, but how to prove it synthetically ?

You use the homothety at $H$ with ratio $\frac{1}{2}$ and use euler circle

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