Angle ODQ=90

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

Angle ODQ=90

G H J

Reveal topic tags

geometric transformation

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: Singapore MO 2011 open round 2 Q1

The post below has been deleted. Click to close.

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

1459 posts

#1 h Jul 2, 2011, 6:50 AM • 2 Y

Y by Adventure10, Mango247

In the acute-angled non-isosceles triangle $ABC$ , $O$ is its circumcenter, $H$ is its orthocenter and $AB>AC$ . Let $Q$ be a point on $AC$ such that the extension of $HQ$ meets the extension of $BC$ at the point $P$ . Suppose $BD=DP$ , where $D$ is the foot of the perpendicular from $A$ onto $BC$ . Prove that $\angle ODQ=90^{\circ}$ .

Z K Y

The post below has been deleted. Click to close.

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

148 posts

#2 h Jul 2, 2011, 8:40 AM • 4 Y

Y by Yunis019, Adventure10, Mango247, and 1 other user

Let the reflection of $H$ in $BC$ be $G$ , and $QD \cap BG=R$ . Its known that $G$ lies on $\odot ABC$ . Since $BD=DP$ , $\overline{HQP}$ is parallel to $\overline{BRK}$ and it follows $QD=DR$ . Let $\odot ABC \cap QR =\{X,Y\}$ . The Butterfly Theorem gives $DX=DY$ so $OD \perp XY = QD \Longrightarrow \angle ODQ =90^{\circ}$ .

Z K Y

The post below has been deleted. Click to close.

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

1343 posts

jgnr

#3 h Jul 2, 2011, 2:21 PM • 2 Y

Y by Adventure10, Mango247

P3 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46146

Z K Y

The post below has been deleted. Click to close.

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

1459 posts

#4 h Jul 2, 2011, 3:06 PM • 2 Y

Y by Adventure10, Mango247

Let $Q'$ be reflection of $Q$ about $AD$ . Then note that $O$ and $Q'$ are isogonal conjugates wrt $\triangle ABD$ . It follows that $\angle ODQ=90$ .

Z K Y

The post below has been deleted. Click to close.

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

1347 posts

#5 h Jul 2, 2011, 4:46 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Johan Gunardi wrote:

P3 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46146

and also here.

Z K Y

The post below has been deleted. Click to close.

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

12 posts

#6 h Jun 25, 2012, 1:03 PM • 2 Y

Y by Adventure10, Mango247

It can be solved by using co-ordinates too and avoiding ugly looking equations. Taking $A(0,a), B(b,0), C(c,0)$ finishes it quite easily and cleanly.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a