Bulgarian National Mathematical Olympiad 2016, Problem 5

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Bulgarian National Mathematical Olympiad 2016, Problem 5

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

fastlikearabbit

27906 posts

fastlikearabbit

#1 h Jun 22, 2017, 6:08 AM • 2 Y

Y by Adventure10, Mango247

Let $\triangle {ABC}$ be isosceles triangle with $AC=BC$ . The point $D$ lies on the extension of $AC$ beyond $C$ and is that $AC>CD$ . The angular bisector of $\angle BCD$ intersects $BD$ at point $N$ and let $M$ be the midpoint of $BD$ . The tangent at $M$ to the circumcircle of triangle $AMD$ intersects the side $BC$ at point $P$ . Prove that points $A,P,M$ and $N$ lie on a circle.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

361 posts

#2 h Jun 22, 2017, 8:57 AM • 3 Y

Y by EntityXY, themathfreak, Adventure10

Consider $L$ as the reflection of $D$ through the perpendicular bisector of $AB$ ,

$ALDB$ is an isosceles trapezoid and $LNMA$ is cyclic, this is a well known property, it can be shown with typical application of thales parallel theorem.

on the other hand $LMPA$ is trivially cyclic and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

237 posts

#3 h Jul 18, 2017, 4:23 PM • 2 Y

Y by Adventure10, Mango247

fastlikearabbit wrote:

Let $\triangle {ABC}$ be isosceles triangle with $AC=BC$ . The point $D$ lies on the extension of $AC$ beyond $C$ and is that $AC>CD$ . The angular bisector of $\angle BCD$ intersects $BD$ at point $N$ and let $M$ be the midpoint of $BD$ . The tangent at $M$ to the circumcircle of triangle $AMD$ intersects the side $BC$ at point $P$ . Prove that points $A,P,M$ and $N$ lie on a circle.

Bulgaria 2016 problem 5

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

69 posts

#4 h Feb 14, 2018, 11:33 PM • 1 Y

Y by Adventure10

Watch $DP$ as the symmedian and the rest of the solution is trivial.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

791 posts

NOLF

#5 h Jan 14, 2019, 2:04 PM • 1 Y

Y by Adventure10

Mate007 wrote:

Watch $DP$ as the symmedian and the rest of the solution is trivial.

Can anybody post complete solution?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2125 posts

#6 h May 25, 2019, 1:44 PM • 2 Y

Y by Adventure10, Mango247

Obviously $CN||AB$ . Reflect $D$ over the perpendicular bisector of $AB$ to $D'$ $\implies$ $ADD'B$ is isosceles trapezium. Let $AD'$ $\cap$ $BD$ $=$ $X$ $-1=(B ,D ;M ,X) \implies XA \cdot XD'=XB \cdot XD = XM \cdot XN \implies AD'MN \text { is cyclic}$ And, $\angle AMP=\angle ADM$ $=$ $\angle AD'P$ $\implies$ $P$ $\in$ $\odot (AD'M)$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a