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.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
such that 
is an integer for every positive integer 
. 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.
has side length 
. Point 
is on 
, and the area of 
is 
. What is 
?
![[asy]
pair A,B,C,D,E;
A=(0,0);
B=(0,50);
C=(50,50);
D=(50,0);
E = (30,50);
draw(A--B);
draw(B--E);
draw(E--C);
draw(C--D);
draw(D--A);
draw(A--E);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E",E,N);
[/asy]](http://latex.artofproblemsolving.com/d/9/a/d9ad7d16a15397560a5daa045621239464589d6e.png)
has side length . Point is on , and the area of is . What is ?
has side length . Point is on , and the area of is . What is ?![[asy]
pair A,B,C,D,E;
A=(0,0);
B=(0,50);
C=(50,50);
D=(50,0);
E = (30,50);
draw(A--B);
draw(B--E);
draw(E--C);
draw(C--D);
draw(D--A);
draw(A--E);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E",E,N);
[/asy]](http://latex.artofproblemsolving.com/7/4/f/74f928acaf0eb1f95e9ff6d944223c023af00a9a.png)
where is the base of the triangle and 
is the height.
,
Hence the answer is 
.
so the answer is 
,
,
,
.
,
.
,
.
which means the answer is 
.
Then 
is 40, so that means the length of BE is 
so 
so the choice is 
