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 .
passes through 
be the Miquel point of 
. We have that 
by spiral similarity so as 
, the two triangles are in fact congruent so 
. As 
clearly lie on the same side of 
, ( lies inside ) we conclude that 
as desired.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(300);
real labelscalefactor = 1; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -9.931619015998317, xmax = 12.35268271700373, ymin = -6.2, ymax = 6.295618999209102; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);
pair A = (-3.023693176494155,4.80720091682328), B = (-5.559967711756501,-2.749982437357092), C = (3.9630069673681714,-2.8168117495423215);
pair M = (-0.7418154036524869,5.291189386209763), P = (-2.5219236580982507,0.48914284080444176), Q = (0.07919666832617143,0.712700419228189);
pair X = (-1.2118272753134591,1.8859515227576646), Y = (-3.6364372221499246,4.089377965597112), Z = (0.8344687186207644,4.067687193302564);
pair O = (-0.78,-0.15), Op = (-1.407153798515055,2.8068644781947234);
path omega = circle(Op, 2.5718758578637173);
pair R = intersectionpoint(X--(2*X-A),omega);
/* draw figures */
draw(Y--X--Z^^X--R,pink);
draw(O--Op,gray(0.7));
draw(circle(O, 5.441323368446319), darkgreen);
draw(B--X--C^^A--X,purple);
draw(omega, red+linetype("4 4"));
draw(circle((-0.8114130509364313,-4.168902949863692), 4.962180682439345), red+linetype("4 4"));
draw(P--A--Q,orange);
draw(A--B--C--cycle, rvwvcq);
/* dots and labels */
dot((-3.023693176494155,4.80720091682328),dotstyle);
label("$A$", A, NW * labelscalefactor);
dot((-5.559967711756501,-2.749982437357092),dotstyle);
label("$B$", B, SW * labelscalefactor);
dot((3.9630069673681714,-2.8168117495423215),dotstyle);
label("$C$", C, SE * labelscalefactor);
dot(M,linewidth(4pt) + dotstyle);
label("$M$", M, N * labelscalefactor);
dot(X,dotstyle);
label("$X$", X, W*1.5);
dot(P,linewidth(4pt) + dotstyle);
label("$P$", P, 1.7*S);
dot(Q,linewidth(4pt) + dotstyle);
label("$Q$", Q, 1.7*S);
dot(Z,linewidth(4pt) + dotstyle);
label("$S$", Z, NE * labelscalefactor);
dot(Y,linewidth(4pt) + dotstyle);
label("$T$", Y, W * labelscalefactor);
dot(R,linewidth(4pt) + dotstyle);
label("$R$", R, S * labelscalefactor);
dot(O,linewidth(4pt) + dotstyle);
dot(Op,linewidth(4pt) + dotstyle);
dot(A^^B^^C,linewidth(4pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/e/5/6/e56d7aa792666714464bbb0b6b7362b20b530a97.png)
is cyclic. Let 
and be the feet of the 
-angle bisectors in triangles 
and 
, respectively. Observe that![\[
\angle PAX = \tfrac12\angle BAX = \angle ABX
\]](http://latex.artofproblemsolving.com/6/a/6/6a6f21593eae6d527a151b3a4c60954591da741e.png)
by the given condition, so 
. This yields 
, and analogous reasoning with 
yields 
. Therefore 
, and so quadrilateral 
is cyclic.
is cyclic. Suppose 
intersects again at point 
(not shown in the diagram). Menelaus and the Angle Bisector Theorem imply![\[
\frac{BY}{YC} = \frac{BP}{PX}\cdot\frac{XQ}{QC} = \frac{BA}{AX}\cdot\frac{XA}{AC} = \frac{BA}{AC}.
\]](http://latex.artofproblemsolving.com/c/1/2/c12c2c55b1f366c82e8702ec45f78587459fe6e6.png)
This means that is the foot of the -external angle bisector of , meaning that , , and 
are concurrent. Now![\[
YA\cdot YM = YB\cdot YC = YP\cdot YQ,
\]](http://latex.artofproblemsolving.com/8/9/9/899d5850b913ba0ffacff298132e450e0a0f8eb9.png)
and so quadrilateral 
is cyclic.
is a center of homothety. Let , 
, and 
intersect 
again at 
, , and respectively. Then a bit of angle chasing yields![\[
\angle RSP = \angle RAP \equiv \angle XAP = \angle ABX,
\]](http://latex.artofproblemsolving.com/b/3/a/b3af3bf26f6696b0b91a591b4f75b4d7ab7f16b8.png)
and so 
. Analogous reasoning yields 
, and furthermore Reim's Theorem tells us 
. Therefore triangles and 
are homothetic, and their center of homothety is . to 
, and so lies on the line connecting the centers of these two circles. But this line is precisely the perpendicular bisector of 
, and so 
as desired.
, this swaps 
, 
and stays, hence swaps 
. Now , which lies on both circles stays, thus lies on and 
.
be the perpendicular bisector of 
and let 
, which means 
is the angle bisector of 
. 
. By symmetry, is equidistant from the perpendicular bisectors of , as desired (sure it could be equidistant to the other arc midpoint but config issues don't matter).
be the circle centered at with radius 
. Consider the "force-overlay polarity" given by doing pole-polar duality wrt , then reflecting over the angle bisector of 
. It is not hard to check that under this transformation, the problem is as follows (angle chase; the two angle conditions act independently):
be an acute triangle, and let the external angle bisector of intersect line at . Let 
be the circle centered at 
through 
; define 
similarly. Suppose is a common tangent of and . Show that 
lies on . is the exsimilicenter of the two circles (compare radii and distances). 
Denote by 
containing 
again at 
so 
so 
respectively such that 
.
which promptly gives us that 
.
and 
to 
,
Thus 
bisects 
and thus 
,
,
denote the reflections of 
,
.
,
.
and 
we see that 
is parallel to the bisector of 
.
,
on 
By easy angle chasing we have 
Then we have 
because we know that 
and 
We conclude that 
is cyclic and we are done.