Equivalence (*****)

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

Equivalence (*****)

G H J

Reveal topic tags

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: own

The post below has been deleted. Click to close.

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

7054 posts

#1 h Jul 6, 2005, 9:17 PM • 2 Y

Y by Adventure10, Mango247

Let $A, B, C, D, E, F$ be some points which belong to the same circle. Prove that the following relations are equivalently;
$\blacksquare \ 1^{\circ}.\ AB.CD.EF=BC.DE.AF$
$\blacksquare \ 2^{\circ}.\ AC.BD.EF=BC.DF.AE$
$\blacksquare \ 3^{\circ}.\ AB.CE.DF=AC.BF.DE$
$\blacksquare \ 4^{\circ}.\ AD\cap BE\cap CF\ne \emptyset$ $WRONG$ Remark: In the other message I will give the correct enunciation !

This post has been edited 5 times. Last edited by Virgil Nicula, Nov 13, 2005, 6:20 PM

Z K Y

The post below has been deleted. Click to close.

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

1906 posts

#2 h Jul 6, 2005, 9:24 PM • 2 Y

Y by Adventure10, Mango247

i remember this beautiful problem in a article you published in Arhimede Magazine a few years ago. I suppose
it follows from V.N. theorem or am I wrong?

Z K Y

The post below has been deleted. Click to close.

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

7054 posts

#3 h Jul 6, 2005, 9:33 PM • 2 Y

Y by Adventure10, Mango247

Yes. It is the my. Ave, C.L !

Z K Y

The post below has been deleted. Click to close.

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

1906 posts

#4 h Jul 6, 2005, 9:37 PM • 1 Y

Y by Adventure10

It's actually your theorem levi,not just a consequence.Am I right?
I promiss I'll study seriously your article.

Z K Y

The post below has been deleted. Click to close.

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

7054 posts

#5 h Nov 13, 2005, 11:44 AM • 2 Y

Y by Adventure10, Mango247

I wrote better in LaTeX. Maybe now you will understand better this problem. I am using it in many next my proposed problem in this section.

Z K Y

The post below has been deleted. Click to close.

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

99 posts

jpe

#6 h Nov 13, 2005, 1:54 PM • 2 Y

Y by Adventure10, Mango247

Unfortunately, I think that it is not true.
Suppose that the three lines AD, BE, CF concur.
I agree with the fact that AB.CE.DF = AC.BF.DE.
Now, let K be the pole of the line BC wrt the given circle; D',E',F' the second intersections of the lines KD, KE, KF with the given circle. We have (*) AB.CE'.D'F' = AC.BF'.D'E' but the lines AD', BE', CF' generally don't concur.

(*) is due to the fact that the inversion wrt the circle (K,KB) fixes B and C and maps D,E,F respectively to D',E',F'.
Hence (CE'.D'F')/(CE.DF) = KB^3/(KD.KE.KF) = (BF'.D'E')/(BF.DE)

Z K Y

The post below has been deleted. Click to close.

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

99 posts

jpe

#7 h Nov 13, 2005, 4:07 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

Here is an easier way to see that your statement is wrong:
D,E being given on the circle
1) There exists obviously only one point F on the circle for which AD, BE, CF concur
2) There exist two points F for which AB.CE.DF = AC.BF.DE (the common points of the given circle with the circle locus of M such as AB.CE.DM = AC.DE.BM), one of them is the point solution of 1) but, for the other one, the three lines cannot concur

Z K Y

The post below has been deleted. Click to close.

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

7054 posts

#8 h Nov 13, 2005, 6:28 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCDEF$ be a convex cyclic hexagon. Prove that:

$AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\Longleftrightarrow AD\cap BE\cap CF\ne \emptyset$ .

Remark. Thanks, Jpe ! Now you see why and where I made a mistake.

Z K Y

The post below has been deleted. Click to close.

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

232 posts

edriv

#9 h Mar 14, 2007, 4:22 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

I'll prove just the easier implication.

Note that the triangles $APB,EPD$ are similar, then
$\frac{AB}{DE}= \frac{AP}{PE}$ . Continuing, we get
$\frac{CD}{FA}= \frac{CP}{PA}$ and
$\frac{EF}{BC}= \frac{EP}{PC}$ . Now multyply them all together and get:
$\frac{AB \cdot CD \cdot EF}{BC \cdot DE \cdot FA}= 1$ , which is what we wanted.

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a