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.
+1 w
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and and 
are on different sides of 
, 
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
. Prove that 
and are collinear.
square is filled with numbers 
.The numbers inside four 
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel., 
, , 
be four points in the plane, with and on the same side of the line , such that 
and 
. Find the ratio![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles and are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles and at the point are perpendicular.)
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
. Let 
be an altitude of triangle , and let 
be the midpoint of the segment . If 
is the common point of the circle and the line 
(distinct from ), then prove that the incircle and the circumcircle of triangle 
are tangent to each other at the point .
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide . (Note: it is possible that has other prime divisors not among .) Prove that for every prime number 
, there exists a lucky number such that 
.
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
is the -excenter of the triangle 
and 
is the circumcircle of this triangle. 
is the middle of arc of which doesn't contain . 
meets at 
. Prove that
and 
be the incenter and the excenters of againts 
Since is 9-point circle of 
then cuts 
at their midpoints 
Since 
(
bisects 
), then 
are concylic hence
which is due to 
bisecting 
than 
be the midpoint of arc containing 
is the circumcenter of 
and 
are tangent to 
is the symmedian of 
lies on the symmedian such that 
bisects 
and the conclusion follows
and since 
is an angle bisector of , 
. Now, let be the second intersection of the circumcircle 
of 
and , so we have 
, which means that 
and 
is tangent to , so 
, so by AA symmetry, 
, and the result follows. be the incenter of . Show that it is possible to construct a right triangle with side lengths 
.
is a triangle with circumcircle 
. Let be the incenter, be the midpoint of arc not including , and let 
be the midpoint of arc 
. If 
, show that 
: Let 
. It's trivial by angle chasing that 
and that 
, now PoP kills it.
be the midpoint of arc 
of and let 
be the -excenter. It is well-known that 
are inscribed in the circle 
of diameter 
with center 
Moreover, note that is the midpoint of arc 
, implying that and 
are the internal and external bisectors of , respectively. Therefore, the inversion with power 
combined with a reflection in swaps and fixes lines and 
It follows that the center of 
(the image of ) is a point on line However, the center of must also lie on the perpendicular bisector of 
, implying that the center is itself. Therefore, is fixed under this inversion, and hence is fixed as well. Thus, 
as desired. 
be the midpoint of arc . And 
is a circle with centre and radius 
. 
.
and 
so 
and it's enough to show that 
We know that 
and 
are tangent to . So we can simply show that 
and .
and 
.It's easy to find the coordinates of by intersecting and .So we finish problem using Distance formula.
and .It's easy to find the coordinates of by intersecting and .So we finish problem using Distance formula.
and .It's easy to find the coordinates of by intersecting and .So we finish problem using Distance formula.
are both tangent to circumcircle of 
by angle chasing.Also 
is cyclic so is the midpoint of symmedian in triangle 
and so two triangles 
is similar to 
from which the relation follows.
is unit circle, 
. So 
and 
. We now want 
. But its easy just solve 
.We get 
And now we need to prove 
. After substituting and 
we see that both sides are equal to 
and we are done.
is the 
dumpty point of 
which implies the result since the dumpty point is the center of spiral similarity sending 
to 
.symmedian and 
, where 
is the circumcenter of . Now by Ceva's theorem
Since 
,
which implies that 
is a symmedian. Moreover,
Hence 
is cyclic which completes the proof.
since 
bisects 
.![[asy] size(5cm); defaultpen(fontsize(10pt));
pair B,A,C,I,M,NN,IB,T,X; B=dir(110); A=dir(210); C=dir(330); I=incenter(A,B,C); M=extension( (B+C)/2,origin,A,I); NN=extension( (C+A)/2,origin,B,I); IB=2NN-I; T=2*foot(origin,M,IB)-M; X=2*foot(origin,IB,C)-C;
draw(circumcircle(A,C,IB),gray); draw(M--A,gray); draw(B--IB,dashed); draw(C--IB,dashed); draw(unitcircle); draw(A--B--C--cycle); draw(M--IB);
dot("\(A\)",A,A); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(M\)",M,M); dot("\(N\)",NN,SW); dot("\(I_B\)",IB,S); dot("\(T\)",T,SE); dot("\(I\)",I,NW); [/asy]](http://latex.artofproblemsolving.com/8/4/7/847e9f4a99fe59c94afa43395984b741d9adc254.png)
![\[\measuredangle I_BTC=\measuredangle MTC=\measuredangle IAC=\measuredangle BI_BC=\measuredangle BI_BT+\measuredangle TI_BC,\]](http://latex.artofproblemsolving.com/4/8/3/483a92707c808d23d140d8b38a45f7729f7c528a.png)
thus 
, so ![\[\triangle TBI_B\sim\triangle TI_BC.\]](http://latex.artofproblemsolving.com/a/5/2/a52d24927e419a6ef011b4a1e19a95c5c22c1f69.png)
It follows that 
, as needed
is the midpoint of the arc, then but notice we want 
and thus we just need one more angle to prove 
. We'll show 
, but 
and as 
and thus 
, proving our claim.
and are tangent to 
gives that is 
dumpty point so this implies the conclusion as
,
,
such that 
coprime, 
and 
.
.
.
.
gives 
,
so 
let 
intersect 
so 
combining with 
we are done because we have 
is diameter of 
so 
and it's well known that 
are tangent to 
centered at 
and 
)
point in triangle 
so Done 
And that would obviously imply the desired.
Also
And
Whence
So
.
We obtain that 
and
hence we conclude that 
.
follows directly from 
are similar, which clearly implies the result. Relabel 
and let 
be the arc midpoint of minor arc 
so 
.
.
,
.
by standard angle facts. So 
,
.
not containing 
is a kite, we get that 
and we know that 
.
.
.
,
.
.
by 
.
.
are tangents to 
so 
is harmonic.Say 
is the midpoint of 
.
,
So 
and is the spiral center taking 
or,
