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.
such that 
for all 
such that 
.
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.
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.
be a triangle inscribed circle 
, orthocenter 
. 
lie on such that 
. 
is midpoint of 
. The line passing though and parallel to 
cuts 
at 
. Prove that 
.
be a circumconic of 
passing through 
and 
. is the isogonal conjugate of the infinity point on 
WRT , is the isogonal conjugate of WRT which is a rectangle hyperbola , is the forth intersection of 
and 
is the center of .
WRT pass through the center 
of (well-known) ,
.
be midpoint of and 
is diameter of thus is midpoint of 
. 
cuts again at . 
is reflection of through 
then lies on . 
so 
is diameter of hence 
. We see 
by parallel lines. Therefore 
.
is projection of on 
. Note that, 
, we have ![\[\angle EAP=\angle EAC-\angle PAC=\angle BAF-\angle PAC=\angle BCF-\angle PAC=\angle ACF-\angle ACB-\angle PAC=\angle TAP-\angle PAC-\angle ACB=\angle TAC-(\angle GAT-\angle TCA)=180^\circ-\angle GTA-\angle ATC=\angle GTS.\]](http://latex.artofproblemsolving.com/f/3/8/f38233c1d044cf427a0f6b7aa2133cba84ce3a03.png)
deduce 
.
thus 
deduce 
.
suy ra 
.
thus 
. Hence if 
is midpoint of then ![\[\angle NMD=\angle NMB+\angle BMD=\angle ACB+\angle MCD+\angle MDC=\angle ACD+\angle QTA=\angle ACD+\angle QAC=\angle QCD.\]](http://latex.artofproblemsolving.com/8/8/b/88bdc2551c05000142da6f669b8d7842afb1f934.png)
be altitude, easily seen 
is cyclic deduce 
. Thus, 
is cyclic 
. We are done.
be the nine point center of .
be the midpoint of 
and 
be the reflection of in .
be the reflection of 
in 
, respectively .
are isogonal conjugate of 
.
,
,
. 
is the reflection of 
in 
, respectively , we get 
.
be the projections of 
on 
and let 
be the midpoint of 
cuts 
at and 
cuts 
at 
Clearly is on 9-point circle 
of 
is cyclic 
Thus since 
with corresponding circumcenters 
Hence 
and 
is cyclic 
, 
, 
, 
as variables and with chord 
being the diameter of parallel to 
, so 
. Then 
Moving points approach Fix and animate 
on the circumcircle. Then varies on line projectively via the map 
Also, varies projectively on the nine-point circle.![[asy] import graph; size(7cm); pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); pair O = (0.,0.), A = (-0.49774,0.86732), B = (-0.86295,-0.50528), C = (0.86211,-0.50672), H = (-0.49858,-0.14468), F = (0.75998,0.64993), D = (-0.62874,0.25325), G = (-0.69648,0.12037), X = (-0.22262,0.58933), Y = (-0.69700,-0.50542); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw((-0.65022,0.21113)--(-0.60809,0.18965)--(-0.58661,0.23178)--D--cycle, linewidth(0.6) + qqwuqq);
draw(circle(O, 1.), linewidth(0.6)); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(O--G, linewidth(0.6)); draw(A--F, linewidth(0.6)); draw((-0.75890,0.65120)--F, linewidth(0.6)); draw(H--(-0.75890,0.65120), linewidth(0.6)); draw(circle((-0.24929,-0.07234), 0.5), linewidth(0.6)); draw(G--D, linewidth(0.6)); draw(C--D, linewidth(0.6)); dot("$O$", O, dir((0.857, 2.328))); dot("$A$", A, dir((0.923, 2.308))); dot("$B$", B, dir((0.887, 2.257))); dot("$C$", C, dir((0.912, 2.178))); dot("$H$", H, dir((0.785, 1.862))); dot("$E$", (-0.75890,0.65120), dir((0.959, 2.297))); dot("$F$", F, dir((0.871, 2.201))); dot("$D$", D, dir((0.872, 1.746))); dot("$G$", G, dir((0.959, 1.882))); // dot("$X$", X, dir((0.829, 1.797))); // dot("$Y$", Y, dir((0.788, 1.825))); // dot("$E'$", (0.75890,-0.65120), dir((0.979, 1.691))); [/asy]](http://latex.artofproblemsolving.com/4/8/5/485fd79f24a032c98f3759833c36f6101c1ab39e.png)
which has degree 
. On the other hand, 
has degree 
. Hence, it suffices to verify the result for 
points, in order for the rotation of 
by 
to coincide with 
. be the foot of the altitude and the midpoint of each of the three sides. (Proof should be written out in a contest, but here it is omitted.)
. I also had posted a proof but it was quite long. I was also surprised that the solution using complex numbers is quite concise compared to the purely geometric method, which is a good example for complex coordinates. I would like to contribute with a general problem. For my original general method, you can see the topic An extension of Nine-point circle. be a triangle inscribed in circle 
. Points and lie on such that 
. A circle 
passes through 
, 
and meets sides 
, at , 
, respectively. 
meets 
at . Let 
be the isogonal of with respect to triangle . is a point on line such that 
. Prove that line 
goes through a intersection of 
and 
.
be a triangle inscribed in circle . Points and lie on such that . A circle passes through , and meets sides , at , , respectively. meets at . Let be the isogonal of with respect to triangle . is a point on line such that . Prove that line goes through a intersection of and .
intersect 
for the second time at and let 
intersect at . it is obvious that 
so 
is cyclic we only need to prove that 
is cyclic.
then it is easy to prove that 
is a line (you can simply use moving point here and check on the lines 
,
and ). is cyclic we have to prove that is the miquel point of 
. So then the requirement is that 
. this is trivial by 
and 
. so we are done.
with 
? I think it must be 
?
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 
.
