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Source: Turkey TST 2020 Day 2 P6

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#1 h Mar 10, 2020, 9:15 AM

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In a triangle $\triangle ABC$ , $D$ and $E$ are respectively on $AB$ and $AC$ such that $DE\parallel BC$ . $P$ is the intersection of $BE$ and $CD$ . $M$ is the second intersection of $(APD)$ and $(BCD)$ , $N$ is the second intersection of $(APE)$ and $(BCE)$ . $w$ is the circle passing through $M$ and $N$ and tangent to $BC$ . Prove that the lines tangent to $w$ at $M$ and $N$ intersect on $AP$ .

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#2 h Mar 10, 2020, 10:18 AM • 1 Y

Y by Said

Let $K$ be the midpoint of $BC$ . It is well-known that $AP$ passes $K$ .

Note that $M$ is the Miquel point of the complete quadrilateral $ADBKCP$ , so we have $M= \odot(ABK) \cap \odot(CPK)$ . Similarly we have $N=\odot(ACK) \cap \odot(BPK)$ .

Now we have the following claim:

Claim. The circle $\odot(MNK)$ is tangent to line $BC$ , and furthermore $KA$ is the $K$ -symmedian of triangle $KMN$ .

Note that this claim immediately kills the problem. So it remains to prove the claim.

Proof. Invert at $K$ . Let us denote the images by an apostrophe.

First we have $KB'=KC'$ , and $M'=A'B' \cap C'P', N'=A'C'\cap B'P'$ . Thus $M'N'$ is parallel to $B'C'$ , hence $\odot(KMN)$ is tangent to $BC$ .
Then $KA'$ is the $K$ -median of triangle $KM'N'$ , thus $KA$ is the $K$ -symmedian of triangle $KMN$ .

So we are done. $\blacksquare$

This post has been edited 4 times. Last edited by lminsl, Apr 6, 2020, 10:15 AM

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#3 h Apr 2, 2020, 8:03 AM • 4 Y

Y by Said, electrovector, FAA2533, aritrads

Let $K$ be the midpoint of $BC$ , $L$ be the midpoint of $AP$ .
It's easy to find that $\triangle MAP$ ~ $\triangle MBC$ and $\triangle NAP$ ~ $\triangle NCB$ ,
so $\triangle MLP$ ~ $\triangle MKC$ and $\triangle NLP$ ~ $\triangle NKB$ .
Then we can see that $M,L,N,K$ are concyclic, and $BC$ is tangent to $(MNK)$ at $K$ .
It is left to prove $\frac {ML}{MK}=\frac {NL}{NK}$ , which is quit obvious.

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#4 h May 4, 2020, 11:01 AM

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Mathlikerchina wrote:

Let $K$ be the midpoint of $BC$ , $L$ be the midpoint of $AP$ .
It's easy to find that $\triangle MAP$ ~ $\triangle MBC$ and $\triangle NAP$ ~ $\triangle NCB$ ,
so $\triangle MLP$ ~ $\triangle MKC$ and $\triangle NLP$ ~ $\triangle NKB$ .
Then we can see that $M,L,N,K$ are concyclic, and $BC$ is tangent to $(MNK)$ at $K$ .
It is left to prove $\frac {ML}{MK}=\frac {NL}{NK}$ , which is quit obvious.

Very Nice!!

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#5 h Jun 28, 2021, 5:12 PM

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Complex Numbers

Let $b=-1$ and $c=1$ , so, $p=a \cdot k$ for some real $k$ , because $0$ , $P$ and $A$ are collinear.
It's easy to see that $MPC \sim MAB \implies m=\dfrac{ac-pb}{a+c-p-b}=\dfrac{a(k+1)}{a(1-k)+2}$ , analogously, $n=-\dfrac{a(k+1)}{a(1-k)-2}$ .

$(MNO)$ is tangent to $BC$ at O $\iff \dfrac{1-0}{n-0} \div \dfrac{0-m}{m-n} \in \mathbb{R} \iff \dfrac{m-n}{mn} \in \mathbb{R} \iff \dfrac{\dfrac{a(k+1)}{a(1-k)+2}+\dfrac{a(k+1)}{a(1-k)-2}}{-\dfrac{a(k+1)}{a(1-k)+2} \cdot \dfrac{a(k+1)}{a(1-k)-2}} = \dfrac{2a(1-k)}{-a(k+1)} = 2\dfrac{k-1}{k+1} \in \mathbb{R}$ . OK!

Now, we just have to prove that $\overleftrightarrow{OP}$ is symmedian of $MON \iff \dfrac{\dfrac{m+n}{2}}{m}=\dfrac{n}{p} \iff \dfrac{m+n}{mn} p= \dfrac{4}{a(k+1)}p = \dfrac{4k}{k+1} \in \mathbb{R}$ . OK!

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#6 h Jul 14, 2024, 6:25 PM • 1 Y

Y by Collatz09

Let $K$ be the midpoint of $BC$ .

Claim $1$ : $(MNK)$ is tangent to $BC$ .
Proof $1$ : Invert from $P$ .
$A'PK'B'N'E'$ is a complete quadrilateral and $C'$ is its miquel point. $D'=C'P\cap (A'B'P)$ and $D'A'M'C'K'PA'$ is a complete quadrilateral where $B'$ is its miquel point. $(PB'C')$ and $(PD'E')$ are tangent at $P$ thus, $D'E'\parallel B'C'$ .
$A'N'C' \sim A'B'M'\sim PB'C'\sim PE'D'$ and $A'N'B'\sim A'C'M'$ hence
$\frac{N'B'}{C'M'}=\frac{A'B'}{A'M'}=\frac{PB'}{PC'}$ Also since $PBK\sim PK'B'$ and $PCK\sim PK'C',$ we have
$\frac{B'K'}{C'K'}=\frac{B'K'}{BK}.\frac{CK}{C'K'}=\frac{PB'}{PK}.\frac{PK}{PC'}=\frac{PB'}{PC'}=\frac{B'N'}{C'M'}$ Thus, $N'M'\parallel B'C'$ . This gives the desired result since $K',B',N'$ and $K',C',M'$ are collinear. $\square$

Claim $2$ : $KP$ is $K-$ symedian in $MNK\iff$ tangents to $(MNK)$ at $M,N$ intersect on $AP$ .
Proof $2$ : Let $MN\cap AK=S$ . We will prove that $\frac{SN}{SM}=\frac{KN^2}{KM^2}$ .
$\frac{SN}{\sin NCA}=\frac{SK}{\sin MKC} \ \ \text{and} \ \ \frac{SM}{\sin MBA}=\frac{SK}{\sin NKB}\implies \frac{SN}{SM}=\frac{\sin NKA}{\sin MKA}.\frac{\sin NKB}{\sin MKC}$ Since $NKP\sim NCE$ and $MKP\sim MBD,$ we have
$\frac{NK}{NC}=\frac{KP}{CE} \ \ \text{and} \ \ \frac{MK}{MB}=\frac{KP}{BD}\implies \frac{KN}{KM}=\frac{NC}{BM}.\frac{BD}{CE}=\frac{NC}{BM}.\frac{AB}{AC}$ $(\frac{NC}{BM}.\frac{AB}{AC})^2=\frac{KN^2}{KM^2}\overset{?}{=}\frac{SN}{SM}=\frac{\sin NKA}{\sin MKA}.\frac{\sin NKB}{\sin MKC}$ $\frac{\sin NKB}{\sin MKC}=\frac{\sin NAC}{\sin MAB}=\frac{NC.\frac{\sin ANC}{AC}}{MB.\frac{\sin AMB}{AB}}=\frac{NC}{NB}.\frac{AB}{AC}$ Also $NAP\sim NCB$ and $MAP\sim MBC$ hence
$\frac{\sin PKM}{\sin MKC}=\frac{\sin ABM}{\sin MAB}=\frac{MA}{MB}=\frac{AP}{BC}=\frac{NA}{NC}=\frac{\sin NCA}{\sin NAC}=\frac{\sin NKP}{\sin NKB}\implies \frac{\sin NKB}{\sin MKC}=\frac{\sin NKA}{\sin AKM}$ Thus,
$\frac{sin NKA}{\sin MKA}.\frac{\sin NKB}{\sin MKC}=(\frac{\sin NKA}{\sin MKA})^2=(\frac{NC}{BM}.\frac{AB}{AC})^2$ As desired. $\blacksquare$

Remark
The diagram of the lemma mentioned in Remark:

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