KMN and PQR are tangent at a fixed point

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KMN and PQR are tangent at a fixed point

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#1 h Mar 19, 2013, 5:04 PM • 1 Y

Y by Adventure10

Let $ABCD$ be cyclic quadrilateral. Let $AC$ and $BD$ intersect at $R$ , and let $AB$ and $CD$ intersect at $K$ . Let $M$ and $N$ are points on $AB$ and $CD$ such that $\frac{AM}{MB}=\frac{CN}{ND}$ . Let $P$ and $Q$ be the intersections of $MN$ with the diagonals of $ABCD$ . Prove that circumcircles of triangles $KMN$ and $PQR$ are tangent at a fixed point.

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#2 h Mar 19, 2013, 6:58 PM • 1 Y

Y by Adventure10

Hint: intersect circles $ACK$ and $BDK$ at point $S$ and prove that $S$ is the tangency point of $PQR$ and $KMN$

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#3 h Mar 24, 2013, 5:17 PM • 2 Y

Y by Adventure10, Mango247

$\frac{AM}{MB}=\frac{CN}{ND} \Rightarrow \frac{AB}{MB}=\frac{CD}{ND}$ [componendo] $\Rightarrow \frac{AB}{CD}=\frac{MB}{ND}$
Let $S=\odot ABR \cap \odot CDR \neq R$ .
Now $\angle ABS=\angle SRC=\angle SDC, \angle BAS=\angle BRS=\angle SCD$ .
So $\triangle SAB \sim \triangle SCD$ .
Hence $\frac{AB}{CD}=\frac{BS}{SD} \Rightarrow \frac{MB}{ND}=\frac{BS}{SD}$
and since $\angle SBM=\angle SDN$ we conclude $\triangle SBM \sim \triangle SDN$ .
So $\frac{SB}{SM}=\frac{SD}{SN}$ .
And $\angle BSD=\angle BSM+\angle MSD=\angle DSN+\angle MSD=\angle MSN$ .
Hence $\triangle SBD \sim \triangle SMN \Rightarrow \angle SBD=\angle SMN$
$\Rightarrow \angle SBP=\angle SMP \Rightarrow SBMP$ is cyclic.
Also $SPDN$ is cyclic by similar reasoning.
Similarly we can prove $SMAQ,SQNC$ are cyclic also.
Now $\angle SQP=\angle SCN=\angle SAM=\angle SRB=\angle SRP \Rightarrow S \in \odot PQR$ .
Now $\angle MKN+\angle MSN=\angle MKN+\angle MSQ+\angle QSN$
$=\angle AKC+\angle KAC+\angle ACK=180^{\circ} \Rightarrow S \in \odot MNK$ .
Now all we have to prove $S$ is the tangency point of $\odot SPQ,\odot SMN$ .
Apply inversion with centre $S$ and power $r$ . Let $X'$ denote the image of $X$ under the inversion.
Hence $\odot SPQ,\odot SMN$ turn into line $P'Q',M'N'$ . To prove $\odot SPQ,\odot SMN$ are tangent we have to prove $P'Q' \parallel M'N'$ .
Now $P'Q' \parallel M'N' \Leftrightarrow \angle P'N'M'=\angle Q'P'N'$
$\Leftrightarrow SN'P'-\angle SN'M'=\angle SP'Q'-\angle SP'N'$
$\Leftrightarrow SPN-\angle SMN=\angle SQP-\angle SNP$
$\Leftrightarrow \angle PSM=\angle QSN \Leftrightarrow \angle PBM=\angle QCN$
$\Leftrightarrow ABD=\angle ACD$ which is actually true since $ABCD$ is cyclic.

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#4 h Apr 15, 2013, 11:30 PM • 1 Y

Y by Adventure10

Hint: fixed point is Mikel point of four lines $AB,AC,DB,DC.$

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