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.
is called pretty if it has finite expression in base
numeral system, 
is a positive integer in ![$[2;2022]$](http://latex.artofproblemsolving.com/f/0/5/f05e359b3b158e7a4a742cbe662e8d1417fd7926.png)
. Prove that there exists finite positive integers 
that with every 
in 
then there is at least one pretty number between 
and 
+1 w
is called a 
if there exists a positive integer , such that 
and the decimal representation of 
consists only of digits among 
. Are there infinitely many 
?
, prove that 
. When we have equality;
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. 
is the orthocenter. 
is the midpoint of segment 
. Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from 
and 
intersect at 
. Prove that 
are colinear
+1 w and 
, define 
as the greatest exponent 
such that 
divides 
. Prove that there are infinitely many such that 
and infinitely many such that 
.
prime and a positive integer. Determine all pairs 
satisfying the equation: 
be positive real numbers with 
. Prove that
and 
satisfy 
, then 
is an even number.
, not all equal to 
, the following inequality holds
for which the equality holds.
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.
such that 
divides![$\sum_{0 \leq i < j \leq 2024} \left[\binom{2024}{i}\binom{2034}{j} - \binom{2024}{j}\binom{2034}{i}\right]^2$](http://latex.artofproblemsolving.com/4/c/4/4c42b9f80506262b06ff41a89c911784d28e9de2.png)
be the point on 
such that 
,
intersect 
intersect 
be the intersections of 
and 
.
,
,
.
,
is the point on infinity along 
.
and we may concur. 
version:-
.
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)