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 written on the board. A move consists of adding either , 
or 
to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is 
, then in a move a player can't choose the number , but he can choose either or ). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
such that 
for all 
such that 
.
identical cars on a circular track. Among all of them, they have just enough gas for one car to complete a lap. Show that there is a car that can complete a lap by collecting gas from the other cars on its way around
, which doesn't have right angle, let 
be its circumcenter. Let 
and 
be orthocenters of triangle 
and 
, respectively. Let 
be circumcenter of triangle 
. If circle 
and circle 
intersect at 
for the second time, prove that lies on 
.
is a polynomial with real coefficients such that![\[ P(x)^2 - 1 = 4 P(x^2 - 4x + 1). \]](http://latex.artofproblemsolving.com/b/4/2/b425a938a1dbbac25e0e0793b30c9e58845d45ab.png)
Find .
and 
be altitudes in an acute triangle 
which meet at 
. Suppose that 
meets the circumcircle of at 
and 
such that lies on the shorter arc of 
and lies on the shorter arc of 
. Let 
and 
meet at 
. Show that the circumcircles of 
and 
and the line 
concur.
be a diameter of a circle 
, and let 
be the tangent at 
. Furthermore, let 
be a fixed, positive real, and consider all pairs of points 
and 
lying on , on opposite sides of , such that 
. The lines 
and 
intersect at points 
and 
, respectively. Show that all the lines 
pass through a common point.
such that
for all 
satisfying 
, your right-neighbor’s is 
, and the next person over is 
, then for every trio in the circle they see
with circumcenter . 
and 
intersect the altitude from 
to at points and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of over . 
is the second intersection of circumcircles of triangles 
and 
. Show that 
are collinear.
and 
, therefore 
...
and the triangles 
are just similar and not congruent. I would add that the last pair is similar, because from similarity of the triangles , we have
have a common vertex and they can be made centrally similar with positive coefficient of similarity by rotating the triangle 
counter-clockwise by the angle 
or centrally similar with negative coefficient of similarity by rotating the triangle clockwise by the angle 
. Hence, they are spirally similar with ceneter of similarity . Let 
be a point on the extended line , such that 
and the points 
follow on the line in this order. Since 
and since the triangles 
are similar, the quadrilateral 
(each with one internal angle equal to 180^o) are also spirally similar with the same center of similarity and the same coefficient of similarity. As the point moves on the arc 
of the circumcircle 
of the triangle 
, the locus of points 
is an arc of a circle 
centered at the vertex 
and radius 
. Similarly, the locus of points 
is an arc of a circle 
centered at the vertex 
and radius 
. When the point coincides with the vertex , the locus arc of the points is limited by the intersection 
of the circumcircle tangent at the vertex with the circle and the locus arc of the points by the intersection 
of the ray 
with the circle . When the point coincides with the vertex , the locus arc of the points is limited by the intersection 
of the circumcircle tangent at the vertex with the circle and the locus arc of the points by the intersection 
of the ray with the circle . Thus the central angles of the locus arcs 
, 
of the points 
are
move in such a way that it is always directly similar to a figure 
and one common point of the similar figures 
does not move at all. Let the locus of a point of the figure be a curve 
. Then the locus of any other point (except ) of the figure is a curve 
similar to the curve . is an arc similar to the arcs , of the points . Since these 2 arcs are centered at the vertices 
of the quadrilateral 
, the locus arc 
of the points is centered at the point corresponding to the point in the spiral similarity of the quadrilaterals and with the same central angle 
. Radius 
of the circle is easily found when the point coincides with the vertex and the triangles and the quadrilaterals become centrally similar with homothety center and homothety coefficient 
. Denote 
. Using the cosine theorem for the triangle 
,
, we get
of the points is limited by the appropriate intersection 
of the line 
with the circle , when the point becomes identical with the vertex and by the appropriate intersection 
of the line 
with the circle , when the point becomes identical with the vertex .
Any geometric ankers (sic) ?
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
and 
.
,
.
and 
to 
,
.
to 
.
are constant now
and it's constant so the locus of Q is a circle by center A.