AoPS Academy
+3 w
[/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.
. Let 
be a point on side 
, and 
be a point on the extension of 
such that 
. Let 
be the circumcenter of 
, and 
be a point on the side extension of 
satisfying 
. Line BP intersects AC at point Q. Prove that 
. The length of the first jump is 
, the second 
, the third 
, and so on. The length of 
jump is equal to 
. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?
over all pairs of positive integers 
.
lamps 
arranged in a circle in that order. At any given time, each lamp is either on or off. Every second, each lamp undergoes a change according to the following rule:
, if 
have the same state in the previous second, then is off right now. (Indices taken mod .) is on right now.
which is on. Prove that for infinitely many integers all the lamps will be off eventually, after a finite amount of time.
and 
, let us consider the equation
[list=a]
and 
, find the least positive integer 
satisfying this equation. and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and .)
for which there exist real numbers 
satisfying 
, 
and
for 
.
such that 
holds for all positive rational numbers 
.
and 
. Prove that
Equality holds when 
Equality holds when 
and are prime-related if 
or 
for some prime 
. Find all positive integers , such that has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related.
and are included as divisors.
is chosen inside a convex quadrilateral 
. Could it happen that
, and 
, a triangle can be formed. Could it happen that from segments of lengths 
a right-angled triangle can be formed?
for which it is possible to arrange all divisors of that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
be a square inscribed in an equilateral triangle 
. Construct another equilateral triangle 
inside the square. Line 
meets 
at 
. Prove that 
is the isosceles right triangle.
.
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](http://latex.artofproblemsolving.com/2/2/e/22e6ec2f98777db80e39333542664e903663bf81.png)
![$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$](http://latex.artofproblemsolving.com/f/c/a/fca343ee701eeff4c309b31beb3426ec96b49b55.png)
Let 
.
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
,
.
.
,
,
.
is a diameter, we have 
,
,
.
.