PA' and B'C' intersect on A-median

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

PA' and B'C' intersect on A-median

G H J

Reveal topic tags

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: Spain MO 2010

The post below has been deleted. Click to close.

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

1347 posts

WakeUp

#1 h Feb 11, 2011, 6:45 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , let $P$ be a point on the bisector of $\angle BAC$ and let $A',B'$ and $C'$ be points on lines $BC,CA$ and $AB$ respectively such that $PA'$ is perpendicular to $BC,PB'\perp AC$ , and $PC'\perp AB$ . Prove that $PA'$ and $B'C'$ intersect on the median $AM$ , where $M$ is the midpoint of $BC$ .

Z K Y

The post below has been deleted. Click to close.

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

2643 posts

yetti

#2 h Feb 11, 2011, 10:16 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be incenter of $\triangle ABC$ . Solutions at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=147114 are good for arbitrary point $P \in AI$ .

Z K Y

The post below has been deleted. Click to close.

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

8 posts

IrvinG

#3 h Apr 6, 2011, 4:39 AM • 1 Y

Y by Adventure10

Let $PA'$ and $B'C'$ meet at $X$ . Now let $E$ and $F$ be points on $AC$ and $BA$ , respectively, such that $X$ lies on $FE$ and $FE$ is parellel to $BC$ . $PXFC'$ is cyclic because $\angle PXF=\angle PC'F=90^{\circ}$ . Using this fact we get $\angle PFX=\angle PC'X$ . Similarly we obtain $\angle PEX=\angle PB'X$ . Now, the quadrilateral $AC'PB'$ is also cyclic, thus $\angle PC'X=\angle PAB'=\angle PAC'=\angle PB'X$ , so $\angle PFX=\angle PEX$ and this implies that $X$ is midpoint of $EF$ , i.e. $AX$ is median of $\Delta ABC \blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

607 posts

#4 h Jan 4, 2014, 7:18 AM • 2 Y

Y by Adventure10, Mango247

Use this lemma- $ABC$ be a triangle, $I$ be its incenter and $IA',IB',IC'$ are perpendicular to $BC,CA,AB$ ,now let $IA'\cap B'C'=F$ then $A,F,M$ are collinear.( $M$ be the middle point of $BC$ )

Z K Y

The post below has been deleted. Click to close.

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

9774 posts

jayme

#5 h Jan 4, 2014, 9:43 AM • 3 Y

Y by vanu1996, Adventure10, Mango247

Dear Mathlinkers,
this is a generalization
of
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=479138
Sincerely
Jean-Louis

Z K Y

The post below has been deleted. Click to close.

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

20 posts

#6 h Feb 2, 2016, 8:27 PM • 1 Y

Y by Adventure10

Sorry to revive this but I wanna know if this is going somewhere:

Let $PA'$ and $B'C'$ meet at $X$ . Let $E$ and $F$ be points on $AC$ and $BA$ , respectively, such that $X$ lies on $BE$ and $CF$ respectively.
Let $D$ be the intersection of $EF$ and the bisector of $\angle BAC$ .

Now by the ceva´s theorem, if $PA'$ and $B'C'$ intersect on the median $AM$ , where $M$ is the midpoint of $BC$ then:
$\frac {AF}{FB}\cdot \frac {BM}{MC}\cdot \frac {CE}{EF}= 1$
And since
$\frac {BM}{MC}=1$
Then we need $\frac {AF}{FB}=\frac {EF}{EC}$
Now it just remains to prove that $EF$ is parallel to $CB$ which I don´t really know how to prove, I've been trying using cyclic quads and the bisector theorem but nothing works, is there anything wrong, if there isn't how should I continue? Thxs

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a