Nice,but not easy

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

Nice,but not easy

G H J

Reveal topic tags

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: From our Preparation olympiad to Zimo

The post below has been deleted. Click to close.

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

1363 posts

Erken

#1 h Jan 8, 2008, 8:23 AM • 1 Y

Y by Adventure10

Let $ABCD$ be a quadrilateral,such that $AB=AC=BD$ .Let $P\in AC\cap BD$ .
Prove that $OI\perp CD$ ,where $O,I$ are the circumcenter and incenter of $\triangle APB$ ,respectively.
Click to reveal hidden text

Z K Y

The post below has been deleted. Click to close.

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

1182 posts

#2 h Jan 8, 2008, 8:54 AM • 1 Y

Y by Adventure10

note that:

$OI\perp CD\iff DI^2 - CI^2 = DO^2 - CO^2$

let the incircle of $\triangle APB$ with radius $r$ touch the sides $PB,PA$ at $D',C'$ respectively,now according to the power of points $D,C$ wrt circle $(I)$ we have:

$DD'^2 = DI^2 - r^2$ and $CC'^2 = CI^2 - r^2$ so we have:

$DI^2 - CI^2 = DD'^2 - CC'^2$ (*)

now let $AB = p,BP = a,PA = b$ and so $AB = AC = BD = p$ now according to some well-known theorems we have:

$DD' = DB - D'B = p - \left(\frac {a + p - b}2\right) = \frac {p + b - a}2$

$CC' = CA = C'A = p - \left(\frac {b + p - a}2\right) = \frac {p + a - b}2$

$\Rightarrow DD'^2 - CC'^2 = p(b - a)$

so by (*) we get that:

$\boxed{DI^2 - CI^2 = p(b - a)} (I)$

now let $R$ be the radius of circle $(O)$ so we have:

$DP.DB = DO^2 - R^2$ and $CP.CA = CO^2 - R^2$ so we have:

$\begin{eqnarray*}DO^2 - CO^2 = DP.DB - CP.CA = \left(DB - PB\right).DB - \left(CA - PA\right).CA = \left(p - a\right)p - \left(p - b\right)p\end{eqnarray*}$

$\boxed{\Rightarrow DO^2 - CO^2 = p(b - a)} (II)$

now from (I),(II) we get that:

$DI^2 - CI^2 = DO^2 - CO^2$

as wanted...

This post has been edited 2 times. Last edited by BaBaK Ghalebi, Jan 8, 2008, 8:59 AM

Z K Y

The post below has been deleted. Click to close.

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

1363 posts

Erken

#3 h Jan 8, 2008, 8:56 AM • 1 Y

Y by Adventure10

Thank you,my idea is the same,but with trigonometry.
P.S:
You are very fast,did you know this problem before?

Z K Y

The post below has been deleted. Click to close.

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

1182 posts

#4 h Jan 8, 2008, 9:05 AM • 1 Y

Y by Adventure10

yes,it's a famous problem,it was given in several olympiads...
there is also a similar problem as follow:

let $D,E$ be points on sides $AC,BC$ of $\triangle ABC$ respectively,s.t $AB=AD=BE$ let $O,I$ be the incenter and cricumcenter of $\triangle ABC$ respecitively,prove that:

$i)DE\perp OI$

$ii)$ the circumradius of $\triangle CDE$ is equal to $OI$ .

Z K Y

The post below has been deleted. Click to close.

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

1363 posts

Erken

#5 h Jan 8, 2008, 9:15 AM • 2 Y

Y by Adventure10, Mango247

OK,thank you.
Our teacher told us,that there exist solution with using inversion,maybe you know it?
In which olympiads exactly?

Z K Y

The post below has been deleted. Click to close.

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

1182 posts

#6 h Jan 8, 2008, 9:21 AM • 2 Y

Y by Adventure10, Mango247

Belarus 2000 I think...

Z K Y

The post below has been deleted. Click to close.

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

1363 posts

Erken

#7 h Jan 8, 2008, 10:20 AM • 2 Y

Y by Adventure10, Mango247

BaBaK Ghalebi wrote:

Belarus 2000 I think...

You are right,thank you.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a