Polish MO Finals 2015, Problem 1

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Polish MO Finals 2015, Problem 1

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Source: Polish MO Finals 2015

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j___d

#1 h Jul 27, 2016, 10:25 PM • 2 Y

Y by jhu08, Adventure10

In triangle $ABC$ the angle $\angle A$ is the smallest. Points $D, E$ lie on sides $AB, AC$ so that $\angle CBE=\angle DCB=\angle BAC$ . Prove that the midpoints of $AB, AC, BE, CD$ lie on one circle.

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#2 h Jul 27, 2016, 11:27 PM • 2 Y

Y by jhu08, Adventure10

Let $M,N,P,Q,R$ be the midpoints of $AC, AB, BE, CD, BC$ . Clearly $RPN, RQM$ are collinear. Note by similarity that $RP \cdot RN = \frac{1}{4} CE \cdot CA = \frac{1}{4} BC^2$ , since $(BEA)$ is tangent to $BC$ . Similarly we deduce $RQ*RM = \frac{1}{4} BC^2 \implies RP\cdot RN = RQ\cdot RM \implies P,Q,M,N \text{ are concyclic.}$ .

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jayme

#3 h Jul 29, 2016, 7:07 AM • 2 Y

Y by jhu08, Adventure10

Dear Mathlinkers,
a proof without any calculation is possible... I am waiting for your suggestion...

Sincerely
Jean-Louis

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jayme

#4 h Jul 29, 2016, 9:49 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Dear Mathlinkers,
any ideas?

Sincerely
Jean-Louis

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#5 h May 14, 2019, 4:42 PM • 6 Y

Y by 62861, AlastorMoody, jhu08, jrsbr, Adventure10, Mango247

This problem was proposed by me

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#6 h Dec 16, 2021, 6:21 AM • 1 Y

Y by jhu08

Let $M,N,K,L,P$ be midpoints of $AB,AC,BE,CD,BC$ .
$\angle BMK = \angle BAC = \angle BMP$ ---> $M,K,P$ are collinear.
$\angle CNL = \angle BAC = \angle CNP$ ---> $N,L,P$ are collinear.
$\angle BMK = \angle PBK$ ---> $BP$ is tangent to $BMK$ circumcircle ---> $PB^2 = PK.PM$
$\angle CNL = \angle LCP$ ---> $CP$ is tangent to $CNL$ circumcircle ---> $PC^2 = PL.PN$
$PB = PC$ ---> $PK.PM$ = $PL.PN$ ---> $NMKL$ is cyclic.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Feb 17, 2022, 10:07 AM

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#7 h May 13, 2022, 1:09 AM • 1 Y

Y by centslordm

Let $F,K,L,M,N$ be the midpoints of $\overline{BC},\overline{AB},\overline{AC},\overline{BE},\overline{CD},$ respectively and notice $\triangle CFN\sim\triangle LFC$ and $\triangle KBF\sim\triangle BMF.$ Hence, $FN\cdot FL=CF^2=BF^2=FM\cdot FK.$ $\square$

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