[/quote]
,
,
,
is a 
is 
,
is a 
is also prime
be positive real numbers.![\[\frac{ab}{a^2+3b^2}+\frac{bc}{b^2+3c^2}+\frac{ca}{c^2+3a^2}\leq \frac 34.\]](http://latex.artofproblemsolving.com/0/3/9/0396d1d0fe304dc632c7c209282cac784da85017.png)
be an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
.
and the circle of radius 
cross 
and 
,
and circle 
are concurrent.
.
problems in a meeting. For each 
,
-
so that the chair can ensure that the longest idle time for any leader does not exceed 
be a finite set of cells in a square grid. On each cell of 
and 
.
Where 
table is filled with numbers from 
to 
as shown in the figure. Is it possible to rearrange some numbers so that there is still one number in each cell, and so that the sum of the numbers does not change in all rectangles of three cells?
in 
?
:o
so 
and since 
is just 
therefore 
so 
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Hi!
I'd like to introduce the Bogus Proof Marathon.
In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution
Use classic Marathon format:
Example posts:
Let's go!! Just don't make it too hard!
I'd like to introduce the Bogus Proof Marathon.
In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution
Use classic Marathon format:
[hide=P#]a1b2c3[/hide] [hide=S#]a1b2c3[/hide]
Example posts:
Let's go!! Just don't make it too hard!
This post has been edited 3 times. Last edited by pifinity, Mar 12, 2018, 7:40 PM
,
,
and
.
MUST equal 
?
which you did flipped the 1/2 because I guess you mistaken it for division
.
,
Checking, 1 and 7 both don't work. So, the cubic doesn't have any rational zeros. However, this is a contradiction! Every cubic -- or for that sake polynomial of odd degree must have at least 1 rational root, because both the irrational 
)
)
doesn't satisfy that! CONTRADICTION. So, I broke math! 
=
to get 
.
,
.
,
.
.
.
for 
!
.
.
or 
.
.Now subtract both sides by 
.
.
.
.
.
,
because that would imply that 
(which is not given).
