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.
(use in a problem for junior) to practice. Can anyone share to me?
be an acute angled scalene triangle with circumcircle 
. Let 
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and , respectively, such that
Prove that 
and 
intersect on the circumcircle of 
.
grid with 
rows and 
columns, and each square of the grid is filled with a non-negative integer. Let 
be the average of all the numbers in the grid. Prove that if 
the there exist two identical rows (meaning same numbers in the same order).
+1 w
is called feirense if it is possible to write on a sheet of paper some integers such that every positive divisor of less than is the difference between two numbers on the sheet, and no other positive number is.
be a square with 
side length and with center 
. A rhombus 
is drawn where point 
lies on line 
. What is the area of ?
numbers 
, calculates the list 
where 
is the number of times the number 
appears in the list. So, for example, for the list 
, the program returns the list 
.
numbers. Then he asked him to apply the program to the resulting list, and so on, until a number greater than or equal to 
appears in the list. Find the largest value of for which, whatever the initial list of positive integers 
, it is possible for Alexander to do what the teacher asked him to do.
table is colored black or white. Every cell on table border is black. It is known, that in every 
square there are cells of two colors. Prove, that exist square that is colored in chess order.
be a non-isosceles triangle with orthocenter , circumcenter . Let 
be midpoints of 
and 
be reflections of across 
. Prove that triangle 
have the same centroid.
be a quadrilateral such that 
and 
are acute and 
. Suppose that 
, prove that 
.
. From the fourth term of the sequence onwards, each term of the sequence is the average of the previous three. Given that the first six terms of the sequence are natural numbers and that the sixth number written was 
, what is the fifth term of the sequence?
be circumscribed about circle 
, and let be the orthocenter of 
. The circle touches line 
at 
. The tangent to the circle 
at meets at 
. Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at . Prove that 
is tangent to .
midpoint resp. of the altitude and 
; it s easy to show that 
so 
thus 
hence 
is the 
- symmedian therefore it goes through .
to meet the tangent at 
to the semicircle at , we see that as 
is right angled and 
, is the circumcenter of , so is the mid point of 
, and as 
, the result follows. Q E D
intersect circle at 
, and let intersect at 
, then 
is a harmonic quadrilateral. onto , noting that is parallel to the tangent at . 
where 
is the point at infinity along , and so is the midpoint of , as required.
, therefore, if is the mid-point of , 
![\[\begin{vmatrix}
a^2 & b^2 & 0 \\
\tan B + \tan C & \tan B & \tan C \\
0 & 0 & 1
\end{vmatrix}
= a^2\tan B-b^2\tan B-b^2\tan C =0
\]](http://latex.artofproblemsolving.com/2/4/b/24b295a3935b9b40230d85fcd98ee73f40d104a2.png)
which is indeed true , hence, 
, Therefore, passes through the mid-point of
, If, 
, 
, Let, 
, hence, combining them, 
![$[\because DA=DB]$](http://latex.artofproblemsolving.com/6/6/3/6632ddce9fb9f9a5dbf310006c60e28ba1411414.png)
amd 
by a well known lemma 
be the point at infinty along 
(AS DESIRED)
and 
, so 
. Now, notice 
which finishes since 
is at infinity. 
are anti-parallel wrt 
.
and 
Prove that 
be the midpoint of 
and 
that of 
.
are isogonals w.r.t 
and 
is the 
-
to be the unit circle. Since 
.
.
from 
.
which is not a difficult tas
、
to simplify calculations much more.