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.
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
be an acute triangle with 
, Let 
be the intouch triangle with 
,
,
,, let 
be the intersecttion of the perpendicular from 
to 
with 
, and 
.
and 
are concylic
the set 
can be partitioned into 
triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.
be a triangle with circumcircle 
, circumcenter 
, and orthocenter 
. Assume that 
and that 
. Let 
and 
be the midpoints of sides and 
, respectively, and let 
and 
be the feet of the altitudes from 
and 
in 
, respectively. Let 
be the intersection of line 
with the tangent line to at 
. Let 
be the intersection point, other than , of with the circumcircle of 
. Let 
be the intersection of lines 
and . Prove that 
.
be the foot of perpendicular from to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle . A circle 
with centre 
passes through and , and it intersects sides and at and 
respectively. Let be the foot of altitude from to 
, and let be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
, 
, 
, 
satisfy 
and 
. Prove that there exists a permutation 
of 
such that 
.
be a cyclic quadrilateral. A circle centered at passes through and and meets lines 
and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
such that 
, where 
are non-negative reals such that 
.
.
a tangencial quadrilateral that is not a rhombus, let 
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
. Let 
be the midpoints of 
resp. Prove that![\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]](http://latex.artofproblemsolving.com/e/7/0/e709acb8b17c0466a385ba8e7a89bb1bf16c72af.png)
is given. It turns out that for every positive integer number 
, for which the numbers 
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
with as its incenter. Let be the intersection of 
and and define 
in a similar way. Furthermore, let 
. Prove that if 
, then 
are concyclic.
for which there exist different integers such that none of them are divisible by either of 
or 
, but the sum of any two of them is divisible by at least one of and .
, where 
are the remainders upon division by 7, 11, 13, respectively. for each remainder 
, let 
denote the complementary remainder; e.g., the complementary remainder modulo 7 of 3 is 4. for two 
-tuples 
, 
of remainders, we say that they "match" in the 
-th position if 
. a set of -tuples will be called "matching" if any two of them match in at least one position; otherwise, we call them "mismatched." so the statement we want to prove is, that a set of matching triples must have size at most 8. first we prove the following lemma.
denote these pairs. assume for the moment that matches 3 of the other pairs - WLOG we may assume they are 
- in the first position. then coincide in the first position, so any two of them must match each other in the second position - but this is impossible. so can match at most 2 of the other pairs in the first position, and similarly it can match at most 2 of the other pairs in the second position. since there are four other pairs, a must match *exactly* 2 in the first position and 2 in the second position. now let 
. by what we've just proved, the other pairs must be of the form 
. now, for 
to match 
, we must have either 
or 
. in both cases, we have 3 pairs which coincide in one position, which as before, is impossible. this completes the proof of the lemma.
.
