All about
flooble
|
fun stuff
|
Get a free chatterbox
|
Free JavaScript
|
Avatars
perplexus
dot
info
Home
>
Shapes
>
Geometry
Four Equal Triangles (
Posted on 2013-06-16
)
Here's a problem I found in the Problems section of a
math journal. The way I read it, it is not true.
Let ABCD be a convex quadrilateral. Prove that
there exists a point P inside ABCD such that
[PAB]=[PBC]=[PCD]=[PDA],
if and only if
the diagonals bisect each other.
Here [XYZ] denotes the area of triangle XYZ.
See The Solution
Submitted by
Bractals
Rating:
5.0000
(1 votes)
Comments: (
Back to comment list
| You must be logged in to post comments.
)
re: Solution
| Comment 2 of 3 |
(In reply to
Solution
by Jer)
I arrived at the same conclusion.
What about restating the problem by replacing
"the diagonals bisect each other"
with
"at least one diagonal bisects the other"?
Posted by
Bractals
on 2013-06-17 13:56:07
Please log in:
Login:
Password:
Remember me:
Sign up!
|
Forgot password
Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ
|
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
blackjack
flooble's webmaster puzzle
Copyright © 2002 - 2024 by
Animus Pactum Consulting
. All rights reserved.
Privacy Information