IMOC 2020 G5 concyclic wanted, parallelogram and concurrent related

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IMOC 2020 G5 concyclic wanted, parallelogram and concurrent related

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Source: https://artofproblemsolving.com/community/c6h2254883p17398793

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#1 h Sep 1, 2020, 8:31 PM

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Let $O, H$ be the circumcentor and the orthocenter of a scalene triangle $ABC$ . Let $P$ be the reflection of $A$ w.r.t. $OH$ , and $Q$ is a point on $\odot (ABC)$ such that $AQ, OH, BC$ are concurrent. Let $A'$ be a points such that $ABA'C$ is a parallelogram. Show that $A', H, P, Q$ are concylic.

(ltf0501).

This post has been edited 1 time. Last edited by parmenides51, Dec 15, 2022, 7:33 PM

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#2 h Sep 1, 2020, 8:48 PM • 1 Y

Y by Gaussian_cyber

Perform a $-\sqrt{HA\cdot HD}$ Inversion around $H$ . We get the following Equivalent Problem.

$\textbf{INVERTED PROBLEM:-}$ $ABC$ be a triangle with Circumcenter $O$ and Incenter $I$ . Let $\overline{OI}\cap\odot(ABC)=T$ and $\odot(AIT)\cap\odot(ABC)=K$ . Let $A'$ be the reflection of $A$ over $\overline{IO}$ . and $D$ be the point where the Incircle of $\Delta ABC$ touches $\overline{BC}$ . Then $A',D,K$ are collinear.

By Radical Axis Theorem on $\odot(ABC),\odot(BIC),\odot(AIT)$ we get that $\overline{AK},\overline{IO},\overline{BC}$ are concurrent. Now the rest follows from #4 here. $\blacksquare$

This post has been edited 1 time. Last edited by amar_04, Sep 1, 2020, 8:48 PM

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#3 h Sep 2, 2020, 6:05 PM • 1 Y

Y by bin_sherlo

Let $a,b,c$ be complex numbers lying on unit circle such that they are vertices of the triangle. Then
$h=a+b+c, a'=b+c-a,\ p=\frac{h\overline{a}}{\overline{h}}=bc\cdot\frac{a+b+c}{ab+bc+ca}.$ From $R$ definition
$\left(\overline{r}=\frac{b+c-r}{bc}\ \wedge\ \frac{\overline{r}}{\overline{h}}=\frac{r}{h}\right)\iff r=\frac{a(b+c)(a+b+c)}{a^2+2a(b+c)+bc}.$ $Q$ definition implies $\overline{r}=\frac{a+q-r}{aq}\iff q=\frac{a-r}{a\cdot\overline{r}-1}=\frac{(a+b)(a+c)-(b+c)^2}{a^2(b+c)^2-(a+b)(a+c)bc}\cdot abc.$ Moving on...
$h-p=\frac{a(b+c)(a+b+c)}{ab+bc+ca},\ h-a'=2a,$ $q-a'=\frac{(a-b)(a-c)(b+c)(ab+bc+ca)}{a^2(b+c)^2-(a+b)(a+c)bc},$ $q-p=\frac{bc(b+c)(a^2-bc)(a^2+2a(b+c)+bc)}{(ab+bc+ca)(a^2(b+c)^2-(a+b)(a+c)bc)}.$ Therefore $\frac{h-p}{h-a'}\cdot\frac{q-a'}{q-p}=\frac{(a+b+c)(a-b)(a-c)(b+c)(ab+bc+ca)}{2bc(a^2-bc)(a^2+2a(b+c)+bc)}=\overline{\left(\frac{h-p}{h-a'}\cdot\frac{q-a'}{q-p}\right)}\square.$ #1726

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#4 h Sep 4, 2020, 6:18 PM • 1 Y

Y by amar_04

amar_04 wrote:

Perform a $-\sqrt{HA\cdot HD}$ Inversion around $H$ . We get the following Equivalent Problem.

$\textbf{INVERTED PROBLEM:-}$ $ABC$ be a triangle with Circumcenter $O$ and Incenter $I$ . Let $\overline{OI}\cap\odot(ABC)=T$ and $\odot(AIT)\cap\odot(ABC)=K$ . Let $A'$ be the reflection of $A$ over $\overline{IO}$ . and $D$ be the point where the Incircle of $\Delta ABC$ touches $\overline{BC}$ . Then $A',D,K$ are collinear.

By Radical Axis Theorem on $\odot(ABC),\odot(BIC),\odot(AIT)$ we get that $\overline{AK},\overline{IO},\overline{BC}$ are concurrent. Now the rest follows from #4 here. $\blacksquare$

That's how this problem was produced.

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#5 h Sep 5, 2020, 1:40 AM • 2 Y

Y by gnoka, SerdarBozdag

I didn't know how to do the original TST problem, but I somehow manage to solve this one

I think my solution is kind of cute so I'm posting it here.

Note that $B,H,C,A'$ are concyclic. Therefore it suffices to show that $BC,A'H,PQ$ are concurrent (by, say, power of a point theorem). Let $A_1$ be the antipedal point of $A$ w.r.t. $\odot(ABC)$ . Since the intersections $AA_1\cap BC$ and $A'H\cap BC$ are symmetric w.r.t. the midpoint $M$ of $BC$ , by the butterfly theorem (or more precisely, a slightly more generalized statement of it), it suffices to show that $AQ\cap BC$ and $PA_1\cap BC$ are symmetric w.r.t. $M$ . Note that $M$ is also a midpoint of $HA_1$ and both $OH, PA_1$ are perpendicular to $AP$ . Therefore the lines $OH$ and $PA_1$ are symmetric w.r.t. $M$ , which gives the desired result.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#6 h Apr 17, 2023, 1:32 PM

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Let $OH$ meet $BC$ at $T$ . Let $AH$ and $AO$ meet $ABC$ at $S$ and $A''$ . Let $M$ be midpoint of $BC$ .
Note that $BHCA'$ is cyclic so by Radical Axis Theorem we need to prove $HA'$ and $PQ$ meet at $BC$ . Let $PQ$ and $A'H$ meet $BC$ at $X$ and $Y$ . Note that $\frac{BX}{XC} = \frac{BQ}{QC}.\frac{BP}{PC}$ and $\frac{BY}{YC} = \frac{BH}{HC}.\frac{BA'}{A'C}$ so we need to prove $\frac{BQ}{QC}.\frac{BP}{PC} = \frac{BS}{SC}.\frac{AC}{AB}$ or $\frac{BQ}{QC}.\frac{AB}{AC} = \frac{BS}{SC}.\frac{PC}{PB}$ or $\frac{TC}{TB} = \frac{CA''}{BA''}.\frac{PC}{PB}$ . Let $T'$ be reflection of $T$ across $M$ . we need to prove $\frac{T'C}{T'B} = \frac{CA''}{BA''}.\frac{PC}{PB}$ or in fact we need to prove $T',A'',P$ are collinear. Note that $M$ is midpoint of $HA''$ so $THT'A''$ is parallelogram so $TO || T'A''$ and $PA'' \perp AP \perp TO$ so $PA'' || TO$ so $P,A'',T'$ are collinear as wanted.
we're Done.

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