a line and a fixed point

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

a line and a fixed point

G H J

Reveal topic tags

geometric transformation

power of a point

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: tuymaada 2006 - problem 3

The post below has been deleted. Click to close.

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

337 posts

maky

#1 h Jul 17, 2006, 12:17 PM • 1 Y

Y by Adventure10

A line $d$ is given in the plane. Let $B\in d$ and $A$ another point, not on $d$ , and such that $AB$ is not perpendicular on $d$ . Let $\omega$ be a variable circle touching $d$ at $B$ and letting $A$ outside, and $X$ and $Y$ the points on $\omega$ such that $AX$ and $AY$ are tangent to the circle. Prove that the line $XY$ passes through a fixed point.

Proposed by F. Bakharev

Z K Y

The post below has been deleted. Click to close.

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

7849 posts

#2 h Jul 17, 2006, 12:45 PM • 2 Y

Y by Adventure10, Mango247

Let $T$ be the midpoint of $XY,\ S=AB\cap XY$ , and $P=d\cap XY$ . If we are to prove that $XY$ passes through a fixed point, then that point lies on the limit of $XY$ as the radius of $\omega$ goes to $0$ , and that line is the perpendicular $\ell$ through $B$ to $AB$ . Proving that $XY\cap\ell$ is constant is equivalent, because $AT\perp XY$ , to proving that $T$ moves on a circle arc passing through $A,B$ (and then the desired fixed point will be the antipode of $A$ in this circle). This means that it suffices to prove that $\angle ATB$ is constant, or, equivalently, that $\angle PTB$ is constant.

$BS$ is the $B$ -symmedian of $BXY$ , so $\angle YBS=\angle TBX$ . This means that $\angle PBS=\angle PBY+\angle YBS$ , which equals $\angle BXT+\angle TBX=\angle PTB$ . This shows that $\angle PTB=\angle PBS=\angle PBA$ , which is, of course, constant.

Z K Y

The post below has been deleted. Click to close.

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

337 posts

maky

#3 h Jul 18, 2006, 10:37 AM • 2 Y

Y by Adventure10, Mango247

during the contest i managed to solve it using coordinates. the computations are very easy...
i took $B$ as origin and $d$ as y-axis, and the equation of $XY$ was a first degree expresion, with one parameter. after i managed to remember the equation of tangent to circle, the problem took me 5 minutes...

Z K Y

The post below has been deleted. Click to close.

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

256 posts

Sailor

#4 h Jul 23, 2006, 11:31 AM • 2 Y

Y by Adventure10, Mango247

Let $Q$ be the intersection of the perpendicular at $B$ to $AB$ with $XY$ .
It's interesting to notice that the midpoint of $AQ$ lies on $d$ .
This follows from the fact that $\triangle{OTB}\sim\triangle{OBA}$ and hence $\angle{AQB}=\angle{BTO}=\angle{OBA=\angle{PBQ}}$ .

Z K Y

The post below has been deleted. Click to close.

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

3 posts

#5 h Jun 14, 2014, 12:04 PM • 1 Y

Y by Adventure10

Let $M$ , $N$ be the midpoints of $AX$ , $AY$ respectively.Analogically $M'$ , $N'$ for $X'Y'$ then $MN$ , $M'N'$ , $d$ intersect at one point ( radical axis for A,w,w') now let the homothety at center A and coefficient 2.then $MN$ goes to $XY$ , $M'N'$ goes to $X'Y'$ and $d$ goes to $d'$
because of $d'$ is constant line and we have that all such $XY$ intersect at the $d'$ then we have that all $XY$ goes through fixed point

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a