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.
If G has a normal subgroup of order 
then prove that 
is abelian without using Sylow Theorems.
be a triangle with circumcircle 
, incenter 
, and 
-excenter 
. Let the incircle and the -excircle hit 
at 
and 
, respectively, and let 
be the midpoint of arc without . Consider the circle tangent to at and arc 
at 
. If 
intersects again at 
, prove that 
and 
meet on . be an acute triangle with 
. be 
the circumcircle circumscribed to the triangle and the midpoint of the smallest arc of this circle. Let and 
points of the segments 
and 
respectively such that 
. Let 
be the second intersection point of the circumcircle circumscribed to 
with . Let and 
be the intersections of lines 
and 
with other than 
, respectively. Let 
and 
be the intersection points of lines 
and 
with lines and respectively. Show that the 
line passes through the midpoint of
are real numbers, find the smallest possible value of the expression:
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
and are primes and is a perfect square, prove that 
is a perfect square.
be a positive integer. Aslı and Zehra are playing a game on an 
chessboard. Initially, 
stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
reals, and 
.
?
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
such that: Among every vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
is differentiable and 
for all 
. Show that for some 
, 
.
such that 
.
then using Mean Value Theorem, there exist an such that 
case, suppose 
and assume 
on the whole interval 
. Then by the mean value theorem, there is some such that 
so 
. That by itself isn't a contradiction, but we can do something similar to get a contradiction. The intuitive idea is that since we're right on the "edge" of a contradiction, we have no room to move the graph of 
. If 
and , that forces 
on the interval ![$[0,1/2]$](http://latex.artofproblemsolving.com/8/2/4/8240ac9beaadf3f2654c3ba09a3987a7305539c5.png)
. Moving the graph up or down at any point makes the slope larger than 
somewhere. Similarly, 
on the interval ![$[1/2,1]$](http://latex.artofproblemsolving.com/5/5/8/5581510e950d8ceb9bcbcfd0dc988d9c4538c684.png)
. for every ![$x\in [0,1/2]$](http://latex.artofproblemsolving.com/0/f/d/0fd16eeab1cb0c53e329c87cea809db55ea5f207.png)
. On any interval ![$[0,a]$](http://latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
with 
, we have 
for some 
, depending on 
. Since we're assuming 
is always bounded by , this means that 
so 
for all ![$a\in[0,1/2]$](http://latex.artofproblemsolving.com/e/0/c/e0c5302f45677b07a169c44aa1085cf1dace8463.png)
.
, then we could use the mean value theorem on the interval ![$[a,1/2]$](http://latex.artofproblemsolving.com/f/7/c/f7c796537cc023b8e3d0dd550cf1c3bdfc0278d4.png)
to get 
But the left side is larger than , and the right side is at most , a contradiction. on the interval , so 
However, this function is not differentiable at 
, so we are done.
:
Assume towards a contradiction that 
.
and 
:![\[\usepackage{mathtools}
\tfrac {1-f(x)}{\frac 12-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\tfrac 12\right):f(x)\ \ge\ 2x\tag{1}\]](http://latex.artofproblemsolving.com/4/6/d/46d07da0b23e15de91ca1912093a406842150f21.png)
LMVT on and 
:![\[\usepackage{mathtools}
\tfrac {f(x)-1}{x-\frac 12}\ge -2\xRightarrow{\hspace{40pt}}\left(\forall \tfrac 12<x< 1\right):f(x)\ \ge\ 2-2x\tag{2}\]](http://latex.artofproblemsolving.com/9/2/c/92c82c921fad2971d842ec2dba87a5e4725c3e97.png)
Differentiability at 
implies 
is not possible.
and 
, reflecting about the line if necessary, we may assume 
.
and 
:![\[\usepackage{mathtools}
\tfrac {2\alpha+\eta-f(x)}{\alpha-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\alpha\right):f(x)\ \ge\ 2x+\eta\tag{3}\]](http://latex.artofproblemsolving.com/8/3/5/83584e1961403c7cf5c6891281a4cbb97346d193.png)
continuity at 
:![\[(\forall \epsilon>0)(\exists\delta>0):0<x<\delta\xRightarrow{\hspace{40pt}}\big|f(x)\big|<\epsilon\tag{4} \]](http://latex.artofproblemsolving.com/9/d/a/9da78481512b68b5c3e9ed771a83552d9878e137.png)
contradicts 
for 
.
is differentiable on . For 
and 
you can hopefully do!! put the value of 
in the two equations,So clearly visible that not differentiable on ....
and 
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and 
2. 
3. 
Bracket 2
2. 
3. 
Bracket 3![$$\int_0^{1}\frac{dx}{1-\sqrt[3]{1-\sqrt{x}}}$$](http://latex.artofproblemsolving.com/d/5/e/d5e8db00985944b2eca341e37a94e34661cb5c86.png)
2. 
3. 
Bracket 4
2. 
3. 
Bracket 5
2. 
3. 
Bracket 6
2. 
3. 
Bracket 7
2. 
3. 
Bracket 8
2. 
3. 
![$$\int_0^{2\pi}\sqrt[4]{\cos^4(x)-\cos(2x)}dx$$](http://latex.artofproblemsolving.com/7/e/e/7eebb4a488fec702e0130a253ef87f0f6dfedb2d.png)
2. ![$$\int_0^1 \frac{x^5}{\sqrt[3]{1-x^3}}dx$$](http://latex.artofproblemsolving.com/6/0/5/605bb9d81d0ccff54a276c9f7bf1c1726c136398.png)
3. 
Bracket 2
2. 
3. 
Bracket 3
2. ![$$\int_0^1 \frac{dx}{(\sqrt{x}+\sqrt[3]{x})^2}$$](http://latex.artofproblemsolving.com/6/9/8/69869c0ed754c0dfacf412ac6411742ffa164a6d.png)
3. 
Bracket 4
2. 
3. 
2. 
3. 
Bracket 2
2. 
3. 
2. 
3. 
2. 
3. 
4. 
5. 
be a function which is continuous on ![\( [0,1] \)](http://latex.artofproblemsolving.com/0/1/0/01026be294289be8f090c7b2586dde64a0d5b479.png)
and differentiable on 
,
.
such that 
.
such that 
