Kazakhstan National Olympiad 2015 Final Round, 11 grade, P4

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Kazakhstan National Olympiad 2015 Final Round, 11 grade, P4

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Source: Kazakhstan National Olympiad 2015 Final Round

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izat

#1 h Mar 22, 2015, 4:57 PM • 1 Y

Y by Adventure10

In triangle $ABC$ , point $N$ - base bisector of $C$ , point $M$ - the midpoint of side $AB$ , and $W$ - circumcircle of triangle $ABC$ . Line $CN$ second time intersects $W$ at point $D$ . At segments $AD$ and $BD$ , points $K$ and $L$ , respectively, so that $\angle ACK = \angle BCL$ . Let circumscribed circles of triangles $ACK$ and $BCL$ second time intersect at point $P$ , and $Q$ - point of intersection of $DM$ and $KL$ . Prove that the points $M, N, P, Q$ lie on a circle.

This post has been edited 1 time. Last edited by izat, Mar 22, 2015, 4:58 PM

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#2 h Mar 22, 2015, 5:31 PM • 3 Y

Y by izat, Adventure10, Mango247

My solution:

Let the external bisector of $\angle C$ cut $AB$ at $T$ .

From $(CT, CD; CA, CK)=(CT,CD; CB, CL)$ we get $T, K, L$ are collinear .
From Miquel theorem ( for $\triangle DKL$ and $P \in KL, B \in LD, A \in DK$ ) we get $P \in KL$ .

Since $\angle APK=\angle ACK=\angle LCB=\angle LPB$ ,
so from $(A,B;T,N)=-1$ we get $PN$ is the internal bisector of $\angle BPA \Longrightarrow NP \perp KL$ ,
hence combine with $DM \perp AB$ we get $M, N, P, Q$ lie on a circle with diameter $NQ$ .

Q.E.D

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PROF65

#3 h Mar 24, 2015, 10:08 PM • 2 Y

Y by Adventure10, Mango247

TelvCohl wrote:

My solution:

Let the external bisector of $\angle C$ cut $AB$ at $T$ .

From $(CT, CD; CA, CK)=(CT,CD; CB, CL)$ we get $T, K, L$ are collinear .
From Miquel theorem ( for $\triangle DKL$ and $P \in KL, B \in LD, A \in DK$ ) we get $P \in KL$ .

Since $\angle APK=\angle ACK=\angle LCB=\angle LPB$ ,
so from $(A,B;T,N)=-1$ we get $PN$ is the internal bisector of $\angle BPA \Longrightarrow NP \perp KL$ ,
hence combine with $DM \perp AB$ we get $M, N, P, Q$ lie on a circle with diameter $NQ$ .

Q.E.D

unless I 'm mistaken ,the proof of the collinearity of K,L and T is incomplete.
i 'll send my solution later

This post has been edited 1 time. Last edited by PROF65, Mar 25, 2015, 9:08 PM
Reason: more accurate

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PROF65

#4 h Mar 25, 2015, 1:27 AM • 3 Y

Y by izat, Adventure10, Mango247

$\widehat{CPL}+\widehat{KPC}=\widehat{CBL}+\widehat{KAC}=\pi$ which means $P,K$ and $L$ are colinear. $\frac{DK}{KA}=\frac{\sin \widehat{DCK}}{\sin \widehat{KCA}}.\frac{DC}{CA},\ \frac{DL}{LB}=\frac{\sin \widehat{DCL}}{\sin \widehat{LCB}}.\frac{DC}{CB} \implies \frac{DK}{KA}\ :\ \frac{DL}{LB} =\frac{CB}{CA} ( *).$
since $DBA$ isoceles $DM$ is bisector and angle bisector then $\widehat{KAQ}=\widehat{LBQ}, AQ=QB$
$\frac{\sin \widehat{KAQ}}{KQ} : \frac{\sin \widehat{LBQ}}{LQ} =\frac{\sin \widehat{AKP}}{AQ}:\frac{\sin \widehat{BLP}}{BQ} \implies \frac{\sin \widehat{AKP}}{\sin \widehat{BLP}} =\frac{LQ}{KQ}= \frac{LD}{KD} (1)$ ,( $DQ$ is $D$ -bisector). besides $\frac{\sin \widehat{BLP}}{BP}:\frac{\sin \widehat{AKP}}{AP}=\frac{\sin \widehat{BPL}}{BL}:\frac{\sin \widehat{APK}}{AK}=\frac{\sin \widehat{BCL}}{BL}:\frac{\sin \widehat{ACK}}{AK}=\frac{AK}{BL}$ $\implies \frac{\sin \widehat{AKP}}{\sin \widehat{BLP}}=\frac{BL}{AK}.\frac{AP}{BP}$ then from $(1) :\frac{BL}{AK}.\frac{AP}{BP}= \frac{LD}{KD}$
combined with $(*)$ leads to $\frac{AP}{BP}=\frac{AC}{BC} \implies PN$ is $P$ -bisector but $\widehat{KPA}=\widehat{LPB}$ thus $PN \perp KL$ therefore $QMNP$ is cyclic

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