AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
in 
colors, there exists an arithmetic progression of size 
. How can we derive Van der Waerden's theorem for 
from this?
be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
is a continuous function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
such that 
, then there exists a 
such that:![\[
F(c) = \frac{1}{c} \int_0^c F(x) \,dx
\]](http://latex.artofproblemsolving.com/8/f/e/8fea4d84bc98b6f4953e19976cfa554945a37ea6.png)
be a nonzero square matrix of order 
satisfying![\[
a_{ik} a_{jk} = a_{kk} a_{ij}, \quad \text{for all } i, j, k.
\]](http://latex.artofproblemsolving.com/7/5/2/752b16a277d9fd18bfda6c3be35d23d42d75ebaa.png)
Denote by 
the trace of 
, which is the sum of the diagonal elements of .
. in terms of .
such that 
holds for all real numbers 
and 
.
be a ring. Let 
such that 
. Prove that if 
is nilpotent, so is 
.
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
be a triangle with 
. Let 
be the midpoint of 
, and let 
be a point on segment such that 
. Let 
be the point of intersection, different from 
, of the circumcircle of triangle 
and line 
. Define 
and 
as the points of intersection of line 
with 
and 
, respectively. Prove that is the midpoint of 
.
and now just note 
and 
so if 
then done, else obvious.
,
.
we have 
and 
, we have
.
and 
are positive.) Therefore we have
,
is constant in the interval ![$[x;z]$](http://latex.artofproblemsolving.com/c/e/d/ced12bea5c9d9ed956234eb6e7ca3c664266ad60.png)
.
is non decreasing
, 
. Thus, we have to prove that ![\[(z-x)(A(z)-A(y)) \geq (z-y)(A(z)-A(x)).\]](http://latex.artofproblemsolving.com/7/9/a/79aad614e7ede2430c859afdf94a0d46bffdb6d9.png)
, we see that the given inequality is equivalent to ![\[A(y) \leq \frac {z-y}{z-x}A(x) + \frac {y-x}{z-x}A(z).\]](http://latex.artofproblemsolving.com/d/9/3/d93cdd0a7ab87b4ec67c9b9505ed7e080ce93971.png)
, there exists a unique 
such that 
Substituting this value in the last one we see that the given inequality is equivalent to proving ![\[A(\alpha x+(1-\alpha) z) \leq \alpha A(x) +(1-\alpha)A(z).\]](http://latex.artofproblemsolving.com/7/8/c/78c28a13196596b1705c8785af60a63d28dcea16.png)
is non-decreasing, this is true due to convexity.
![\[(1)\,\,\,\,\frac{1}{z-y}\int_{y}^{z}f(u)\,du \ge \frac{1}{z-x}\int_{x}^{z}f(u)\,du.\]](http://latex.artofproblemsolving.com/7/7/9/779f0966c9d9901f381e15452f1408465fdb39d3.png)
This is the comparison of two averages. The second average is over a larger range of -values. And each additional -value is 
every -value in the first average. Intuitively the second average is less than or equal to the first. (And if I were grading this "solution", I would give it a perfect score.)
On the right side of (1), let 
(1) becomes![\[\int_0^1 f(y+t(z-y))\,dt \ge \int_0^1 f(x+t(z-x))\,dt.\]](http://latex.artofproblemsolving.com/7/a/8/7a8e99dd8a4079e9ec9b2b3922adebc1487508ac.png)
Is this true? Sure, simply because 
for all ![$t \in [0,1].$](http://latex.artofproblemsolving.com/8/f/5/8f5910ef016e9517702212753728d80b112b7568.png)
is non-decreasing, this is true due to convexity. imply that its integral is convex??
a non-decreasing function. Then show this inequality holds for all 
such that 
.
With 
Let 
and 
for some 
by Lagrange Mean Value Theorem
for some 
and 
is non-decreasing we have 
![\begin{align*}
\mathfrak{I}
&=(z-x)\int_y^z f -(z-y)\int_x^z f\\
&=(z-x)\int_y^zf-\Big[(z-x)-(y-x)\Big]\int_x^z f\\
&=(y-x)\int_x^zf-(z-x)\int_x^y f\\
&=\Big[(y-x)-(z-x)\Big]\int_x^yf+(y-x)\int_y^z f\\
&=(y-x)\int_y^z f-(z-y)\int_x^y f\\
\end{align*}](http://latex.artofproblemsolving.com/5/e/d/5ed1044a6ca45650b1eac763c762510ae192d22a.png)
Now, it's given that is non-decreasing, so 
. Therefore, ![\begin{align*}
(y-x)\int_y^z f&\geqslant (y-x)(z-y)f(y)\\
&=(z-y)\Big[(y-x)f(y)\Big]\\
&\geqslant (z-y)\int_x^y f\\
\end{align*}](http://latex.artofproblemsolving.com/c/e/b/cebabb246ac9d3932fbea9db3227ac274609b571.png)
This implies that 
, i.e. 
Equality holds when 
and 
, i.e. is a constant function. 
and now just note and so if then done, else obvious.
? From your solution it looks like you have used the mean value theorem for integrals, obviously a very clever step towards the problem. But you have to use different variables for that, since, the LHS lies in some different domain than the RHS (obviously, or else what's the use of inequality here ). and it's given that function is non-decreasing it immediately follows that 
, where 
is in 
and in 
.
a non-decreasing function. Then show this inequality holds for all such that .
Now as it is given that is continuous So By Fundamental theorem of Calculus 
and 
such that
as is non decreasing and 
So 
. This implies the result.
a non-decreasing function. Then show this inequality holds for all such that .
(Well known) Let 
and 
be two converging sequences such that 
, then 
.![\begin{align*}
(z-x)\int_{y}^{z} f(u) \mathrm{du}\ge (z-y)\int_{x}^{z} f(u)\mathrm{du}
&\iff (z-y)\int_{y}^{z} f(u)\mathrm{du} +(y-x)\int_{y}^{z}f(u)\mathrm{du}\ge (z-y)\int_{x}^{y}f(u)\mathrm{du} +(z-y)\int_{y}^{z} f(u)\mathrm{du}\\
&\iff (y-x)\int_{y}^{z}f(u)\mathrm{du}\ge (z-y)\int_{x}^{y}f(u)\mathrm{du}\\
&\iff\left(\frac{\int_{y}^{z}f(u)\mathrm{du}}{z-y}\right)\ge\left(\frac{\int_{x}^{y}f(u)\mathrm{du}}{y-x}\right)\qquad [\text{as } z-y>0 \text { and } y-x>0]\\
&\iff \lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^{n} f\left(y+\left(\frac{z-y}{n}\right)i\right)\ge\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^{n} f\left(x+\left(\frac{y-x}{n}\right)i\right) \cdots\cdots\cdots(\bigstar) (\text{ Riemann sum})
\end{align*}](http://latex.artofproblemsolving.com/7/e/b/7eb797f5878478b5d67f1433083bb925c92dd011.png)
is non-decreasing, so 
we have, 
is true, and hence the original inequality is true. 
substitution which gives 
. Now we see that we essentially want to compare the two integrals 
and 
which is simple since we are given 
and the fact that is non decreasing.
is continuous! Mean Value Theorems has an assumption that is continuous on the closed interval. So indeed applying Fundamental Theorem of Calculus is flawed. Continuity is a very strong condition which everyone should acknowledge. which is a sufficient criteria for integrability. As a matter of fact, via Daboux's equivalent definition of Riemann Integrals, the proof is fairly simple as well!
is continuous! Mean Value Theorems has an assumption that is continuous on the closed interval. So indeed applying Fundamental Theorem of Calculus is flawed. Continuity is a very strong condition which everyone should acknowledge. which is a sufficient criteria for integrability. As a matter of fact, via Daboux's equivalent definition of Riemann Integrals, the proof is fairly simple as well! is explicitly given to be continuous, but not in this post. So applying MVT is justified.
in the most brutal way mankind has ever witnessed, some people making substitutions, differentiating, Mean Value theorem, like wtf. Riemann be crying in the coffin. Anyways here is a solution (if my solution is wrong, ppl have complete authority to criticize me), anyways, lets not talk BS. were continuous, then intuitively this makes sense as it reminds us of the continuous version of `Averages'. This idea is the main theme of the solution.
for any 
, we have
Rearranging the above yields the following inequality which is easier to deal with, you'll soon find out why,
Replace the integrals by a silly appoximation in such a way that it preserves the condition, which is achieved as follows
This follows from the fact that is non-decreasing. 
such that 
,
,
.
,
.
is 
.
.
.
in domain o
.
is non decreasing.
for 
.
being nonnegative? Is it just to make them stay in the domain?