[/quote]
,
for which 
is the square of an integer
of natural numbers with 
of 
if the product of the elements of 
![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be an integer greater than zero. Then, the value of the sum of the reciprocals of the cubes of the roots of the equation![\[
mx^4 + 8x^3 - 139x^2 - 18x + 9 = 0
\]](http://latex.artofproblemsolving.com/a/3/a/a3a9bca412e927632e4921c51f2a6593409031d9.png)
is equal to:
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
and 
be on segment 
of an acute triangle 
such that 
and 
.
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
.
)
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
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I was hoping this wouldn't be necessary, but it seems like it is.
The tl;dr is that you should not discuss cheating on public forums like this, full stop. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, no accusations that so-and-so is cheating, and so on.
To elaborate a bit more:
More bluntly: don't tempt people into unethical behavior by shouting from the rooftops about how everyone else is doing it. Otherwise, you will find that after a while, you are correct.
In the words of Steven Dunbar: Cheating at the AMC contests is pointless and stupid, and those who engage in it, or even discuss it have completely missed our purpose, the purpose of education, and the joy of learning. Lift your heads up.
The tl;dr is that you should not discuss cheating on public forums like this, full stop. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, no accusations that so-and-so is cheating, and so on.
To elaborate a bit more:
- There are many posts which are essentially users bragging about how easily they can cheat on the exams if they wanted to. This isn't cool; don't do this.
- If you believe an individual or a group of students is cheating on the exams, and you have evidence or reasonable suspicion, please do the responsible thing and report this to your Competition Manager or to the AMC HQ, not here.
- If you don't have reasonable suspicion and are just speculating on the extent of cheating, you should stop.
- Don't discuss methods of cheating. It is impossible to do so without inadvertently providing advice about how to cheat, so it is better avoided altogether.
- If you want to try and convince the AMC to change their procedures for better test security, that is reasonable, and you can write to them with (as usual) well-thought, constructive feedback. (I have done this myself, for example.) This forum is not the place for such feedback; the AMC HQ does not monitor it (AFAIK).
More bluntly: don't tempt people into unethical behavior by shouting from the rooftops about how everyone else is doing it. Otherwise, you will find that after a while, you are correct.
In the words of Steven Dunbar: Cheating at the AMC contests is pointless and stupid, and those who engage in it, or even discuss it have completely missed our purpose, the purpose of education, and the joy of learning. Lift your heads up.
This post has been edited 6 times. Last edited by v_Enhance, Apr 20, 2021, 6:23 PM
