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 four given points on the plane, so that no three of them are collinear. Let 
be the set of conics passing through . Consider a fixed point 
, for each 
, suppose there are distinct points 
, so that 
. Let 
be the line joining the intersection of 
and 
with the intersection of 
and 
. does not depend on the choice of . passes through a fixed point when 
varies."
such that:![\[f(1+xf(y))=yf(x+y)\]](http://latex.artofproblemsolving.com/f/2/a/f2a9e8eb184c921d5f1fa59137d99e38b29c6962.png)
be an acute scalene triangle , 
be the orthogonal projection of 
on 
, 
and 
are the midpoints of 
and 
respectively. Let and 
are points on the minor arcs 
and 
of the circumcircle of triangle respectively such that 
. Show that the circumcircles of triangles 
and 
are tangent if and only if lies on 
.
, where 
, and 
are positive integers that satisfy
and 
.
be real positive numbers such that: 
. Prove that: 
+1 w
: 
such that:
for all x,y E 
, 
, 
, 
be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, 
. Prove that 
for all 
.
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
for every pair 
be a trapezoid inscribed in circle 
, 
. Let is the symmetric point of across , 
intersects at 
. Let is midpoint of and 
is orthocenter of triangle . On 
take a point 
so that 
. Prove that: 
coincide
be integer numbers 
satisfy that:
. Prove that: are even
is a perfect square number
can’t be any power 
of a positive integer number
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
is given where are positive reals. Prove that:
is the circumcenter of 
where 
are the feet of 
.let 
the orthocenter and the intersection of 
and we have 
is the polar of WRT so 
so 
since 
where 
is the foot of then 
therefore midpoint of 
hence 
and the altitude from 
goes through 
and 
medians in 
and 
and using that 
we obtain that 
. In 
we have 
. (*)
. That`s from Pappus in 
and in 
combined with (*)
be an acute triangle with orthocenter . Let be the midpoint of . Rays 
and 
intersect the line through perpendicular to 
at 
and 
respectively. Prove that 
.
be the orthic triangle of . Let be the intersection of and the line through parallel to 
. Let be the foot from to . We have 
, which we want to show is 
. This is equivalent to 
collinear. Now, radical axis on (1) 
, (2) 
, the 9-pt circle, (3) 
, the circle of diameter 
yields that 
concur. Since 
, 
is the line through parallel to , so we are done.
. Define instead 
. By Brokard, we know is the pole of 
, so 
.
. Then 
.
Let be the coordinate of point 
such that 
. We will prove that is the midpoint of segment . From there it's clear that 
.
in formula for 
we get 
Essential part:![$$e+f=\frac{(c-a)[b^3+2b^2(c-a)+b(a^2+c^2-5ac)+3a^2c-ac^2]-(b-a)[c^3+2c^2(b-a)+c(a^2+b^2-5ab)+3a^2b-ab^2]}{(c-a)(b-a)(b-c)}=2a$$](http://latex.artofproblemsolving.com/d/8/9/d89c5ac1cefa4ae3f9e8f9820125e326644fdd48.png)
so it's all true.
and the median 
Draw 
The line which goes through and is perpendicular to intersects at 
On the opposite ray of the ray 
choose such that 
Prove that 
be an acute triangle with orthocenter . Let be the midpoint of . Rays and intersect the line through perpendicular to at points and , respectively. Prove that 
.
and 
is a cyclic quadrilateral, hence, by Brokard's is the orthocenter of 
Thus, 
and 
. Since, 
i.e. 
thus, they intersect at 
, the point at infinity along 
denote the midpoint of 
Now, by Cevians Induce Harmonic Bundles Lemma 
But, by Midpoints and Parallel Lines Lemma 
, thus 
and we're done. 
and 
that intersect the line through perpendicular to at points nd .Define 
,
e can easily prove that 
are collinear and from definition we know that 
is perpendicular to .
,and we are done.
be an acute triangle with orthocenter . Let be the midpoint of . Rays and intersect the line through perpendicular to at points and , respectively. Prove that .
and be the feet of the 
-altitude and -altitude respectively, and define 
.
and 
are tangent to 
. Now, it's easy to see 
.
., so 
. But 
from the problem statement, and the result follows easily. 
denote the point at infinity on . Observe 
which finishes the problem. 
be the --Humpty point and 
be the feet of the altitude from 
. By radical axis theorem on 
we get that 
and are concurrent. Now notice that
However 
, so 
are colinear and the conclusion follows.
be an acute triangle with orthocenter . Let be the midpoint of 
. Rays and intersect the line through perpendicular to 
at points and , respectively. Prove that .
, 
, and 
s.t. 
, 
, and 
. Since 
lie on the circle 
with diameter . Since 
lie on the circle 
with diameter and center . Therefore, 
so is tangent to . Similarly, is tangent to , so is the symmedian of 
and 
so 
is harmonic. Therefore 
so , as desired.
be the orthic triangle. Then let 
be the -Humpty point. Notice that lies on and lies on 
so 
. Then we will show that 
is cyclic. Let 
. It is well known that 
. Combining the facts that 
from Ceva-Menelaus and that 
we get that 
which implies the desired. So then 
which then implies that 
. Now projecting 
implies that is the midpoint of segment so we are done.
Prove that![$$\frac{a}{b}+ \frac{kb^2}{c^2} + \frac{c}{a}\geq 5\sqrt[5]{\frac{k}{16}}$$](http://latex.artofproblemsolving.com/8/b/e/8bed6605f3988f59b8395b091b92a8b9e8b40419.png)
Where 
be the point on 
,
intersect 
be the intersections of 
and 
.
,
,
.
,
is the point on infinity along 
.
and we may concur. 
version:-
and the foot of 
and since 
so projecting onto 
.
,![\[-1 = (DK; BC) \overset{H}{=} (A \infty, XY) \implies AX = AY. \quad \blacksquare\]](http://latex.artofproblemsolving.com/5/9/0/590e12974a5d2249fb9b2c56f1384c4c7065f8f3.png)