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 .
and 
be altitudes in . Let 
be the perpendicular lines from vertices 
to 
respectively. Prove that are concurrent lines.
are passing through the circumcenter (the case when is right-angled triangle is a special case). The 2nd official solution makes use of the Carnot theorem
and 
are symmetric with respect to the angle bisector of 
, and similary for the other ones. Hence, the lines
meet at the isogonal conjugate of the orthocentre, which is exactly the circumcentre of .
![[asy]
size(12cm);
pair C=(0,0);
pair B=(9,0);
pair A=(3,7);
pair Ha=foot(A,B,C);
pair Hb=foot(B,C,A);
pair Hc=foot(C,A,B);
pair H=orthocenter(A,B,C);
pair O=circumcenter(A,B,C);
pair Hap=foot(A,Hb,Hc);
D(A--B--C--cycle);
D(A--Ha);
D(B--Hb);
D(C--Hc);
D(circumcircle(B,C,Hc));
D(Hb--Hc);
D(A--extension(A,foot(A,Hb,Hc),B,C));
D(rightanglemark(Hc,foot(A,Hb,Hc),A));
D(rightanglemark(A,Ha,C));
D(rightanglemark(C,Hb,B));
D(rightanglemark(C,Hc,B));
D(MA(C,A,Ha,.8,blue));
D(MA(O,A,Hc,.8,blue));
pair point = origin;
pair[] p={A,B,C,Ha,Hb,Hc,H};
string s = "A,B,C,H_A,H_B,H_C,H";
int size = p.length;
real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;}
string[] k= split(s,",");
for(int i = 0;i<p.length;++i) {
dot("$"+k[i]+"$",p[i],dir(O--p[i]),red);
}
dot("$O$",O,dir(0),red);
dot("$H_{A'}$",Hap,dir(240),red);
[/asy]](http://latex.artofproblemsolving.com/3/9/9/39906e76317bbe1797a300a7da32e60bec4a84bf.png)
so 
is cyclic, hence 
. Let 
be the foot of 
on 
. It follows from the similarity that 
, so 
and 
are reflections over the angle bisector. The same is true for 
. Since 
concur at , their reflections over their respective angle bisectors must concur at the isogonal conjugate of , which happens to be the circumcenter of .
are isogonal and 
is anti-parallel to 
be the intersection of lines 
with , respectively. By Carnot's theorem for 
and points , it is enough to prove that 
. However, the Pythagorean theorem gives us the following relations:
, 
, 
, 
, 
and 
.
, which is true by Carnot's theorem for and points 
.
has unit length, and set the origin at point 
. Then 
, where and 
are real parameters. Now, the equations of 
and 
are 
, 
and 
, respectively. Next, we obtain that the equations of and are 
, 
and 
, respectively. By solving two systems of linear equations, it follows that 
and 
. Lines 
and 
have slopes 
, 
and 
. Finally, the equations of lines are 
, 
and 
, so it is clear that they pass through a common point 
.
are isogonal lines and hence their intersection, ie, orthocenter is the isogonal conjugate of circumcenter which is indeed where the lines, 
concur
is orthic triangle, hence, 
are angle bisectors in 
And, 
Hence, 
Combining them, 
Similar procedure for, 
and, 
, we can either asiign barycentric coordinates for 
and get the concurrence point as circumcenter or we can use ceva and prove that 
are concurrent lines
and be altitudes in . Let be the perpendicular lines from vertices to respectively. Prove that are concurrent lines.
and 
where 
respectively.
is the 
point of the 
excircle of .
are the extouch point of the 
excircles of the the triangle 
respectively.
concur on the Nagel Point of the orthic triangle of . also 
are concurrent. are concurrent. 
.
pass through circumcentre of triangle 