Serbia JBMO TST 2021 problem 4

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

Serbia JBMO TST 2021 problem 4

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Serbia JBMO TST 2021

The post below has been deleted. Click to close.

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

53 posts

#1 h Jan 3, 2022, 7:36 PM • 1 Y

Y by MilleniumX01

On sides $AB$ and $AC$ of an acute triangle $\Delta ABC$ , with orthocenter $H$ and circumcenter $O$ , are given points $P$ and $Q$ respectively such that $APHQ$ is a parallelogram. Prove the following equality:
$\begin{align*} \frac{PB\cdot PQ}{QA\cdot QO}=2 \end{align*}$

Z K Y

The post below has been deleted. Click to close.

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

53 posts

#2 h Jan 4, 2022, 8:02 PM • 6 Y

Y by Jalil_Huseynov, StarLex1, ilovepizza3000, sttsmet, egxa, MilleniumX01

Let $(ABC)$ be $\Omega$ and $r$ be radius of $\Omega$
Let $AD, BE$ and $CF$ be altitudes of $\Delta ABC$

We will first prove that $OP=OQ$
$\angle BHP=90=\angle CHQ$ ( $PH\parallel AC$ and $QH\parallel AB$ )
$\angle PBH=\angle FBE=\angle FCE=\angle HCQ$ because $BFEC$ is cyclic
$\implies \Delta BHP\sim \Delta CHQ$
$\frac{BP}{HP}=\frac{CQ}{HQ}$
$\frac{BP}{AQ}=\frac{CQ}{AP}$
$BP\cdot PA=CQ\cdot QA$ so powers of $P$ and $Q$ with respect to $\Omega$ are equal
$OP^2-r^2=pow_\Omega(P)=pow_\Omega(Q)=OQ^2-r^2$
$\implies OP=OQ$

We will prove that $OP$ and $OQ$ are tangents to $(APQ)$
Case 1: $AB=AC$
$A, H, O$ are collinear
$\angle BOD=\frac{\angle BOC}{2}=\angle BAC=\angle BPH \implies BPHO$ cyclic
$\angle OPQ=180-\angle APQ-\angle BPO=180-\angle ABC-\angle BHO=180-2\cdot (90-\frac{\angle BAC}{2})=\angle BAC$
$\implies OP$ is tangent to $(APQ)$ , but by symmetry same follows for $OQ$

Case 2: $AB\neq AC$
It is known that $AH$ and $AO$ are isogonal
$AH$ is the $A$ -median of $\Delta APQ$ so $AO$ is it's $A$ -symmedian
Thus $AO$ passes through intersection of tangents to $(APQ)$ at $P$ and $Q$ , say at point $O'$
We see that $O'P=O'Q$ so $O'=O$ or rather $OP$ and $OQ$ are tangents to $(APQ)$

Let $X$ be midpoint of $PB$
$\angle OPQ=\angle BAC=\angle XPH$
$\angle PHB=90 \implies XP=XH$
Combining that with $OP=OQ$ we get $\Delta XPH\sim \Delta OPQ$
So $\frac{PX}{PH}=\frac{PO}{PQ}$
$\frac{PX\cdot PQ}{PH\cdot PO}=1$
$\frac{PX\cdot PQ}{QA\cdot QO}=1$
$\frac{PB\cdot PQ}{QA\cdot QO}=2$

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

9775 posts

jayme

#3 h Jul 31, 2022, 7:59 AM

Y by

Dear Mathlinkers,

here Problem 4

Sincerely
Jean-Louis

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a