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.
be an acute-angled triangle with circumcircle 
and circumcentre 
. Points 
and 
lie on such that 
and 
. Let 
meet 
at 
, and 
meet 
at 
.
and 
have an intersection lie on line 
.
be a positive integer. Alex plays on a row of 9 squares as follows. Initially, all squares are empty. In each turn, Alex must perform exactly one of the following moves:
Choose a number of the form 
, with 
a non-negative integer, and place it in an empty square.
Choose two (not necessarily consecutive) squares containing the same number, say . Replace the number in one of the squares with 
and erase the number in the other square.
, while the other squares are empty. Determine, as a function of , the maximum number of turns Alex can make.
be a convex hexagon with 
and 
.
which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
, show that 
.
be a point in the plane, distinct from the vertices of 
. Consider 
the reflections of with respect to lines 
and 
, in this order.
are collinear if and only if lies on the circumcircle of . does not lie on the circumcircle of and the centroids of triangles and 
coincide, prove that is equilateral.
given that there exist positive integers 
satisfying
and , we define 
as follows:
where 
are all the positive divisors of . For example,
.
Find all positive integers such that 
.
Find all positive integers such that 
.
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
be real numbers satisfying
be a fixed square and 
a variable point on segment 
. The square 
is constructed such that 
is on segment 
and 
is on segment 
. Let 
be the intersection point of lines 
and 
. Find the locus of as varies along segment .
be a positive integer. In an 
board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place queens on the board such that no two threaten each other when: 
. 
.
denote the number of positive integers less than that are relatively prime to . Prove that there exists a positive integer for which the equation 
has at least 
solutions in .
that can be written in the form
where 
are pairwise coprime positive integers such that 
, 
, and 
.
be a circle and a given point in the plane. Each line through which intersects determines a chord of . Show that the midpoints of these chords lie on a circle.
be . is midpoint of the chord (specifically, diameter) through 
, so it is one of the points of interest. Let the midpoint of be . Let the midpoint of any one of the described chords be . Then 
because is the midpoint of one of the chords of , which has center . Then 
is right, meaning 
. Since is the center of an arbitrary chord, which contains when extended, all the midpoints of such chords lie on the circle with center and radius 
.
be a circle and a given point in the plane. Each line through which intersects determines a chord of . Show that the midpoints of these chords lie on a circle. is midpoint of 
, we get that 
where is midpoint of such chords lie on a circle center at with radius 
. Could someone please elaborate on that?
is a part of a chord on circle 
which we can let be segment for 
on the circle. Since 
is isosceles and thus the median 
is also an altitude, from which the result follows.
![[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(12pt));
pair o=origin,p=(1.2,3.6),m=(-1.889,4.63),l=(4.289,2.57),po=(0,4),m2=(-3,4),l2=(3,4);
dot(po);
dot(p);
dot((0,0));
draw(m--l,dashed);
draw(m2--l2);
draw(o--p,dashed);
draw(po--o);
draw(circle(o,5));
label("$P_o$",po,SW);
label("$P_i$",p,SE);
label("$\omega$",o,SE);
label("$r_o$",0.5*po,W);
label("$r_i$",0.5*p,E);
label(scale(0.75)*"$\theta_i$",(0.26,1.65));
add(pathticks(po--m2,2,0.5,4,5));
add(pathticks(po--l2,2,0.5,4,5));
add(pathticks(p--m,3,0.5,4,5));
add(pathticks(p--l,3,0.5,4,5));
draw(anglemark(p,o,po,25));
draw(rightanglemark(po,p,o,12));
[/asy]](http://latex.artofproblemsolving.com/4/1/5/415c1abf15d262d6a3698085947e852b47f94be4.png)
(that fulfills the conditions of the problem), center , and 
is 
, the distance between and center is
(where 
is the distance between and the center of ), which is quite clearly a circle when graphed in polar coordinates.
and the center of the circle is the center of the locus?
. Consider an arbitrary chord with endpoints 
and on circle and let be the midpoint of . By definition, 
. Hence, we see that 
is a right triangle with hypotenuse . However, since this is an arbitrary chord, , all the chords passing through will form infinitely many right triangles with hypotenuse . Finally, to finish, we note that any triangle with hypotenuse has a circumcircle of diameter of . Since there are infinitely many right triangles with hypotenuse , this is the definition of a circle with diameter of and we are done. 
be the the centre of the circle and 
be the mid-points of the chords of passing through P.
If is the centre of a circle and is the mid-point of a chord XY of that circle, then :
and let it's mid-point be 
.
by the above mentioned lemma.
and similarly consider the mid-point of another chord 
. let this this mid-point be some 
. SO now by the lemme we have 
.
, it implies that 
is cyclic. Similarly we can go on taking distinct and proving that 
is cyclic. we still have 
and in common. Thus all the mid-points of the chords of lie on the circumcircle of 
. and the chord perpendicular to passing through . The midpoints of these two chords are respectively and . Now consider any other chord, , passing through and let be its midpoint. It is clear that 
. Thus, points 
lie on a circle with diameter . We can do the same thing for all other chords of passing through which finishes the problem. 
denote the centre of the circle is the midpoint itself becomes the midpoint
and lie on the circle be a chord of circle with on it and midpoint 
(perpendiculars from centre bisect the chord) too lies on the circle with diameter 
, we choose four points from the set of chords of that contain - 
. Let be the center of .
are cyclic.
lies on a circle. Which implies that every element from the set of chords of that contain must lie on 
.
, which we will denote as .
be the midpoints of the chords that go through point .
for all , so they all lie on a circle with diameter . 
be 
Let be the midpoint of an arbitrary chord 
It is well-known that 
(
by SSS). Therefore, 
implying that lies on the circle with diameter by intercepted arcs. Since the properties of apply to the midpoint of any chord passing through a point in the interior of 
all midpoints must lie on a circle with diameter 
so we are done.
as . Name the midpoints of two arbitrary cord that passes through as and which are not the diameter. Consider the quadrilateral 
. Since 
are midpoints we know that 
and 
. This means that is cyclic. Thus the circle with diameter contains all the midpoints of the chords that pass through .
be the midpoint of the chord determined by the line 
. Because intersects the circle , we know that the segment is a diameter of the circle. Because all diameters of a circle are congruent, it follows that all the midpoints of the chords determined by lines through are congruent. Therefore, all of these midpoints lie on a circle, which we will call 
. is indeed a circle, we must show that it satisfies the three properties of a circle: are congruent, it follows that all of the points on are equidistant from its center, which is the midpoint of the diameter defined by . are equidistant from its center, it follows that all points in the plane that are equidistant from its center are also on . are equidistant from its center, which is a fixed point, it follows that is the locus of all points in the plane that are equidistant from this fixed point. is a circle.
of any chord containing satisfies 
, where is the center of . Hence, all lie on the circle with diameter 
, which finishes. 
and getting nowhere ("is this some crazy homothety thing?"), before realizing that I could fix two of the points. The only points that one knows actually anything about are and (which are indeed midpoints), and substituting yields 
. At the time of solving I was in the car and consequently had to freedraw, which delayed my noticing: "wait, they're both right!"
with 
collinear, with the midpoint of being 
from congruent triangles. Hence, lies on the circle with diameter 
(where is the center of the circle) serving as the diameter.. We see that, no matter where the chord is, we can draw the line segment 
to bisect the chord. Since lies on the chord, 
for any such . By the inscribed angle theorem, we can see all such form a circle with diameter , just like we had suspected. 
and let be the centre. Consider the diametre passing through . Draw a random chord 
containing but not . Name the midpoint of as . Notice that 
is isosceles and bisects and isosceles triangles have a cool property by which we obtain 
. Now by simply staring at the drawing, you realise that lies on the circle with diametre (due to Thales ). 
be the center of . We claim that all the midpoints lie on the circle with diameter . For any midpoint , 
, proving our claim, as desired.
for all midpoints. Therefore, it lies on the circle with diameter 
and the midpoints of all chords passing through be 
.
.
is cyclic 
(By Inscribed Angle Theorem).
lie on a circle with diameter .
, let 
and 
be two arbitrary fixed points on the conic, and let be a point in the plane. Let be a moving point on , let be the second intersection of 
with . Let 
, and let 
be the harmonic conjugate of in segment . Let be the pole of line 
in . Prove that moves on a conic through 
, and that 
is a harmonic quadrilateral on this conic." moves on a conic. Move with degree 
on , then by projecting through we get that moves with degree , and moves with degree 
. Consider the polar of in the fixed conic ; this line moves with degree by Plucker's formulas, and passes through the pole of line = by pole-polar duality. Therefore the intersection of the polar of with line 
(degree ) moves with degree 
, so moves on a fixed conic. = 
, and the tangent to in the locus of (let this tangent line be 
), we get that this conic passes through . through , since ![\[P(I,J;O,P) = (I, J, OP \cap IJ, \ell \cap IJ) = -1\]](http://latex.artofproblemsolving.com/d/f/c/dfcf3a454a0ba2263fea6a00e36bf7963c89c519.png)
(the final harmonicity follows by 
being the polar of ), we are done.
to the circle points, and we get that the midpoint of the circle chord moves on a conic 
such that 
, implying that 
, which is the circle with diameter .
be the centre of circle and , be the two feet of tangent WRT . THE CIRCLE in question is the circumcircle of 
.
is concyclic due to two right angles, giving that is the diameter of THE CIRCLE.
intersecting circle at and and the midpoint of is called 
. In we have 
. Now move your focus to THE CIRCLE. Basically, we have shown that 
, meaning 
(or 
) is concyclic. 
. be the midpoints of any two chords passing through . It's well known that lies on the perpendicular bisector of and , meaning that 
, so 
lie on a circle with diameter , which is true for any . 
and![\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/c/8/9/c895ae7fdf8d7f284ac9fc94cc077d6edad6cbf0.png)
is right. Let the midpoints of all the other chords through 
.
is right, so quadrilateral 
is cyclic. Since we have all the circles based on 
and 
We claim 
lies on the circle with diameter 
as 
by SSS. 
since this immediately implies that all such midpoints lie on the circle with diameter 
congruent to 
which immediately implies the result.
so ![[asy] pair p, a, b, m; a = dir(50); b = dir(150); p = 0.4 * a + 0.6 * b; m = (a+b)/2;
draw(circle(0, 1),blue); draw(circle(p/2, abs(p/2)),green);
draw(a--b, blue); draw(m--(0,0), blue);
dot("$P$", p, dir(100)); dot("$A$", a, dir(30)); dot("$B$", b, dir(150)); dot("$M$", m, dir(100)); dot("$O$", (0,0)); rightanglemark(p, m, (0,0));
[/asy]](http://latex.artofproblemsolving.com/5/a/e/5ae2f5b0119dda8705975bf1a68c1cfe4dc8fec5.png)
,
,
be the chords intersecting 
,
,
,
,
,
,
,
.
,
,
where 
.
.