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 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 .
be positive real numbers satisfying 
. Prove that![\[
3(x^{\frac{2}{3}} + y^{\frac{2}{3}}) \geq 4 + 2x^{\frac{1}{3}}y^{\frac{1}{3}}.
\]](http://latex.artofproblemsolving.com/f/3/c/f3cd114f909f388eacee142d3f90d07f6b149131.png)
be the maximal number of the vertices of a convex lattice polygon with exactly 
lattice points in the interior. Show that:
for 
for some constant 
.
and a set consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
, for 
in at least 
different ways.
be real numbers such that 
. Prove that
Let be real numbers such that 
. Prove that
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 a convex quadrilateral. Suppose that the circles with diameters 
and 
intersect at points and 
. Let 
and 
. Prove that the points 
, 
, and are concyclic. and are not the diagnols)
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
. Prove that there exists a positive integer satisfying 
.
?
be the midpoints of 
respectively , 
the circumcentre of triangle 
, and the circumcircle of triangle 
intersect 
the second time at , we want to show that 
or equivalently 
or 
( 
) .
and 
, we have 
.
be 
respectively. Let 
. Let 
. Note that since 
are collinear, the converse of Pascal's on 
implies 
lies on 
. Thus:![\[\angle BAC=\angle BA_2C=\pi-\angle C_2BC-\angle B_2CB=\angle BC_2E+\angle CB_2D-\pi\]](http://latex.artofproblemsolving.com/e/9/5/e9553cbe75b3a0fd49a71fa7e7d0949a753f4b54.png)
, we have 
. Thus, from 
similarity we see that 
. Thus:![\[\angle BAC=\angle BC_2E+\angle CB_2D-\pi=\angle M_3M_1E+\angle M_4M_1D-\pi=\angle M_4M_1M_3\]](http://latex.artofproblemsolving.com/0/a/f/0afb900897dde9c09586995bfa5d678728628e88.png)
we have:![\[\angle BAC=\angle M_4M_2M_3\]](http://latex.artofproblemsolving.com/8/c/0/8c0da98f1892da6c66680935b6181514afe6b831.png)
are concyclic as required.
as a beginning observation.
.
)
and 
to be 
, so we take the intersection and claim that it lies on the circumcircle, which is simply the converse of Pascal. yay
by 
. 
is the circumcenter of 
. 
reflects with about . So we need to prove that 
be the midpoints of 
and 
. We have 
and 
. On the other hand, 
. So, 
. It follows that 
. Then, 
so . Then 
passes through , it means that 
is cyclic
is closer to .Let the midpoints of 
, 
, , and 
be 
, 
, , . Let 
. Let 
be a point on 
such that 
. Let 
be a point on such that 
and 
concur on at point 
.
Now we have that is the midpoint of 
and 
, so we have that 
and 
is 
, respectively.
, hence we only need to prove 
.
be the reflection of the circumcenter w.r.t 
(to eliminate fractions). Obviously, 
. Now, let 
denoted as 
. Then, 
. Then, 
and 
. Also, 
.
is real. After some calculation this equals to 
, which could be easily vertified is a real number. We are done.
![[asy]
unitsize(2.5inches);
pair A=dir(115);
pair B=dir(220);
pair C=dir(-40);
pair X=dir(172);
pair Y=dir(45);
pair D=extension(A,B,X,Y);
pair E=extension(A,C,X,Y);
pair DD=(C+D)/2;
pair EE=(B+E)/2;
pair M=(X+Y)/2;
pair N=(D+E)/2;
pair D1=2*M-D;
pair E1=2*M-E;
pair P=2*foot(0,C,D1)-C;
draw(circumcircle(A,B,C));
draw(X--Y);
draw(B--A--C);
draw(C--D);
draw(B--E);
draw(EE--M--DD);
draw(B--P,dotted);
draw(C--P,dotted);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$X$",X,dir(X));
dot("$Y$",Y,dir(Y));
dot("$D$",D,dir(150));
dot("$E$",E,dir(75));
dot("$D'$",DD,dir(250));
dot("$E'$",EE,dir(300));
dot("$M$",M,dir(90));
dot("$N$",N,dir(N));
dot("$D_1$",D1,1.6*dir(D1));
dot("$E_1$",E1,dir(E1));
dot("$P$",P,dir(P));
[/asy]](http://latex.artofproblemsolving.com/8/e/c/8ec1430b404508dda924f098d6bf46035a30b756.png)
be 
and the midpoint of be 
. Let be the midpoint of and 
the midpoint of . Note that 
and 
, so it suffices to show that 
. To this end, let 
and 
be on such that 
and 
. Note that this definition immediately implies that is the midpoint of 
and 
. It suffices to show that 
and 
concur on 
.
, and let 
be the unique conic passing through 
. Our goal is to show that 
, because the only conic passing through 
is the circle . We will show this by Desargues involution theorem on line and quadrilateral 
. The theorem implies that there is a unique involution on line swapping 
, 
, and 
. The first two pairs imply that the involution is reflection in , so we have that 
, completing the proof.
![[asy]
unitsize(100);
pair A = dir(110), B = dir(210), C = dir(330), X = dir(160), Y = dir(30);
pair D = extension(A, B, X, Y), E = extension(A, C, X, Y);
pair K = (B+E)/2, L = (C+D)/2, M = (D+E)/2, N = (X+Y)/2;
pair R = orthocenter(A, D, E), T = -A, S = (R+T)/2;
draw(D--R--E^^K--S--L^^B--T--C, gray(0.5));
draw(S--N^^A--R, gray(0.5));
draw(R--T, gray(0.5)+dashed);
draw(unitcircle);
draw(A--B--C--cycle);
draw(X--Y);
draw(B--E^^C--D);
draw(K--M--L);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$T$", T, dir(T));
dot("$D$", D, dir(dir(A-D)+dir(X-D)));
dot("$E$", E, dir(dir(A-E)+dir(Y-E)));
dot("$K$", K, rotate(90) * dir(E-B));
dot("$L$", L, rotate(90) * dir(C-D));
dot("$M$", M, dir(150));
dot("$N$", N, dir(N));
dot("$R$", R, dir(50));
dot("$S$", S, dir(220));
[/asy]](http://latex.artofproblemsolving.com/5/c/4/5c4879e6d510549ef8c07bde3817984b1bfde078.png)
, 
, , be the midpoints of 
, 
, 
, 
, 
be the orthocenter of 
, be the antipode of 
on , and be the midpoint of 
.
.
and 
are perpendicular to 
, 
is perpendicular to as well. As 
, we obtain 
. Likewise 
. 
. and are antipodes on , the midpoint of 
is equidistant from and . Now 
, so the midpoint of is also equidistant from and , which means . , , , , are concyclic.
as the unit circle centered at 
. Let the midpoints of , , , be 
, 
, 
, , respectively. BY chord intersection formula 
and 
, 
, 
, 
, 
. Then 
is cyclic if and only if 
, which is true by conjugating (it took me 15 minutes at most to compute 
on paper). We're done. 
be such that 
is a parallelogram and be the point so that 
and 
are isotomic wrt 
. By homothety at , the problem is equivalent to showing 
concyclic. Observe that as 
moves, the perpendicular bisector of 
is a moving line with 
, the perpendicular bisector of 
is a moving line with , and the perpendicular bisector of 
is a moving line with 
, thus the desired concyclicity is a 
condition, and we need to check 
values of . For 
the result is obvious. For 
, 
so the result is again obvious. Finally, we check 
, which will finish because of continuity in the case 
.
concyclic. Let line 
intersect 
again at . Clearly 
is an isosceles trapezoid, so 
is an isosceles trapezoid. Then, the result follows from the observation that![\[\measuredangle KPE = \measuredangle KBA=\measuredangle KAE.\]](http://latex.artofproblemsolving.com/e/0/1/e014f9adc3399fb3c70836b08826ff92f0755540.png)
, 
, , and in that order. Let vary around . Now, we see that and vary projectively about with respect to . The midpoints of , , and do, too, so we need to show three points of on the circumcircle that satisfy that , , , and 's midpoints are concyclic. approaches infinitely close to (e.g. 
), we have a homothety of centered about with ratio 
that gives us the circumcircle of the midpoints of , , , and . Similarly, as approaches infinitely close to (e.g. 
), we have a homothety of centered about with ratio that gives us the circumcircle of the midpoints of , , , and . Finally, with being the point on such that the midpoints of and coincide, we see that this is true, too, because all triangles have circumcircles. As a result, our condition is true and we are done.
and and move in such that 
so now from this 
and 
, now let 
the midpoints of respectivily, note that moves on the B-midbase of 
and moves on the C-midbase of 
so from here and the degrees of 
we have that 
and 
, now let 
a point in the C-midbase of such that 
now that 
is projective since its just rotation over a fixed point by a fixed angle so the degrees are preserved so 
hence 
is a 
degree condition.
is such that 
but this means that satisfies condition so as desired. holds always so 
is cyclic as desired thus we are done 
.
denote directed angles modulo 
., , and be the midpoints of segments , , and , respectively, and let the circumcircle 
of 
intersect again at . Let a homothety of scale factor 
centered at and map to 
and 
, respectively, and let and be the centers of and .
. and over lie on and respectively, so lies on the radical axis of and . Since and have twice the radius of , we have by symmetry. 
.
. Indeed, we have 
, 
, and, noticing that 
because 
and 
,
so by SAS, as desired. and are the reflections of and over the center of , so 
is a parallelogram. Thus, the perpendicular bisector 
of 
is perpendicular to , so is the midpoint of , as desired. 
and move on where 
Let be the midpoints of 
respectively. Note that 
moves on the midline 
of 
respectively. Note that 
due to the homotheties at and 
Thus, if we show 
we're done. Let 
There is a map from 
because 
because it's just rotating about by 
which is fixed. Thus, we just need to check 
for three cases and we're done.
because 
duh.
. as the reflection of over . Define as the reflection of over the midpoint of .
is a cyclic quadrilateral and that lies on the line 
through parallel to .
is cyclic by a homothety.
is cyclic and thus 
is an isoceles trapezoid and is thus cyclic.
. It remains to show that quadrilateral 
is cyclic, which follows as 
.
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
be reals such that 
and 
.
at 
and vary 
projectively on the circle. Let 
be the midpoints of 
.![\[T_4 \overset{B}{\mapsto} E \overset{C}{\mapsto} A \overset{B}{\mapsto} D \overset{C}{\mapsto} T_3\]](http://latex.artofproblemsolving.com/1/c/8/1c823aa6120d04212bc304b8a084d15e9fdb24c1.png)
is projective.
and 
to be the 
-
and the 
-
.
is the map sending each point 
on 
on 
(rotation), so it suffices to verify this for three values of 